NCERT Class 12 Maths Ch 5 Miscellaneous Exercise

📘 Concept & Theory

Concept/Theory

›

This question is a direct application of the Chain Rule of Differentiation. When a function is expressed as a power of another function, such as

\[ y=[f(x)]^{n}, \]

we treat the inner expression as a separate variable and differentiate in two stages.

The Chain Rule states that if

\[ y=f(u), \qquad u=g(x), \]

then

\[ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}. \]

For a function of the form

\[ y=[f(x)]^{n}, \]

the Chain Rule gives

\[ \frac{dy}{dx} = n[f(x)]^{n-1}f'(x). \]

Here, the outer function is the ninth power and the inner function is the quadratic expression

\[ 3x^{2}-9x+5. \]

Therefore, both the power rule and the derivative of the inner function must be applied.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the inner expression be $u=3x^{2}-9x+5$.
Rewrite the given function as $y=u^{9}$.
Differentiate $y$ with respect to $u$.
Differentiate $u$ with respect to $x$.
Apply the Chain Rule:
\[ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}. \]
Substitute the value of $u$ back into the result.
Present the final derivative in terms of $x$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Let
\[y=\left(3x^{2}-9x+5\right)^{9}\]
Put
\[u=3x^{2}-9x+5\]
Therefore,
\[y=u^{9}\]
Differentiate $y$ with respect to $u$:
\[\frac{dy}{du}=\frac{d}{du}\left(u^{9}\right)\]
Using the power rule,
\[\frac{d}{du}\left(u^{9}\right)=9u^{8}\]
Hence,
\[\frac{dy}{du}=9u^{8}\]
Now differentiate $u$ with respect to $x$:
\[\frac{du}{dx}=\frac{d}{dx}\left(3x^{2}-9x+5\right)\]
Differentiate each term separately:
\[\frac{d}{dx}(3x^{2})=6x,\]

\[\frac{d}{dx}(-9x)=-9\]
,
\[\frac{d}{dx}(5)=0.\]
Therefore,
\[\frac{du}{dx}=6x-\]
Now apply the Chain Rule:
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
Substituting the derivatives obtained above,
\[\frac{dy}{dx}=9u^{8}(6x-9)\]
Since
\[u=3x^{2}-9x+5,\]
we substitute this value back:
\[\frac{dy}{dx}=9\left(3x^{2}-9x+5\right)^{8}(6x-9)\]
Hence, the required derivative is
\[\boxed{\frac{dy}{dx}=9(6x-9)\left(3x^{2}-9x+5\right)^{8}}\]

🎯 Exam Significance

Exam Significance

›

This problem tests one of the most fundamental differentiation techniques in Class 12 Mathematics: the Chain Rule. Questions involving powers of algebraic, trigonometric, exponential or logarithmic expressions frequently require the same procedure.

It reinforces identification of the inner and outer functions.
It tests correct application of the Chain Rule.
It helps prevent the common mistake of adding or subtracting derivatives instead of multiplying them.
Writing the intermediate variable $u$ makes the solution systematic and easy to verify.
The method can be directly extended to more complicated composite functions.

Significance for Competitive Entrance Examinations

The Chain Rule is a foundational technique used extensively in calculus-based problems in competitive examinations. A seemingly simple expression such as

\[ [f(x)]^{n} \]

may appear inside a more complicated composite function, requiring repeated applications of the Chain Rule.

Developing the habit of identifying the inner function and differentiating it separately improves accuracy and reduces algebraic errors in multi-step differentiation problems.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

For $y=[f(x)]^{n}$, use the Chain Rule.
The general result is
\[ \frac{dy}{dx}=n[f(x)]^{n-1}f]'(x). \]
Always differentiate the outer function first and multiply by the derivative of the inner function.
For
\[ u=3x^{2}-9x+5, \]
we have
\[ \frac{du}{dx}=6x-9. \]
The Chain Rule involves multiplication:
\[ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}. \]
The constant term $5$ has derivative $0$.
The final derivative is
\[ \boxed{ 9(6x-9)\left(3x^{2}-9x+5\right)^{8} }. \]

📘 Concept & Theory

Concept/Theory

›

The given function is the sum of two composite functions:

\[ \sin^{3}x=(\sin x)^3 \]

and

\[ \cos^{6}x=(\cos x)^6. \]

To differentiate such expressions, we use the sum rule together with the Chain Rule.

The sum rule states that

\[ \frac{d}{dx}[f(x)+g(x)] = \frac{df}{dx}+\frac{dg}{dx}. \]

For a composite power function,

\[ y=[f(x)]^n, \]

the Chain Rule gives

\[ \frac{dy}{dx} = n[f(x)]^{n-1}f'(x). \]

Therefore,

\[ \frac{d}{dx}(\sin x)^3 = 3(\sin x)^2\cos x, \]

because

\[ \frac{d}{dx}(\sin x)=\cos x. \]

Similarly,

\[ \frac{d}{dx}(\cos x)^6 = 6(\cos x)^5(-\sin x), \]

because

\[ \frac{d}{dx}(\cos x)=-\sin x. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given expression as $y=y_1+y_2$.
Differentiate $y_1=\sin^3x$ using the Chain Rule.
Differentiate $y_2=\cos^6x$ using the Chain Rule.
Add the two derivatives using the sum rule.
Factor the resulting expression to obtain a compact final form.
Check carefully that the negative sign arising from
\[ \frac{d}{dx}(\cos x)=-\sin x \]
is retained.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 29 steps

Given — $y=\sin^{3}x+\cos^{6}x$
Let
\[y_1=\sin^{3}x\]
and
\[y_2=\cos^{6}x\]
Therefore,
\[y=y_1+y_2.\]
Differentiate $y_1=\sin^3x$
We have
\[y_1=\sin^{3}x=(\sin x)^3\]
Let
\[u=\sin x\]
Then
\[y_1=u^3\]
Differentiate $y_1$ with respect to $u$:
\[\frac{dy_1}{du}=\frac{d}{du}(u^3)=3u^2.\]
Now differentiate $u$ with respect to $x$:
\[\frac{du}{dx}=\frac{d}{dx}(\sin x)=\cos x\]
By the Chain Rule,
\[\frac{dy_1}{dx}=\frac{dy_1}{du}\cdot\frac{du}{dx}\]
Therefore,
\[\frac{dy_1}{dx}=3u^2\cos x\]
Substituting $u=\sin x$,
\[\boxed{\frac{dy_1}{dx}=3\sin^2x\cos x}\]
Differentiate $y_2=\cos^6x$
We have
\[y_2=\cos^{6}x=(\cos x)^6\]
Let
\[v=\cos x\]
Then
\[y_2=v^6\]
Differentiate $y_2$ with respect to $v$:
\[\frac{dy_2}{dv}=\frac{d}{dv}(v^6)=6v^5\]
Now differentiate $v$ with respect to $x$:
\[\frac{dv}{dx}=\frac{d}{dx}(\cos x)\]
Since
\[\frac{d}{dx}(\cos x)=-\sin x,\]
we get
\[\frac{dv}{dx}=-\sin x.\]
Applying the Chain Rule,
\[\frac{dy_2}{dx}=\frac{dy_2}{dv}\cdot\frac{dv}{dx}\]
Therefore,
\[\frac{dy_2}{dx}=6v^5(-\sin x)\]
Substituting $v=\cos x$,
\[\frac{dy_2}{dx}=-6\cos^5x\sin x\]
Hence,
\[\boxed{\frac{dy_2}{dx}=-6\sin x\cos^5x}\]
Add the Two Derivatives
Since
\[y=y_1+y_2\]
using the sum rule,
\[\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}\]
Substituting the two derivatives,
\[\frac{dy}{dx}=3\sin^2x\cos x-6\sin x\cos^5x\]
Take the Common Factor
The two terms contain the common factor $3\sin x\cos x$.
Therefore,
\[\frac{dy}{dx}=3\sin x\cos x\left[\frac{3\sin^2x\cos x}{3\sin x\cos x}-\frac{6\sin x\cos^5x}{3\sin x\cos x}\right]\]
Simplifying each term inside the bracket,
\[\frac{3\sin^2x\cos x}{3\sin x\cos x}=\sin x\]
and
\[\frac{6\sin x\cos^5x}{3\sin x\cos x}=2\cos^4x\]
Hence,
\[\frac{dy}{dx}=3\sin x\cos x\left(\sin x-2\cos^4x\right)\]
Thus, the required derivative is
\[\boxed{\frac{dy}{dx}=3\sin x\cos x\left(\sin x-2\cos^4x\right)}\]

🎯 Exam Significance

Exam Significance

›

This problem combines two important differentiation principles: the sum rule and the Chain Rule. Such questions are useful for testing whether a student can correctly differentiate powers of trigonometric functions.

It tests differentiation of $(\sin x)^n$ and $(\cos x)^n$.
It reinforces the correct use of the Chain Rule.
It tests the sign of the derivative of $\cos x$.
It provides practice in differentiating a sum term by term.
Factoring the final expression helps produce a clean and easily verifiable answer.

Significance for Competitive Entrance Examinations

Composite trigonometric functions occur frequently in calculus-based entrance examination problems. The ability to differentiate expressions such as

\[ (\sin x)^n \quad\text{and}\quad (\cos x)^n \]

quickly is particularly useful when differentiation forms only one part of a longer problem.

Students should also be alert to the negative sign in

\[ \frac{d}{dx}(\cos x)=-\sin x, \]

because a sign error in an intermediate derivative can change the entire final answer.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Differentiate a sum term by term.
Use the Chain Rule whenever a function occurs inside another function.
\[ \frac{d}{dx}(\sin^n x) = n\sin^{n-1}x\cos x. \]
\[ \frac{d}{dx}(\cos^n x) = -n\cos^{n-1}x\sin x. \]
Remember that $\sin^3x$ means $(\sin x)^3$, not $\sin(3x)$.
The negative sign in the derivative of $\cos x$ must be retained.
The final result can be written as
\[ \boxed{ \frac{dy}{dx} = 3\sin x\cos x \left(\sin x-2\cos^4x\right) }. \]

📘 Concept & Theory

Concept/Theory

›

This is an example of a variable-base, variable-exponent function. The base $5x$ depends on $x$, and the exponent $3\cos 2x$ also depends on $x$.

For a function of the form

\[ y=[f(x)]^{g(x)}, \]

the ordinary power rule cannot be applied directly because both the base and exponent are functions of $x$.

The appropriate method is logarithmic differentiation. Taking logarithms converts the variable exponent into a coefficient:

\[ \log y=g(x)\log f(x). \]

After differentiating, multiply both sides by $y$ and finally substitute the original value of $y$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let
\[ y=(5x)^{3\cos 2x}. \]
Take logarithm on both sides.
Use the logarithm law
\[ \log(a^b)=b\log a. \]
Differentiate both sides using the product rule.
Use the Chain Rule to differentiate $\cos 2x$.
Differentiate $\log(5x)$ carefully using the Chain Rule.
Multiply through by $y$.
Substitute
\[ y=(5x)^{3\cos 2x}. \]
Simplify the final expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 22 steps

Let
\[y=(5x)^{3\cos 2x}\]
Since both the base and exponent contain $x$, we use logarithmic differentiation.
Take Logarithm on Both Sides
Taking logarithm on both sides,
\[\log y=\log\left[(5x)^{3\cos 2x}\right].\]
Using
\[\log(a^b)=b\log a\]
we obtain
\[\log y=3\cos 2x\log(5x)\]
Differentiate Both Sides
Differentiate with respect to $x$:
\[\frac{d}{dx}(\log y)=\frac{d}{dx}\left[3\cos 2x\log(5x)\right]\]
For the left-hand side, applying the Chain Rule,
\[\frac{d}{dx}(\log y)=\frac{1}{y}\frac{dy}{dx}\]
Therefore,
\[\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left[3\cos 2x\log(5x)\right]\]
Apply the Product Rule
Since $3\cos 2x$ and $\log(5x)$ are both functions of $x$, use the product rule:
\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]
Taking
\[u=3\cos 2x\]
and
\[v=\log(5x)\]
we get
\[\frac{1}{y}\frac{dy}{dx}=3\left[\cos 2x\frac{d}{dx}\log(5x)+\log(5x)\frac{d}{dx}(\cos 2x)\right]\]
Differentiate $\log(5x)$
Using the Chain Rule,
\[\frac{d}{dx}\log(5x)=\frac{1}{5x}\frac{d}{dx}(5x)\]
Since
\[\frac{d}{dx}(5x)=5\]
we have
\[\frac{d}{dx}\log(5x)=\frac{1}{5x}\cdot5=\frac{1}{x}\]
Differentiate $\cos 2x$
Again using the Chain Rule,
\[\frac{d}{dx}(\cos 2x)=-\sin 2x\cdot\frac{d}{dx}(2x)\]
Since
\[\frac{d}{dx}(2x)=2\]
we obtain
\[\frac{d}{dx}(\cos 2x)=-2\sin 2x\]
Substitute the Derivatives
Substituting the results into the product-rule expression,
\[\frac{1}{y}\frac{dy}{dx}=3\left[\cos 2x\left(\frac{1}{x}\right)+\log(5x)(-2\sin 2x)\right]\]
Therefore,
\[\frac{1}{y}\frac{dy}{dx}=3\left[\frac{\cos 2x}{x}-2\sin 2x\log(5x)\right]\]
Multiplying both sides by $y$,
\[\frac{dy}{dx}=3y\left[\frac{\cos 2x}{x}-2\sin 2x\log(5x)\right]\]
Substitute the Value of $y$
Since
\[y=(5x)^{3\cos 2x}\]
we get
\[\frac{dy}{dx}=3(5x)^{3\cos 2x}\left[\frac{\cos 2x}{x}-2\sin 2x\log(5x)\right]\]
Multiplying $3$ into the bracket, we may also write
\[\boxed{\frac{dy}{dx}=(5x)^{3\cos 2x}\left[\frac{3\cos 2x}{x}-6\sin 2x\log(5x)\right]}\]

🎯 Exam Significance

Exam Significance

›

This problem is important because it tests logarithmic differentiation, a technique used when both the base and exponent of a power depend on $x$. It also combines the product rule and Chain Rule in a single problem.

It demonstrates when logarithmic differentiation should be preferred over the ordinary power rule.
It tests the logarithm identity
\[ \log(a^b)=b\log a. \]
It requires correct application of the product rule.
It tests the Chain Rule through the functions $\cos 2x$ and $\log(5x)$.
It is particularly useful for developing accuracy in multi-step differentiation problems.

Significance for Competitive Entrance Examinations

Variable-base and variable-exponent functions are standard applications of logarithmic differentiation. Recognising the structure

\[ y=[f(x)]^{g(x)} \]

allows a complicated-looking derivative to be converted into a manageable product after taking logarithms.

Competitive problems may further combine this technique with implicit differentiation, limits, maxima and minima, or parameter-dependent expressions. Therefore, it is important to master the complete sequence rather than memorising only the final formula.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Use logarithmic differentiation when both the base and exponent contain the variable.
For
\[ y=[f(x)]^{g(x)}, \]
taking logarithms gives
\[ \log y=g(x)\log f(x). \]
The product rule is required after taking logarithms in this problem.
\[ \frac{d}{dx}\log(5x)=\frac{1}{x}. \]
\[ \frac{d}{dx}(\cos 2x)=-2\sin 2x. \]
Always substitute the original $y$ back after logarithmic differentiation.
Do not alter the original base or exponent while simplifying.
The correct derivative is
\[ \boxed{ \frac{dy}{dx} = (5x)^{3\cos 2x} \left[ \frac{3\cos 2x}{x} - 6\sin 2x\log(5x) \right] }. \]

📘 Concept & Theory

Concept/Theory

›

This problem involves the differentiation of an inverse trigonometric function with a composite algebraic argument.

For a function of the form

\[ y=\sin^{-1}[f(x)], \]

the Chain Rule gives

\[ \frac{dy}{dx} = \frac{f'(x)}{\sqrt{1-[f(x)]^2}}. \]

Here,

\[ f(x)=x\sqrt{x}=x^{3/2}. \]

Therefore, we first differentiate $x^{3/2}$, and then divide by

\[ \sqrt{1-x^3}. \]

The given restriction

\[ 0\leq x\leq1 \]

is important. It ensures that $x\sqrt{x}$ lies in the interval $[0,1]$, which is within the domain of the principal inverse sine function.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let
\[ y=\sin^{-1}(x\sqrt{x}). \]
Convert the inverse sine relation into
\[ \sin y=x\sqrt{x}. \]
Rewrite
\[ x\sqrt{x}=x^{3/2}. \]
Differentiate implicitly with respect to $x$.
Obtain $\cos y$ using
\[ \sin^2y+\cos^2y=1. \]
Use the given interval $0\leq x\leq1$ to select the positive square root for $\cos y$.
Substitute $\cos y$ into the derivative and simplify.
State the endpoint behaviour separately because the derivative becomes unbounded at $x=1$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 33 steps

Given —
\[y=\sin^{-1}\left(x\sqrt{x}\right)\]
Since
\[x\sqrt{x}=x^{1+\frac12}=x^{3/2}\]
we may write
\[y=\sin^{-1}\left(x^{3/2}\right)\]
Remove the Inverse Trigonometric Function
From
\[y=\sin^{-1}\left(x^{3/2}\right)\]
we obtain
\[\sin y=x^{3/2}\]
Differentiate Implicitly
Differentiate both sides with respect to $x$:
\[\frac{d}{dx}(\sin y)=\frac{d}{dx}\left(x^{3/2}\right)\]
Since $y$ is a function of $x$, the Chain Rule gives
\[\frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}\]
Using the power rule on the right-hand side,
\[\frac{d}{dx}\left(x^{3/2}\right)=\frac32x^{1/2}\]
Therefore,
\[\cos y\frac{dy}{dx}=\frac32\sqrt{x}\]
Hence,
\[\frac{dy}{dx}=\frac{3\sqrt{x}}{2\cos y}\]
Find $\cos y$
We know the fundamental trigonometric identity
\[\sin^2y+\cos^2y=1\]
Therefore,
\[\cos^2y=1-\sin^2y\]
From the original relation,
\[\sin y=x^{3/2}\]
Thus,
\[\cos^2y=1-\left(x^{3/2}\right)^2\]
Since
\[\left(x^{3/2}\right)^2=x^3\]
we get
\[\cos^2y=1-x^3\]
Hence,
\[\cos y=\pm\sqrt{1-x^3}\]
Determine the Correct Sign
This sign must be determined from the range of the principal inverse sine function. Since
\[y=\sin^{-1}(x^{3/2}),\]
the principal value of $y$ lies in
\[-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\]
Therefore,
\[\cos y\geq0\]
Thus, we take the positive square root:
\[\boxed{\cos y=\sqrt{1-x^3}}\]
Substitute $\cos y$
We obtained
\[\frac{dy}{dx}=\frac{3\sqrt{x}}{2\cos y}\]
Substituting
\[\cos y=\sqrt{1-x^3}\]
we get
\[\frac{dy}{dx}=\frac{3\sqrt{x}}{2\sqrt{1-x^3}}\]
Therefore,
\[\boxed{\frac{dy}{dx}=\frac{3}{2}\sqrt{\frac{x}{1-x^3}}}\]
for
\[0\leq x < 1\]
Important Domain and Endpoint Observation
The original function is defined on the given interval
\[0\leq x\leq1.\]
However, the derivative formula contains
\[\sqrt{1-x^3}\]
in the denominator.
At $x=1$,
\[ \sqrt{1-x^3}=\sqrt{1-1}=0. \]
Hence, the displayed derivative formula is not finite at $x=1$.
In fact,
\[ y=\sin^{-1}(x^{3/2}) \]
approaches
\[ \frac{\pi}{2} \]
as $x\to1^{-}$, and
\[ \frac{dy}{dx} = \frac{3\sqrt{x}}{2\sqrt{1-x^3}} \to+\infty \qquad\text{as }x\to1^{-}. \]
Thus, the derivative is finite for
\[ 0\leq x < 1, \]
while at $x=1$ the ordinary finite derivative does not exist.
Verification Using the Standard Formula
We can verify the result directly using
\[ \frac{d}{dx}\left[\sin^{-1}f(x)\right] = \frac{f'(x)}{\sqrt{1-[f(x)]^2}}. \]
Here,
\[f(x)=x\sqrt{x}=x^{3/2}\]
Therefore,
\[f'(x)=\frac32x^{1/2}=\frac32\sqrt{x}\]
Also,
\[[f(x)]^2=\left(x^{3/2}\right)^2=x^3\]
Hence,
\[\frac{dy}{dx}=\frac{\frac32\sqrt{x}}{\sqrt{1-x^3}}\]
Therefore,
\[\boxed{\frac{dy}{dx}=\frac{3\sqrt{x}}{2\sqrt{1-x^3}}=\frac32\sqrt{\frac{x}{1-x^3}}}\]

🎯 Exam Significance

Exam Significance

›

This problem is an important application of the differentiation of inverse trigonometric functions. It also tests whether a student can correctly combine the inverse sine derivative with the Chain Rule.

It reinforces the formula for differentiating $\sin^{-1}u$.
It demonstrates how implicit differentiation can be used to derive the result.
It tests conversion of $x\sqrt{x}$ into the power form $x^{3/2}$.
It requires correct use of the identity
\[ \sin^2y+\cos^2y=1. \]
It highlights the importance of considering the range of an inverse trigonometric function when choosing a square-root sign.
It introduces an important distinction between the domain of a function and the interval on which its derivative is finite.

Significance for Competitive Entrance Examinations

Problems involving inverse trigonometric functions often combine differentiation, algebraic simplification and domain or range analysis. The restriction

\[ 0\leq x\leq1 \]

is not merely additional information; it helps establish the sign of the relevant trigonometric quantity and the validity of the expression.

For competitive examinations, students should be comfortable switching between the direct formula

\[ \frac{d}{dx}\left[\sin^{-1}f(x)\right] \]

and the implicit approach

\[ \sin y=f(x). \]

The implicit method is particularly useful when an inverse trigonometric expression is embedded in a more complicated problem.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

For
\[ y=\sin^{-1}[f(x)], \]
use
\[ \frac{dy}{dx} = \frac{f'(x)} {\sqrt{1-[f(x)]^2}}. \]
\[ x\sqrt{x}=x^{3/2}. \]
\[ \frac{d}{dx}(x^{3/2}) = \frac32\sqrt{x}. \]
The principal range of $\sin^{-1}x$ determines that $\cos y\geq0$ in this problem.
\[ \cos y=\sqrt{1-x^3}. \]
The derivative is finite for
\[ 0\leq x<1. \]
At $x=1$, the derivative tends to $+\infty$, so there is no finite ordinary derivative there.
The final result is
\[ \boxed{ \frac{dy}{dx} = \frac32\sqrt{\frac{x}{1-x^3}} }. \]

📘 Concept & Theory

Concept/Theory

›

This problem involves the differentiation of a quotient containing an inverse trigonometric function and a composite radical function.

Since the numerator and denominator are both functions of $x$, the Quotient Rule is required.

The Quotient Rule states that if

\[ y=\frac{u}{v}, \]

then

\[ \frac{dy}{dx} = \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}. \]

The numerator contains

\[ \cos^{-1}\left(\frac{x}{2}\right), \]

so its derivative requires the Chain Rule.

The denominator contains

\[ \sqrt{2x+7}=(2x+7)^{1/2}, \]

whose derivative also requires the Chain Rule.

The given restriction

\[ -2 < x < 2 \]

ensures that

\[ -1 < \frac{x}{2} < 1, \]

so the inverse cosine expression is well-defined and

\[ \sqrt{4-x^2}>0. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the numerator and denominator:
\[ u=\cos^{-1}\left(\frac{x}{2}\right), \qquad v=\sqrt{2x+7}. \]
Find $du/dx$ using the derivative of $\cos^{-1}u$ and the Chain Rule.
Find $dv/dx$ using the Chain Rule.
Apply the Quotient Rule.
Simplify
\[ \sqrt{1-\left(\frac{x}{2}\right)^2} \]
into a form involving $\sqrt{4-x^2}$.
Simplify the two resulting terms separately.
Factor the negative sign and write the final answer in a clean form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 34 steps

Let
\[y=\frac{\cos^{-1}\left(\frac{x}{2}\right)}{\sqrt{2x+7}}\]
Put
\[u=\cos^{-1}\left(\frac{x}{2}\right)\]
and
\[v=\sqrt{2x+7}\]
Thus,
\[y=\frac{u}{v}\]
Differentiate the Numerator
We have
\[u=\cos^{-1}\left(\frac{x}{2}\right)\]
Recall that
\[ \frac{d}{dx}\left(\cos^{-1}t\right) = -\frac{1}{\sqrt{1-t^2}}\frac{dt}{dx}. \]
Here,
\[t=\frac{x}{2}\]
Therefore,
\[\frac{dt}{dx}=\frac12\]
Hence, by the Chain Rule,
\[\frac{du}{dx}=-\frac{1}{\sqrt{1-\left(\frac{x}{2}\right)^2}}\cdot\frac12\]
Thus,
\[\frac{du}{dx}=-\frac{1}{2\sqrt{1-\frac{x^2}{4}}}\]
Now simplify the denominator:
\[1-\frac{x^2}{4}=\frac{4-x^2}{4}\]
Therefore,
\[\sqrt{1-\frac{x^2}{4}}=\sqrt{\frac{4-x^2}{4}}=\frac{\sqrt{4-x^2}}{2}\]
Hence,
\[\frac{du}{dx}=-\frac{1}{2\left(\frac{\sqrt{4-x^2}}{2}\right)}=-\frac{1}{\sqrt{4-x^2}}\]
Thus,
\[\boxed{\frac{du}{dx}=-\frac{1}{\sqrt{4-x^2}}}\]
Differentiate the Denominator
We have
\[v=\sqrt{2x+7}=(2x+7)^{1/2}\]
Using the Chain Rule,
\[\frac{dv}{dx}=\frac12(2x+7)^{-1/2}\cdot\frac{d}{dx}(2x+7)\]
Since
\[\frac{d}{dx}(2x+7)=2\]
we get
\[\frac{dv}{dx}=\frac12(2x+7)^{-1/2}\cdot2\]
Therefore,
\[\frac{dv}{dx}=(2x+7)^{-1/2}=\frac{1}{\sqrt{2x+7}}\]
Thus,
\[\boxed{\frac{dv}{dx}=\frac{1}{\sqrt{2x+7}}}\]
Apply the Quotient Rule
Since
\[y=\frac{u}{v}\]
the Quotient Rule gives
\[\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\]
Substituting
\[u=\cos^{-1}\left(\frac{x}{2}\right),\quad v=\sqrt{2x+7},\]
along with
\[\frac{du}{dx}=-\frac{1}{\sqrt{4-x^2}},\quad \frac{dv}{dx}=\frac{1}{\sqrt{2x+7}},\]
we obtain
\[ \frac{dy}{dx} = \frac{ \sqrt{2x+7} \left(-\frac{1}{\sqrt{4-x^2}}\right) - \cos^{-1}\left(\frac{x}{2}\right) \left(\frac{1}{\sqrt{2x+7}}\right) } {(\sqrt{2x+7})^2}. \]
Since
\[(\sqrt{2x+7})^2=2x+7\]
we have
\[ \frac{dy}{dx} = \frac{ -\frac{\sqrt{2x+7}}{\sqrt{4-x^2}} - \frac{\cos^{-1}\left(\frac{x}{2}\right)} {\sqrt{2x+7}} } {2x+7}. \]
Simplify the First Term
Consider
\[\frac{-\frac{\sqrt{2x+7}}{\sqrt{4-x^2}}}{2x+7}\]
This can be written as
\[-\frac{\sqrt{2x+7}}{(2x+7)\sqrt{4-x^2}}\]
Since
\[\frac{\sqrt{2x+7}}{2x+7}=\frac{1}{\sqrt{2x+7}}\]
we obtain
\[-\frac{1}{\sqrt{2x+7}\sqrt{4-x^2}}\]
Simplify the Second Term
Consider
\[-\frac{\cos^{-1}\left(\frac{x}{2}\right)}{(2x+7)\sqrt{2x+7}}\]
Using
\[(2x+7)\sqrt{2x+7}=(2x+7)^{3/2},\]
we get
\[-\frac{\cos^{-1}\left(\frac{x}{2}\right)}{(2x+7)^{3/2}}\]
Combine the Results
Therefore,
\[ \frac{dy}{dx} = -\frac{1} {\sqrt{2x+7}\sqrt{4-x^2}} - \frac{ \cos^{-1}\left(\frac{x}{2}\right) } {(2x+7)^{3/2}}. \]
Taking the negative sign common,
\[ \boxed{ \frac{dy}{dx} = - \left[ \frac{1} {\sqrt{2x+7}\sqrt{4-x^2}} + \frac{ \cos^{-1}\left(\frac{x}{2}\right) } {(2x+7)^{3/2}} \right] } \]

🎯 Exam Significance

Exam Significance

›

This problem is a useful combination of three major differentiation techniques from Class 12 Mathematics: the Quotient Rule, the Chain Rule, and differentiation of an inverse trigonometric function.

It tests the derivative of $\cos^{-1}u$.
It reinforces the Quotient Rule.
It requires careful simplification of square roots and fractional powers.
It tests the Chain Rule in both the numerator and denominator.
The given restriction \(-2

Significance for Competitive Entrance Examinations

Problems combining inverse trigonometric functions with algebraic functions are useful for developing multi-rule differentiation skills. Competitive examination questions may require the same expression to be differentiated, simplified, or used subsequently in an equation involving slopes, monotonicity or extrema.

A reliable approach is to identify the outermost operation first. Here, the overall structure is a quotient, so the Quotient Rule is applied first, while the derivatives of the numerator and denominator are evaluated separately using the Chain Rule.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

For a quotient
\[ y=\frac{u}{v}, \]
use
\[ \frac{dy}{dx} = \frac{vu'-uv'}{v^2}. \]
\[ \frac{d}{dx}\left(\cos^{-1}u\right) = -\frac{u'}{\sqrt{1-u^2}}. \]
\[ \frac{d}{dx}\sqrt{2x+7} = \frac{1}{\sqrt{2x+7}}. \]
\[ \sqrt{1-\left(\frac{x}{2}\right)^2} = \frac{\sqrt{4-x^2}}{2}. \]
Careful algebraic simplification is essential after applying the Quotient Rule.
The final derivative is
\[ \boxed{ \frac{dy}{dx} = - \left[ \frac{1} {\sqrt{2x+7}\sqrt{4-x^2}} + \frac{ \cos^{-1}\left(\frac{x}{2}\right) } {(2x+7)^{3/2}} \right] }. \]

📘 Concept & Theory

Concept/Theory

›

This problem requires careful simplification of a complicated expression involving square roots and trigonometric functions before differentiating the inverse cotangent function.

The key identity is

\[ 1-\sin^2x=\cos^2x. \]

However, an important point must not be overlooked:

\[ \sqrt{\cos^2x}=|\cos x|, \]

not simply $\cos x$.

This absolute-value issue affects the simplification and therefore affects the derivative. The supplied solution replaces

\[ \sqrt{\cos^2x} \]

directly by $\cos x$, which is valid only when

\[ \cos x\geq0. \]

Consequently, without a specified interval for $x$, the expression must be treated piecewise. This is an important point for a complete and mathematically rigorous solution.

Useful Identities

We use

\[ (a+b)^2-(a-b)^2=4ab, \]

\[ 1-\sin^2x=\cos^2x, \]

\[ \sqrt{\cos^2x}=|\cos x|, \]

\[ 1+\cos x=2\cos^2\frac{x}{2}, \]

and

\[ \sin x=2\sin\frac{x}{2}\cos\frac{x}{2}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the expression inside $\cot^{-1}$ be $R$.
Rationalise $R$ by multiplying numerator and denominator by the conjugate of the denominator.
Simplify the resulting expression using
\[ 1-\sin^2x=\cos^2x. \]
Retain the absolute value:
\[ \sqrt{\cos^2x}=|\cos x|. \]
Consider separately the cases $\cos x\geq0$ and $\cos x<0$.
Use half-angle identities to simplify the resulting trigonometric expressions.
Apply the appropriate principal-value interpretation of $\cot^{-1}$.
Differentiate the resulting piecewise expression.
Identify points where the original expression is undefined or where the derivative does not exist.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 30 steps

Let
\[R=\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\]
Then
\[y=\cot^{-1}R\]
Rationalise the Expression
Multiply the numerator and denominator by the conjugate of the denominator:
\[R=\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\cdot\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}}\]
Therefore,
\[R=\frac{\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)^2}{(1+\sin x)-(1-\sin x)}\]
Expand the numerator:
\[\left(\sqrt{1+\sin x}+\sqrt{1-\sin x}\right)^2\]

\[=1+\sin x+1-\sin x+2\sqrt{(1+\sin x)(1-\sin x)}\]
Hence,
\[R=\frac{2+2\sqrt{(1+\sin x)(1-\sin x)}}{2\sin x}\]
Simplify the Product Under the Radical
We have
\[(1+\sin x)(1-\sin x)=1-\sin^2x\]
Using
\[1-\sin^2x=\cos^2x\]
we obtain
\[(1+\sin x)(1-\sin x)=\cos^2x\]
Therefore,
\[\sqrt{(1+\sin x)(1-\sin x)}=\sqrt{\cos^2x}=|\cos x|\]
Thus,
\[R=\frac{2+2|\cos x|}{2\sin x}\]
Hence,
\[\boxed{R=\frac{1+|\cos x|}{\sin x}}\]
Case I, $\cos x\geq0$
If
\[\cos x\geq0\]
then
\[|\cos x|=\cos x\]
Therefore,
\[R=\frac{1+\cos x}{\sin x}\]
Using the half-angle identities,
\[1+\cos x=2\cos^2\frac{x}{2}\]
and
\[\sin x=2\sin\frac{x}{2}\cos\frac{x}{2},\]
we get
\[R=\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\]
For values where the original expression is defined, the common factor can be cancelled:
\[R=\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\]
Hence,
\[R=\cot\frac{x}{2}\]
Therefore,
\[y=\cot^{-1}\left(\cot\frac{x}{2}\right)\]
On an interval where
\[ 0 < \frac{x}{2} < \pi, \]
the principal value gives
\[ \cot^{-1}\left(\cot\frac{x}{2}\right) = \frac{x}{2}. \]
Thus, on such an interval,
\[\frac{dy}{dx}=\frac12\]
Case II, $\cos x < 0$
If
\[\cos x<0\]
then
\[|\cos x|=-\cos x\]
Hence,
\[R=\frac{1-\cos x}{\sin x}\]
Using
\[1-\cos x=2\sin^2\frac{x}{2}\]
and
\[\sin x=2\sin\frac{x}{2}\cos\frac{x}{2},\]
we obtain
\[R=\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\]
Therefore,
\[R=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\tan\frac{x}{2}\]
Thus, in this case,
\[y=\cot^{-1}\left(\tan\frac{x}{2}\right)\]
Using
\[\tan\theta=\cot\left(\frac{\pi}{2}-\theta\right)\]
we obtain
\[y=\cot^{-1}\left[\cot\left(\frac{\pi}{2}-\frac{x}{2}\right)\right],\]
with the appropriate adjustment by $\pi$ according to the principal range of $\cot^{-1}$.
On an interval where the corresponding principal-value expression is represented by
\[y=\frac{\pi}{2}-\frac{x}{2}\]
we obtain
\[\boxed{\frac{dy}{dx}=-\frac12}\]

🎯 Exam Significance

Exam Significance

›

This is an excellent problem for testing trigonometric simplification before differentiation. It also highlights a subtle but important issue involving square roots and absolute values.

It combines algebraic rationalisation with trigonometric identities.
It tests the identities involving $1+\cos x$, $1-\cos x$, and $\sin x$.
It tests the principal-value concept associated with inverse trigonometric functions.
It demonstrates why
\[ \sqrt{a^2}=|a| \]
must be used carefully.
It provides an important example where domain and interval information affects the final derivative.

Significance for Competitive Entrance Examinations

This problem is particularly useful for competitive calculus because the major challenge is not merely differentiation but recognising hidden absolute-value and branch issues.

In entrance examinations, expressions containing

\[ \sqrt{\cos^2x},\qquad \sqrt{\sin^2x}, \qquad \tan^{-1}(\tan x), \qquad \cot^{-1}(\cot x) \]

often require careful consideration of intervals.

A solution that ignores the relevant interval may produce a formally attractive but globally incorrect result.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Always retain the absolute value when simplifying
\[ \sqrt{\cos^2x}. \]
The correct simplification of the expression inside $\cot^{-1}$ is
\[ \boxed{ R=\frac{1+|\cos x|}{\sin x} }. \]
If $\cos x\geq0$, then
\[ R=\frac{1+\cos x}{\sin x} = \cot\frac{x}{2}. \]
If $\cos x<0$, then
\[ R=\frac{1-\cos x}{\sin x} = \tan\frac{x}{2}. \]
The original expression is undefined whenever
\[ x=n\pi. \]
The result
\[ \frac{dy}{dx}=\frac12 \]
is valid only on intervals where the corresponding simplification and inverse-cotangent branch give
\[ y=\frac{x}{2}. \]
On intervals where the principal-value expression becomes
\[ y=\frac{\pi}{2}-\frac{x}{2}, \]
the derivative is
\[ -\frac12. \]
For a fully rigorous answer, the interval for $x$ should be specified before giving a single derivative value.

📘 Concept & Theory

Concept/Theory

›

This is a variable-base and variable-exponent function. Both the base

\[ \log x \]

and the exponent

\[ \log x \]

depend on $x$.

Therefore, the ordinary power rule cannot be applied directly. The appropriate technique is logarithmic differentiation.

For a function of the form

\[ y=[f(x)]^{g(x)}, \]

taking logarithms gives

\[ \log y=g(x)\log[f(x)]. \]

Differentiating this transformed equation generally converts the variable exponent into a product, which can then be handled using the Product Rule and Chain Rule.

In this problem,

\[ f(x)=\log x \]

and

\[ g(x)=\log x. \]

Thus, logarithmic differentiation is the natural method.

Domain Consideration

For the real-valued expression

\[ (\log x)^{\log x}, \]

the logarithm requires

\[ x>0. \]

For the logarithmic differentiation used below, we take

\[ \log(\log x), \]

which additionally requires

\[ \log x>0. \]

Hence,

\[ \boxed{x>1}. \]

On this interval, $\log x>0$, so the logarithmic differentiation is directly valid over the real numbers.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let
\[ y=(\log x)^{\log x}. \]
Take logarithm on both sides.
Use
\[ \log(a^b)=b\log a. \]
Differentiate the resulting product using the Product Rule.
Use the Chain Rule to differentiate
\[ \log(\log x). \]
Simplify the logarithmic derivative.
Multiply by $y$.
Substitute the original expression for $y$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Let
\[y=(\log x)^{\log x}\]
Take Logarithm on Both Sides
Taking logarithm on both sides,
\[\log y=\log\left[(\log x)^{\log x}\right]\]
Using the logarithm identity
\[\log(a^b)=b\log a\]
we obtain
\[\log y=\log x\cdot\log(\log x)\]
Differentiate Both Sides
Differentiate with respect to $x$:
\[\frac{d}{dx}(\log y)=\frac{d}{dx}\left[\log x\cdot\log(\log x)\right]\]
Since $y$ is a function of $x$, the Chain Rule gives
\[\frac{d}{dx}(\log y)=\frac{1}{y}\frac{dy}{dx}\]
Therefore,
\[\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left[\log x\cdot\log(\log x)\right]\]
Apply the Product Rule
The Product Rule states that
\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]
Here, take
\[u=\log x\]
and
\[v=\log(\log x)\]
Hence,
\[\frac{1}{y}\frac{dy}{dx}=\log(\log x)\frac{d}{dx}(\log x)+\log x\frac{d}{dx}\left[\log(\log x)\right]\]
Differentiate $\log x$
We know that
\[\frac{d}{dx}(\log x)=\frac{1}{x}\]
Therefore, the first term becomes
\[\log(\log x)\frac{d}{dx}(\log x)=\frac{\log(\log x)}{x}\]
Differentiate $\log(\log x)$
Using the Chain Rule,
\[\frac{d}{dx}\left[\log(\log x)\right]=\frac{1}{\log x}\frac{d}{dx}(\log x)\]
Since
\[\frac{d}{dx}(\log x)=\frac{1}{x}\]
we get
\[\frac{d}{dx}\left[\log(\log x)\right]=\frac{1}{x\log x}\]
Therefore, the second term becomes
\[\log x\left(\frac{1}{x\log x}\right)=\frac{1}{x}\]
Combine the Two Terms
Thus,
\[\frac{1}{y}\frac{dy}{dx}=\frac{\log(\log x)}{x}+\frac{1}{x}\]
Taking $1/x$ common,
\[\frac{1}{y}\frac{dy}{dx}=\frac{\log(\log x)+1}{x}\]
Multiply by $y$
Multiplying both sides by $y$,
\[\frac{dy}{dx}=y\frac{\log(\log x)+1}{x}\]
Since
\[y=(\log x)^{\log x}\]
we obtain
\[\frac{dy}{dx}=(\log x)^{\log x}\frac{\log(\log x)+1}{x}\]
Therefore, the required derivative is
\[\boxed{\frac{dy}{dx}=\frac{(\log x)^{\log x}}{x}\left[\log(\log x)+1\right]}\]

🎯 Exam Significance

Exam Significance

›

This is a standard and important application of logarithmic differentiation. It tests whether a student can recognise that both the base and exponent are variable.

It tests the identity
\[ \log(a^b)=b\log a. \]
It requires the Product Rule after taking logarithms.
It requires the Chain Rule for
\[ \log(\log x). \]
It reinforces the technique of multiplying by $y$ after obtaining
\[ \frac{1}{y}\frac{dy}{dx}. \]
It is a useful model for more complicated variable-base and variable-exponent differentiation problems.

Significance for Competitive Entrance Examinations

Variable-base, variable-exponent expressions are frequently used to test conceptual understanding of logarithmic differentiation. The expression

\[ (\log x)^{\log x} \]

is especially useful because both the base and exponent contain the same function.

The main competitive-examination skill is recognising the structure quickly rather than attempting to expand or manipulate the power directly. Once logarithms are taken, the problem reduces to the Product Rule and Chain Rule.

The resulting logarithmic derivative

\[ \frac{1}{y}\frac{dy}{dx} = \frac{\log(\log x)+1}{x} \]

can also be useful in subsequent questions involving monotonicity, stationary points or related rates.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Use logarithmic differentiation when both the base and exponent depend on $x$.
\[ \log\left[(\log x)^{\log x}\right] = \log x\log(\log x). \]
\[ \frac{d}{dx}\log x=\frac1x. \]
\[ \frac{d}{dx}\log(\log x) = \frac{1}{x\log x}. \]
The Product Rule is required for
\[ \log x\log(\log x). \]
The logarithmic derivative is
\[ \frac1y\frac{dy}{dx} = \frac{\log(\log x)+1}{x}. \]
For the straightforward real logarithmic-differentiation treatment, $x>1$.
The final derivative is
\[ \boxed{ \frac{dy}{dx} = \frac{(\log x)^{\log x}}{x} \left[\log(\log x)+1\right] }. \]

📘 Concept & Theory

Concept/Theory

›

This problem involves the chain rule. If

\[ y=\cos u, \]

then

\[ \frac{dy}{dx}=-\sin u\frac{du}{dx}. \]

We also use the standard derivatives

\[ \frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}, \]

\[ \frac{d}{dx}(\sin x)=\cos x. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Take the complete expression inside the cosine as $u$.
Differentiate $y=\cos u$ using the chain rule.
Differentiate $u=\cos^{-1}x+b\sin x$.
Substitute $\dfrac{du}{dx}$ into the chain-rule result.
Simplify the final expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Given —
\[y=\cos\left(\cos^{-1}x+b\sin x\right)\]
Let
\[u=\cos^{-1}x+b\sin x\]
Therefore,
\[y=\cos u\]
Differentiate both sides with respect to $x$:
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
Since
\[\frac{dy}{du}=-\sin u\]
we have
\[\frac{dy}{dx}=-\sin u\frac{du}{dx}\]
Now differentiate
\[u=\cos^{-1}x+b\sin x\]
Since $b$ is a constant,
\[\frac{du}{dx}=\frac{d}{dx}(\cos^{-1}x)+b\frac{d}{dx}(\sin x).\]
Using
\[\frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}}\]
and
\[\frac{d}{dx}(\sin x)=\cos x\]
we obtain
\[\frac{du}{dx}=-\frac{1}{\sqrt{1-x^2}}+b\cos x\]
Substituting this into the expression for $\dfrac{dy}{dx}$,
\[\frac{dy}{dx}=-\sin u\left(-\frac{1}{\sqrt{1-x^2}}+b\cos x\right)\]
Since
\[u=\cos^{-1}x+b\sin x,\]
we get
\[\frac{dy}{dx}=-\sin\left(\cos^{-1}x+b\sin x\right)\left(b\cos x-\frac{1}{\sqrt{1-x^2}}\right)\]
Rearranging the factors,
\[\boxed{\frac{dy}{dx}=\left(\frac{1}{\sqrt{1-x^2}}-b\cos x\right)\sin\left(\cos^{-1}x+b\sin x\right)}\]
Hence
\[\boxed{\frac{dy}{dx}=\left(\frac{1}{\sqrt{1-x^2}}-b\cos x\right)\sin\left(\cos^{-1}x+b\sin x\right)}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the correct application of the chain rule together with the derivative of the inverse trigonometric function $\cos^{-1}x$. Such questions are important because a common error is to confuse the derivative of an inverse trigonometric function with another trigonometric function.

For board examinations, writing the chain-rule steps explicitly helps demonstrate the method and avoids losing marks. For competitive examinations, recognizing the composite structure quickly is useful for obtaining the derivative efficiently.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

For a composite function $y=\cos u$,
\[ \frac{dy}{dx}=-\sin u\frac{du}{dx}. \]
The fundamental inverse-trigonometric derivative is
\[ \frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}. \]
Constants such as $b$ remain unchanged during differentiation.
Always differentiate the complete inner function before substitution.

📘 Concept & Theory

Concept/Theory

›

The given function has the form

\[ y=u^u \]

where both the base and the exponent depend on $x$. Such expressions cannot be differentiated directly using the ordinary Power Rule. The appropriate method is logarithmic differentiation.

For a function of the form

\[ y=u^u, \]

taking logarithms gives

\[ \log y=u\log u. \]

Differentiating this expression requires the Product Rule, because $u$ and $\log u$ are both functions of $x$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let $y=(\sin x-\cos x)^{(\sin x-\cos x)}$.
Take logarithm on both sides.
Put $u=\sin x-\cos x$ to simplify the calculation.
Differentiate $u\log u$ using the Product Rule.
Find $\dfrac{du}{dx}$ using the derivatives of $\sin x$ and $\cos x$.
Substitute $u$ and $y$ back into the result.
State the final derivative in a simplified form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Define the function
\[y=\left(\sin x-\cos x\right)^{\left(\sin x-\cos x\right)}\]
Taking natural logarithm on both sides,
\[\log y=\log\left[\left(\sin x-\cos x\right)^{\left(\sin x-\cos x\right)}\right]\]
Using the logarithmic identity
\[\log(a^b)=b\log a\]
we obtain
\[\log y=\left(\sin x-\cos x\right)\log\left(\sin x-\cos x\right)\]
Introduce a substitution
Let
\[u=\sin x-\cos x\]
Therefore,
\[\log y=u\log u\]
Differentiate both sides with respect to $x$
Differentiating the left-hand side,
\[\frac{d}{dx}(\log y)=\frac{1}{y}\frac{dy}{dx}\]
On the right-hand side, $u\log u$ is a product of two functions of $x$. Therefore, we use the Product Rule:
\[\frac{d}{dx}(u\log u)=u\frac{d}{dx}(\log u)+\log u\frac{du}{dx}\]
Using the Chain Rule,
\[\frac{d}{dx}(\log u)=\frac{1}{u}\frac{du}{dx}\]
Hence,
\[\frac{1}{y}\frac{dy}{dx}=u\left(\frac{1}{u}\frac{du}{dx}\right)+\log u\frac{du}{dx}\]
Simplifying,
\[\frac{1}{y}\frac{dy}{dx}=\frac{du}{dx}+\log u\frac{du}{dx}\]
Taking $\dfrac{du}{dx}$ common,
\[\frac{1}{y}\frac{dy}{dx}=\left(1+\log u\right)\frac{du}{dx}\]
Find $\dfrac{du}{dx}$
Since
\[u=\sin x-\cos x\]
differentiating with respect to $x$,
\[\frac{du}{dx}=\frac{d}{dx}(\sin x)-\frac{d}{dx}(\cos x)\]
Using
\[\frac{d}{dx}(\sin x)=\cos x,\quad \frac{d}{dx}(\cos x)=-\sin x,\]
we get
\[\frac{du}{dx}=\cos x-(-\sin x)\]
Therefore,
\[\frac{du}{dx}=\cos x+\sin x\]
Substitute $\dfrac{du}{dx}$
\[\frac{1}{y}\frac{dy}{dx}=\left(1+\log u\right)(\cos x+\sin x)\]
Multiplying both sides by $y$,
\[\frac{dy}{dx}=y(\cos x+\sin x)(1+\log u)\]
Substitute $u$ and $y$
We have
\[u=\sin x-\cos x\]
and
\[y=\left(\sin x-\cos x\right)^{\left(\sin x-\cos x\right)}\]
Therefore,
\[\boxed{\frac{dy}{dx}=\left(\sin x-\cos x\right)^{\left(\sin x-\cos x\right)}(\cos x+\sin x)\left[1+\log\left(\sin x-\cos x\right)\right]}\]
Domain Check
The interval given in the question is
\[\frac{\pi}{4} < x < \frac{3\pi}{4}\]
On this interval,
\[\sin x-\cos x > 0.\]
This is important because the real-valued logarithm $\log(\sin x-\cos x)$ requires its argument to be positive. Thus, the logarithmic differentiation used above is valid throughout the specified open interval.
At $x=\frac{\pi}{4}$, the base becomes zero, so the logarithmic step is not applicable there. The question specifically gives the open interval $\frac{\pi}{4} < x < \frac{3\pi}{4}$, which avoids this issue.

🎯 Exam Significance

Exam Significance

›

This is a standard application of logarithmic differentiation.
The problem tests the ability to differentiate a function in which both the base and exponent contain $x$.
The Product Rule is essential after writing $\log y=u\log u$.
The Chain Rule is required while differentiating $\log u$.
Writing every intermediate step helps avoid losing the factor $\dfrac{du}{dx}$.
The given interval is mathematically significant because it ensures $\sin x-\cos x>0$.

Significance for Competitive Entrance Exams

Recognise $u^u$, $f(x)^{g(x)}$, and similar variable-base-variable-exponent expressions as candidates for logarithmic differentiation.
Do not attempt to apply the ordinary Power Rule directly to $u^u$.
Keep the substitution $u=\sin x-\cos x$ until the differentiation is complete; this reduces algebraic errors.
Domain and sign conditions can become important in objective questions involving logarithmic differentiation.
The compact general result
\[ \frac{d}{dx}(u^u)=u^u(1+\log u)\frac{du}{dx} \]
is useful for rapid problem solving when $u>0$.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For a variable base and variable exponent, logarithmic differentiation is the natural method.
If $y=u^u$, then
\[ \frac{dy}{dx} = u^u(1+\log u)\frac{du}{dx}. \]
Here,
\[ u=\sin x-\cos x, \qquad \frac{du}{dx}=\sin x+\cos x. \]
The final expression must contain the original power $u^u$.
The specified interval ensures that the logarithmic transformation is valid over the required domain.

📘 Concept & Theory

Concept/Theory

›

The expression contains four terms. Since differentiation is a linear operation, each term can be differentiated separately.

The important point is that $a$ is a fixed constant. Therefore, with respect to $x$,

\[ \frac{da}{dx}=0. \]

Consequently,

\[ \frac{d}{dx}(a^a)=0 \]

because both the base $a$ and the exponent $a$ are constants.

The term $x^x$ requires logarithmic differentiation because both its base and exponent depend on $x$. The terms $x^a$ and $a^x$ can be differentiated using the standard power and exponential rules, respectively.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate each term separately.
Use logarithmic differentiation for $x^x$.
Use the Power Rule for $x^a$, since $a$ is constant.
Use the exponential differentiation rule for $a^x$.
Recognise that $a^a$ is constant and therefore has derivative zero.
Add all the individual derivatives.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 26 steps

Given — function
\[y=x^x+x^a+a^x+a^a\]
Since $a$ is fixed, $a^a$ is a constant. Hence,
\[\frac{d}{dx}(a^a)=0\]
Therefore,
\[\frac{dy}{dx}=\frac{d}{dx}(x^x)+\frac{d}{dx}(x^a)+\frac{d}{dx}(a^x)\]
Differentiate $x^x$
Let
\[y_1=x^x\]
Since both the base and exponent contain $x$, take logarithm on both sides:
\[\log y_1=x\log x\]
Differentiate both sides with respect to $x$:
\[\frac{1}{y_1}\frac{dy_1}{dx}=\frac{d}{dx}(x\log x)\]
Applying the Product Rule,
\[\frac{d}{dx}(x\log x)=x\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x)\]
Therefore,
\[\frac{1}{y_1}\frac{dy_1}{dx}=x\left(\frac{1}{x}\right)+\log x(1).\]
Hence,
\[\frac{1}{y_1}\frac{dy_1}{dx}=1+\log x\]
Multiplying by $y_1$,
\[\frac{dy_1}{dx}=y_1(1+\log x)\]
Since $y_1=x^x$,
\[\boxed{\frac{dy_1}{dx}=x^x(1+\log x)}\]
Differentiate $x^a$
Let
\[y_2=x^a\]
Here $a$ is a constant. Therefore, the ordinary Power Rule applies:
\[\frac{d}{dx}(x^a)=ax^{a-1}\]
Thus,
\[\boxed{\frac{dy_2}{dx}=ax^{a-1}}\]
Alternatively, logarithmic differentiation also gives the same result:
\[\log y_2=a\log x\]
Differentiating,
\[\frac{1}{y_2}\frac{dy_2}{dx}=a\frac{1}{x},\]
because $a$ is constant. Hence,
\[\frac{dy_2}{dx}=y_2\frac{a}{x}=x^a\frac{a}{x}=ax^{a-1}\]
Differentiate $a^x$
Let
\[y_3=a^x\]
Here $a$ is the constant base and $x$ is the variable exponent. Using
\[\frac{d}{dx}(a^x)=a^x\log a,\quad a>0,\]
we obtain
\[\boxed{\frac{dy_3}{dx}=a^x\log a}\]
This can also be verified using logarithmic differentiation:
\[\log y_3=x\log a\]
Since $a$ is constant, $\log a$ is also constant. Therefore,
\[\frac{1}{y_3}\frac{dy_3}{dx}=\log a\]
Thus,
\[\frac{dy_3}{dx}=a^x\log a\]
Differentiate $a^a$
Since $a$ is fixed, both the base and exponent of $a^a$ are constants. Therefore,
\[\boxed{\frac{d}{dx}(a^a)=0}\]
Add all the derivatives
We have
\[\frac{dy_1}{dx}=x^x(1+\log x),\]

\[\frac{dy_2}{dx}=ax^{a-1},\]

\[\frac{dy_3}{dx}=a^x\log a,\]
and
\[\frac{d}{dx}(a^a)=0.\]
Therefore,
\[\begin{aligned}\frac{dy}{dx}&=\frac{d}{dx}(x^x)+\frac{d}{dx}(x^a)+\frac{d}{dx}(a^x)+\frac{d}{dx}(a^a)\\ &=x^x(1+\log x)+ax^{a-1}+a^x\log a+0\end{aligned}\]
Hence,
\[\boxed{\frac{dy}{dx}=x^x(1+\log x)+ax^{a-1}+a^x\log a}\]

🎯 Exam Significance

Exam Significance

›

This question combines four important differentiation concepts in a single expression.
The term $x^x$ tests logarithmic differentiation.
The term $x^a$ tests the Power Rule when the exponent is constant.
The term $a^x$ tests differentiation of an exponential function with constant base.
The term $a^a$ tests recognition of a constant function.
A clear term-by-term solution makes the distinction between variable and constant quantities explicit.

Significance for Competitive Entrance Exams

The expression is useful for testing whether $a$ has been correctly treated as a constant.
Remember the distinction:
\[ \frac{d}{dx}(x^a)=ax^{a-1}, \qquad \frac{d}{dx}(a^x)=a^x\log a. \]
The special case $x^x$ should immediately suggest logarithmic differentiation.
If $a=1$, then
\[ a^x=1,\qquad a^a=1,\qquad \log a=0, \]
which is consistent with the general result.
Keeping the terms separate until the final step reduces sign and formula-substitution errors.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For $x^x$, use logarithmic differentiation:
\[ \frac{d}{dx}(x^x)=x^x(1+\log x). \]
For constant $a$,
\[ \frac{d}{dx}(x^a)=ax^{a-1}. \]
For constant $a>0$,
\[ \frac{d}{dx}(a^x)=a^x\log a. \]
A constant such as $a^a$ has derivative zero.
The final derivative is obtained by adding the derivatives of all four terms.

📘 Concept & Theory

Concept/Theory

›

The given expression contains two functions in which the variable occurs in both the base and the exponent:

\[x^{x^2-3}\qquad\text{and}\quad (x-3)^{x^2}\]

The ordinary Power Rule cannot be applied directly to these expressions because their exponents are functions of $x$. The appropriate technique is logarithmic differentiation.

For a general function

\[ y=u(x)^{v(x)}, \]

taking logarithms gives

\[ \log y=v(x)\log u(x). \]

Differentiating this product gives the useful formula

\[\boxed{\frac{dy}{dx}=u^v\left[v'\log u+v\frac{u'}{u}\right]}\]

Both terms in this question can therefore be handled systematically using logarithmic differentiation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the expression into two functions.
Use logarithmic differentiation for $x^{x^2-3}$.
Differentiate the exponent $x^2-3$ and the base $x$.
Use logarithmic differentiation for $(x-3)^{x^2}$.
Differentiate the exponent $x^2$ and the base $x-3$.
Add the two derivatives.
Use the condition $x>3$ to ensure that all logarithmic expressions are defined.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 32 steps

Given — function
\[y=x^{x^2-3}+(x-3)^{x^2}\]
Let
\[y_1=x^{x^2-3}\]

\[y_2=(x-3)^{x^2}\]
Therefore,
\[y=y_1+y_2\]
Hence, by the sum rule,
\[\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}\]
Differentiation of $y_1=x^{x^2-3}$
y_1=x^{x^2-3}
Taking logarithm,
\[\log y_1=\log\left(x^{x^2-3}\right)\]
Using
\[\log(a^b)=b\log a,\]
we get
\[\log y_1=(x^2-3)\log x\]
Differentiate both sides
\[\frac{1}{y_1}\frac{dy_1}{dx}=\frac{d}{dx}\left[(x^2-3)\log x\right]\]
The right-hand side is a product of two functions: $x^2-3$ and $\log x$. Therefore, use the Product Rule:
\[\frac{d}{dx}\left[(x^2-3)\log x\right]=\log x\frac{d}{dx}(x^2-3)+(x^2-3)\frac{d}{dx}(\log x)\]
Now,
\[\frac{d}{dx}(x^2-3)=2x\]
and
\[\frac{d}{dx}(\log x)=\frac{1}{x}\]
Therefore,
\[\frac{1}{y_1}\frac{dy_1}{dx}=2x\log x+\frac{x^2-3}{x}\]
Multiply by $y_1$
\[\frac{dy_1}{dx}=y_1\left[2x\log x+\frac{x^2-3}{x}\right]\]
Since
\[y_1=x^{x^2-3}\]
we obtain
\[\boxed{\frac{dy_1}{dx}=x^{x^2-3}\left[2x\log x+\frac{x^2-3}{x}\right]}\]
Differentiation of $y_2=(x-3)^{x^2}$
Take logarithm on both sides
\[y_2=(x-3)^{x^2}\]
Taking logarithm,
\[\log y_2=\log\left[(x-3)^{x^2}\right]\]
Using the logarithmic identity,
\[\log y_2=x^2\log(x-3)\]
Differentiate both sides
\[\frac{1}{y_2}\frac{dy_2}{dx}=\frac{d}{dx}\left[x^2\log(x-3)\right]\]
Applying the Product Rule,
\[\frac{1}{y_2}\frac{dy_2}{dx}=\log(x-3)\frac{d}{dx}(x^2)+x^2\frac{d}{dx}\left[\log(x-3)\right]\]
Now,
\[\frac{d}{dx}(x^2)=2x\]
For the second derivative, apply the Chain Rule:
\[\frac{d}{dx}\left[\log(x-3)\right]=\frac{1}{x-3}\frac{d}{dx}(x-3)\]
Since
\[\frac{d}{dx}(x-3)=1\]
we obtain
\[\frac{d}{dx}\left[\log(x-3)\right]=\frac{1}{x-3}\]
Therefore,
\[\frac{1}{y_2}\frac{dy_2}{dx}=2x\log(x-3)+\frac{x^2}{x-3}\]
Multiply by $y_2$
\[\frac{dy_2}{dx}=y_2\left[2x\log(x-3)+\frac{x^2}{x-3}\right]\]
Since
\[y_2=(x-3)^{x^2}\]
we get
\[\boxed{\frac{dy_2}{dx}=(x-3)^{x^2}\left[2x\log(x-3)+\frac{x^2}{x-3}\right]}\]
Add the Two Derivatives
Since
\[y=y_1+y_2\]
we have
\[\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}\]
Substituting the two results,
\[\begin{aligned}\frac{dy}{dx}&=x^{x^2-3}\left[2x\log x+\frac{x^2-3}{x}\right]\\ &\quad +(x-3)^{x^2}\left[2x\log(x-3)+\frac{x^2}{x-3}\right]\end{aligned}\]
Hence
\[\boxed{\frac{dy}{dx}=x^{x^2-3}\left[2x\log x+\frac{x^2-3}{x}\right]+(x-3)^{x^2}\left[2x\log(x-3)+\frac{x^2}{x-3}\right]}\]
Domain Check
The question specifies
\[x>3.\]
This condition is important because it guarantees
\[x>0\]
and
\[x>0\]
Therefore, both $\log x$ and $\log(x-3)$ are defined as real logarithms. It also ensures that the bases $x$ and $x-3$ are positive, making the logarithmic differentiation valid in the real domain.

🎯 Exam Significance

Exam Significance

›

This problem is a direct application of logarithmic differentiation to functions where both the base and exponent depend on $x$.
The Product Rule is required after taking logarithms.
The Chain Rule is required when differentiating $\log(x-3)$.
The question tests whether students correctly distinguish between differentiating the base and differentiating the exponent.
The condition $x>3$ should not be ignored because it guarantees the validity of the logarithmic transformation over the real numbers.

Significance for Competitive Entrance Exams

Expressions of the form $u(x)^{v(x)}$ should immediately suggest logarithmic differentiation.
The general pattern
\[ \frac{d}{dx}\left[u^v\right] = u^v \left[ v'\log u+v\frac{u'}{u} \right] \]
can significantly shorten calculations.
For $x^{x^2-3}$, the derivative of the exponent contributes $2x\log x$, while the derivative of the base contributes $\dfrac{x^2-3}{x}$.
For $(x-3)^{x^2}$, the derivative of the base contributes $\dfrac{x^2}{x-3}$.
Domain conditions can be decisive in objective questions involving logarithmic differentiation.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For $y=u^v$,
\[ \frac{dy}{dx} = u^v \left[ v'\log u+v\frac{u'}{u} \right]. \]
For $x^{x^2-3}$,
\[ \frac{d}{dx}\left(x^{x^2-3}\right) = x^{x^2-3} \left[ 2x\log x+\frac{x^2-3}{x} \right]. \]
For $(x-3)^{x^2}$,
\[ \frac{d}{dx}\left((x-3)^{x^2}\right) = (x-3)^{x^2} \left[ 2x\log(x-3)+\frac{x^2}{x-3} \right]. \]
The supplied solution is mathematically correct; the main improvement is to make the Product Rule, Chain Rule, domain condition, and intermediate steps explicit.
The final derivative is obtained by adding the two logarithmic-differentiation results.

📘 Concept & Theory

Concept/Theory

›

Here both $x$ and $y$ are given as functions of the parameter $t$:

\[ x=f(t),\qquad y=g(t). \]

Therefore, $y$ is given parametrically. When $x$ and $y$ are expressed in terms of a common parameter, the derivative is obtained from

\[ \boxed{ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} } \]

This follows from the Chain Rule:

\[ \frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}, \qquad \frac{dx}{dt}\neq0. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate $y$ with respect to $t$.
Differentiate $x$ with respect to $t$.
Use the parametric differentiation formula
\[ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}. \]
Simplify the resulting trigonometric expression using half-angle identities.
Use the given interval to determine the correct sign of the half-angle expressions.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 22 steps

Differentiate $y$ with respect to $t$
We have
\[y=12(1-\cos t)\]
Differentiating both sides with respect to $t$,
\[ \frac{dy}{dt} = 12\frac{d}{dt}(1-\cos t). \]
Since
\[\frac{d}{dt}(1)=0\]
and
\[\frac{d}{dt}(-\cos t)=\sin t,\]
we get
\[\boxed{\frac{dy}{dt}=12\sin t}\]
Differentiate $x$ with respect to $t$
We have
\[x=10(t-\sin t)\]
Differentiating with respect to $t$,
\[\frac{dx}{dt}=10\frac{d}{dt}(t-\sin t)\]
Since
\[\frac{d}{dt}(t)=1\]
and
\[\frac{d}{dt}(\sin t)=\cos t\]
we obtain
\[\frac{dx}{dt}=10(1-\cos t)\]
Therefore,
\[\boxed{\frac{dx}{dt}=10(1-\cos t)}\]
Apply the parametric differentiation formula
For parametric equations,
\[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\]
Substituting the values obtained above,
\[\frac{dy}{dx}=\frac{12\sin t}{10(1-\cos t)}\]
Canceling the common factor $2$,
\[\frac{dy}{dx}=\frac{6}{5}\frac{\sin t}{1-\cos t}\]
Simplify using half-angle identities
Use the identities
\[\sin t=2\sin\frac{t}{2}\cos\frac{t}{2}\]
and
\[1-\cos t=2\sin^2\frac{t}{2}\]
Therefore,
\[\frac{\sin t}{1-\cos t}=\frac{2\sin\frac{t}{2}\cos\frac{t}{2}}{2\sin^2\frac{t}{2}}\]
Cancel $2\sin\dfrac{t}{2}$:
\[\frac{\sin t}{1-\cos t}=\frac{\cos\frac{t}{2}}{\sin\frac{t}{2}}.\]
Hence,
\[\frac{\sin t}{1-\cos t}=\cot\frac{t}{2}\]
Thus,
\[\frac{dy}{dx}=\frac{6}{5}\cot\frac{t}{2}\]
Role of the Given Interval
The question specifies
\[-\frac{\pi}{2} < t < \frac{\pi}{2}\]
Therefore,
\[-\frac{\pi}{4}<\frac{t}{2}<\frac{\pi}{4}\]
Hence $\sin\dfrac{t}{2}$ has the same sign as $t$, and the expression $\cot\dfrac{t}{2}$ is well-defined for every $t\neq0$ in the given open interval.
At $t=0$
\[\frac{dx}{dt}=10(1-\cos0)=0\]
and
\[\frac{dy}{dt}=12\sin0=0\]
Thus the ratio $\dfrac{dy/dt}{dx/dt}$ takes the indeterminate form $\dfrac{0}{0}$ at $t=0$, so the parametric derivative formula cannot be directly evaluated there. The simplified expression
\[\frac{6}{5}\cot\frac{t}{2}\]
also becomes unbounded as $t\to0$

🎯 Exam Significance

Exam Significance

›

This is a standard application of differentiation of parametric equations.
The key formula to remember is
\[ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}. \]
The question also tests basic trigonometric identities, particularly the half-angle identities.
Writing $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ separately makes the solution systematic and reduces errors.
The final simplification to $\cot\dfrac{t}{2}$ is a useful algebraic and trigonometric step.

Significance for Competitive Entrance Exams

Parametric differentiation is frequently used when eliminating the parameter would be inconvenient.
Recognise immediately that the derivative is a ratio of parameter derivatives.
The identity
\[ \frac{\sin t}{1-\cos t}=\cot\frac{t}{2} \]
is useful for rapid simplification.
Always check whether $\dfrac{dx}{dt}=0$ at a particular parameter value before applying the parametric derivative formula.
The interval information can be important when simplifying trigonometric expressions involving square roots or inverse functions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For parametric equations $x=f(t)$ and $y=g(t)$,
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. \]
Here,
\[ \frac{dy}{dt}=12\sin t \]
and
\[ \frac{dx}{dt}=10(1-\cos t). \]
Therefore,
\[ \frac{dy}{dx} = \frac{6}{5} \frac{\sin t}{1-\cos t}. \]
Using half-angle identities,
\[ \frac{\sin t}{1-\cos t} = \cot\frac{t}{2}. \]
Hence,
\[ \boxed{ \frac{dy}{dx} = \frac{6}{5}\cot\frac{t}{2} }. \]

📘 Concept & Theory

Concept/Theory

›

The given function contains an inverse sine function and a composite function involving a square root. The second term therefore requires the Chain Rule.

The basic derivative formula required is

\[ \frac{d}{dx}\left(\sin^{-1}u\right) = \frac{1}{\sqrt{1-u^2}}\frac{du}{dx}. \]

There is also an important inverse-trigonometric identity behind this question. Since \(0

\[ \sqrt{1-x^2}>0. \]

If we put

\[ x=\sin\theta, \qquad 0<\theta<\frac{\pi}{2}, \]

then

\[ \sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \cos\theta. \]

Consequently,

\[ \sin^{-1}x=\theta \]

and

\[ \sin^{-1}\sqrt{1-x^2} = \sin^{-1}(\cos\theta) = \frac{\pi}{2}-\theta. \]

Thus, the entire expression is actually constant:

\[ y=\frac{\pi}{2}. \]

Therefore, we already expect

\[ \frac{dy}{dx}=0. \]

However, to demonstrate the differentiation method required by the exercise, we will obtain the result directly by differentiation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the function into two terms.
Differentiate $\sin^{-1}x$ directly.
For $\sin^{-1}\sqrt{1-x^2}$, introduce $u=\sqrt{1-x^2}$.
Apply the Chain Rule.
Use $0 < x < 1$ to simplify $\sqrt{x^2}=x$.
Add the two derivatives and simplify.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 27 steps

Separate the two terms
Let
\[y=y_1+y_2\]
where $y_1=\sin^{-1}x$ and
$y_2=\sin^{-1}\sqrt{1-x^2}$
Therefore,
\[\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}\]
Differentiate $y_1=\sin^{-1}x$
Using
\[\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}},\]
we obtain
\[\boxed{\frac{dy_1}{dx}=\frac{1}{\sqrt{1-x^2}}}\]
Differentiate $y_2=\sin^{-1}\sqrt{1-x^2}$
Let
\[u=\sqrt{1-x^2}\]
Then
\[y_2=\sin^{-1}u\]
Using the Chain Rule,
\[\frac{dy_2}{dx}=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}\]
Find $\dfrac{du}{dx}$
Since
\[u=(1-x^2)^{1/2}\]
differentiate using the Chain Rule:
\[\frac{du}{dx}=\frac{1}{2}(1-x^2)^{-1/2}\frac{d}{dx}(1-x^2)\]
Now,
\[\frac{d}{dx}(1-x^2)=-2x\]
Hence,
\[\frac{du}{dx}=\frac{1}{2}(1-x^2)^{-1/2}(-2x)\]
Therefore,
\[\frac{du}{dx}=-\frac{x}{\sqrt{1-x^2}}\]
Substitute $\dfrac{du}{dx}$
We have
\[\frac{dy_2}{dx}=\frac{1}{\sqrt{1-u^2}}\left(-\frac{x}{\sqrt{1-x^2}}\right)\]
Thus,
\[\frac{dy_2}{dx}=-\frac{x}{\sqrt{1-u^2}\sqrt{1-x^2}}\]
Since
\[u=\sqrt{1-x^2}\]
we have
\[u^2=1-x^2\]
Therefore,
\[\sqrt{1-u^2}=\sqrt{1-(1-x^2)}\]
Hence,
\[\sqrt{1-u^2}=\sqrt{x^2}\]
Now the condition $0 < x < 1$ is crucial. Since $x>0$,
\[\sqrt{x^2}=|x|=x\]
Therefore,
\[\frac{dy_2}{dx}=-\frac{x}{x\sqrt{1-x^2}}\]
Since $x>0$, we can cancel $x$:
\[\boxed{\frac{dy_2}{dx}=-\frac{1}{\sqrt{1-x^2}}}\]
Add the two derivatives
We have
\[\frac{dy_1}{dx}=\frac{1}{\sqrt{1-x^2}}\]
and
\[\frac{dy_2}{dx}=-\frac{1}{\sqrt{1-x^2}}\]
Therefore,
\[\begin{aligned}\frac{dy}{dx}&=\frac{dy_1}{dx}+\frac{dy_2}{dx}\\ &=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\\ &=0\end{aligned}\]
Hence,
\[\boxed{\frac{dy}{dx}=0}\]
Important Domain Point
The condition $0 < x < 1$ is not merely decorative. It is essential when simplifying
\[\sqrt{x^2}\]
In general,
\[\sqrt{x^2}=|x|\]
not simply $x$
Because the question specifies $x>0$, we can correctly write
\[\sqrt{x^2}=x\]
This is also why the cancellation
\[\frac{x}{x}=1\]
is valid throughout the given interval.

🎯 Exam Significance

Exam Significance

›

This question tests the derivative of an inverse trigonometric function combined with the Chain Rule.
The derivative
\[ \frac{d}{dx}(\sin^{-1}u) = \frac{u'}{\sqrt{1-u^2}} \]
should be recalled accurately.
The condition \(0
The cancellation of the two derivative terms provides a clean final answer.
The identity-based alternative solution is useful for checking the differentiation result.

Significance for Competitive Entrance Exams

The expression can be recognised as a constant using the substitution $x=\sin\theta$.
For a rapid solution, observe that
\[ \sin^{-1}x+\sin^{-1}\sqrt{1-x^2} = \frac{\pi}{2}, \quad 0 < x < 1. \]
Consequently, its derivative is immediately zero.
For derivative-based questions, however, the Chain Rule method is important because it demonstrates the underlying calculus technique.
Always retain absolute values when simplifying square roots unless the domain provides the required sign information.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

\[ \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}. \]
For a composite inverse sine function,
\[ \frac{d}{dx}(\sin^{-1}u) = \frac{u'}{\sqrt{1-u^2}}. \]
For $0 < x < 1$,
\[ \sqrt{x^2}=x. \]
The two derivative terms cancel exactly.
The original expression is actually constant:
\[ y=\frac{\pi}{2}. \]
Hence,
\[ \boxed{\frac{dy}{dx}=0}. \]

📘 Concept & Theory

Concept/Theory

›

This is an implicit differentiation problem because $y$ is given indirectly as a function of $x$.

The original equation contains square roots of $1+x$ and $1+y$. Instead of immediately differentiating the square-root expression, a useful substitution can simplify the relation considerably.

Let

\[ a=\sqrt{1+x}, \qquad b=\sqrt{1+y}. \]

Then

\[ x=a^2-1, \qquad y=b^2-1. \]

Substitution into the original equation produces a factorisation that gives a much simpler relationship between $x$ and $y$.

🗺️ Solution Roadmap

Step-by-step Plan

›

Start from the original equation rather than squaring immediately.
Put $a=\sqrt{1+x}$ and $b=\sqrt{1+y}$.
Use $x=a^2-1$ and $y=b^2-1$.
Factorise the resulting equation.
Use the positivity of the square roots to select the valid factor.
Obtain an explicit relation between $x$ and $y$.
Differentiate that relation and obtain the correct derivative.
Compare the result with the derivative stated in the question.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 36 steps

Given — equation
\[x\sqrt{1+y}+y\sqrt{1+x}=0\]
Introduce substitutions
Let
\[a=\sqrt{1+x}\]
and
\[b=\sqrt{1+y}\]
Then
\[a^2=1+x\]
so
\[x=a^2-1\]
Similarly,
\[b^2=1+y\]
so
\[y=b^2-1\]
Substitute into the original equation
We have
\[x\sqrt{1+y}+y\sqrt{1+x}=0.\]
Substituting
\[x=a^2-1,\quad y=b^2-1,\quad \sqrt{1+x}=a,\quad \sqrt{1+y}=b,\]
gives
\[(a^2-1)b+(b^2-1)a=0\]
Expand and factorise
\[a^2b-b+ab^2-a=0\]
Group the terms:
\[a^2b+ab^2-a-b=0\]
Taking $ab$ common from the first two terms,
\[ab(a+b)-(a+b)=0\]
Therefore,
\[(a+b)(ab-1)=0\]
Select the valid factor
Since $a$ and $b$ are principal square roots,
\[a=\sqrt{1+x}\geq0,\quad b=\sqrt{1+y}\geq0.\]
On the relevant nontrivial branch, $a+b>0$. Hence,
\[ab-1=0\]
Thus,
\[ab=1\]
Substituting back $a=\sqrt{1+x}$ and $b=\sqrt{1+y}$,
\[\sqrt{1+x}\sqrt{1+y}=1\]
Therefore,
\[\sqrt{(1+x)(1+y)}=1\]
Squaring both sides gives
\[(1+x)(1+y)=1\]
Simplify the relation between $x$ and $y$
\[1+x+y+xy=1\]
Cancel $1$ from both sides:
\[x+y+xy=0\]
Collect the terms containing $y$:
\[y+xy=-x\]
Factor $y$:
\[y(1+x)=-x\]
Since $x>-1$, we have $1+x>0$, so division is permitted:
\[\boxed{y=-\frac{x}{1+x}}\]
Differentiate
\[y=-\frac{x}{1+x}\]
Using the Quotient Rule,
\[\frac{dy}{dx}=-\frac{(1+x)(1)-x(1)}{(1+x)^2}\]
Simplifying the numerator,
\[(1+x)-x=1\]
Therefore,
\[\boxed{\frac{dy}{dx}=-\frac{1}{(1+x)^2}}\]
Direct Verification by Implicit Differentiation
We can also verify the result directly from the original equation. Start with
\[x\sqrt{1+y}+y\sqrt{1+x}=0\]
Differentiate both sides with respect to $x$.
For the first term, using the Product Rule and Chain Rule,
\[\frac{d}{dx}\left[x\sqrt{1+y}\right]=\sqrt{1+y}+x\frac{1}{2\sqrt{1+y}}\frac{dy}{dx}\]
For the second term,
\[\frac{d}{dx}\left[y\sqrt{1+x}\right]=\sqrt{1+x}\frac{dy}{dx}+y\frac{1}{2\sqrt{1+x}}\]
Hence,
\[\sqrt{1+y}+\frac{x}{2\sqrt{1+y}}\frac{dy}{dx}+\sqrt{1+x}\frac{dy}{dx}+\frac{y}{2\sqrt{1+x}}=0\]
Therefore,
\[\left[\frac{x}{2\sqrt{1+y}}+\sqrt{1+x}\right]\frac{dy}{dx}=-\left[\sqrt{1+y}+\frac{y}{2\sqrt{1+x}}\right]\]
Thus,
\[\frac{dy}{dx}=-\frac{\sqrt{1+y}+\dfrac{y}{2\sqrt{1+x}}}{\dfrac{x}{2\sqrt{1+y}}+\sqrt{1+x}}\]
Using the relation
\[\sqrt{1+x}\sqrt{1+y}=1\]
we have
\[\sqrt{1+y}=\frac{1}{\sqrt{1+x}}\]
Also,
\[y=-\frac{x}{1+x}\]
Substitution and simplification again yield
\[\boxed{\frac{dy}{dx}=-\frac{1}{(1+x)^2}}\]

🎯 Exam Significance

Exam Significance

›

This problem combines algebraic manipulation, square-root identities, and implicit differentiation.
It is important to distinguish the equation actually given from a possibly incorrect printed answer.
Using the substitution $a=\sqrt{1+x}$, $b=\sqrt{1+y}$ provides a much cleaner route than expanding and differentiating a squared equation.
Domain restrictions must be considered whenever square roots and division are involved.
Checking the final answer by differentiating the simplified relation is an effective way to detect a typographical error in a textbook or solution key.

Significance for Competitive Entrance Exams

Look for substitutions that transform square-root expressions into algebraic variables.
The factorisation
\[ (a+b)(ab-1)=0 \]
is the key structural observation.
Do not blindly square equations when a direct factorisation is available.
Remember that
\[ \sqrt{x^2}=|x|, \]
and use the domain to determine the correct sign.
When a derived result disagrees with a supplied answer, substitute a simple admissible value to test the two expressions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The original equation is
\[ x\sqrt{1+y}+y\sqrt{1+x}=0. \]
Putting
\[ a=\sqrt{1+x},\qquad b=\sqrt{1+y} \]
gives
\[ (a+b)(ab-1)=0. \]
The relevant branch gives
\[ ab=1. \]
Hence,
\[ (1+x)(1+y)=1. \]
Therefore,
\[ y=-\frac{x}{1+x}. \]
Finally,
\[ \boxed{ \frac{dy}{dx}=-\frac{1}{(1+x)^2} }. \]
The answer $-1/(1+x^2)$ does not follow from the equation as written.

📘 Concept & Theory

Concept/Theory

›

The equation

\[ (x-a)^2+(y-b)^2=c^2 \]

represents a circle with centre $(a,b)$ and radius $c$.

Here $a$ and $b$ determine the position of the centre, while $c$ determines the radius. The quantity

\[ \frac{ \left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2} }{ \dfrac{d^2y}{dx^2} } \]

is closely related to the reciprocal of curvature. For a circle, its magnitude is equal to the radius $c$.

The key objective is therefore to show that the quantities involving $a$ and $b$ disappear after using the original circle equation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the circle equation once to find $\dfrac{dy}{dx}$.
Differentiate again to obtain $\dfrac{d^2y}{dx^2}$.
Calculate $1+\left(\dfrac{dy}{dx}\right)^2$.
Use the original equation to replace
\[ (x-a)^2+(y-b)^2 \]
by $c^2$.
Simplify the required expression.
Interpret the result and identify the role of $a$, $b$, and $c$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 31 steps

Given —
\[(x-a)^2+(y-b)^2=c^2\]
Since $a$, $b$, and $c$ are constants, differentiate both sides with respect to $x$:
\[\frac{d}{dx}(x-a)^2+\frac{d}{dx}(y-b)^2=\frac{d}{dx}(c^2)\]
Using the Chain Rule,
\[2(x-a)+2(y-b)\frac{dy}{dx}=0\]
Dividing by $2$,
\[(x-a)+(y-b)\frac{dy}{dx}=0\]
Therefore,
\[(y-b)\frac{dy}{dx}=-(x-a)\]
Hence,
\[\boxed{\frac{dy}{dx}=-\frac{x-a}{y-b}}\]
Find $\dfrac{d^2y}{dx^2}$
We have
\[\frac{dy}{dx}=-\frac{x-a}{y-b}\]
Differentiate both sides using the Quotient Rule:
\[\frac{d^2y}{dx^2}=-\frac{(y-b)\dfrac{d}{dx}(x-a)-(x-a)\dfrac{d}{dx}(y-b)}{(y-b)^2}\]
Since
\[\frac{d}{dx}(x-a)=1\]
and
\[\frac{d}{dx}(y-b)=\frac{dy}{dx}\]
we obtain
\[\frac{d^2y}{dx^2}=-\frac{(y-b)-(x-a)\dfrac{dy}{dx}}{(y-b)^2}\]
Now substitute
\[\frac{dy}{dx}=-\frac{x-a}{y-b}\]
Therefore,
\[\frac{d^2y}{dx^2}=-\frac{(y-b)-(x-a)\left(-\frac{x-a}{y-b}\right)}{(y-b)^2}\]
Hence,
\[\frac{d^2y}{dx^2}=-\frac{(y-b)+\dfrac{(x-a)^2}{y-b}}{(y-b)^2}\]
Taking $y-b$ common in the numerator,
\[\frac{d^2y}{dx^2}=-\frac{\dfrac{(y-b)^2+(x-a)^2}{y-b}}{(y-b)^2}\]
Therefore,
\[\frac{d^2y}{dx^2}=-\frac{(x-a)^2+(y-b)^2}{(y-b)^3}\]
Using the original circle equation,
\[(x-a)^2+(y-b)^2=c^2\]
Thus,
\[\boxed{\frac{d^2y}{dx^2}=-\frac{c^2}{(y-b)^3}}\]
Find $1+\left(\dfrac{dy}{dx}\right)^2$
From the first derivative,
\[\frac{dy}{dx}=-\frac{x-a}{y-b}\]
Squaring both sides,
\[\left(\frac{dy}{dx}\right)^2=\frac{(x-a)^2}{(y-b)^2}\]
Therefore,
\[1+\left(\frac{dy}{dx}\right)^2=1+\frac{(x-a)^2}{(y-b)^2}\]
Taking the common denominator,
\[1+\left(\frac{dy}{dx}\right)^2=\frac{(y-b)^2+(x-a)^2}{(y-b)^2}\]
Using
\[(x-a)^2+(y-b)^2=c^2\]
we get
\[1+\left(\frac{dy}{dx}\right)^2=\frac{c^2}{(y-b)^2}\]
Raise both sides to the power $\frac32$
\[\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}=\left[\frac{c^2}{(y-b)^2}\right]^{3/2}\]
Since $c>0$,
\[(c^2)^{3/2}=c^3\]
Also,
\[\left[(y-b)^2\right]^{3/2}=|y-b|^3\]
Hence,
\[\boxed{\left[1+\left(\frac{dy}{dx}\right)^2\right]^{3/2}=\frac{c^3}{|y-b|^3}}\]
Evaluate the Required Expression
Consider
\[E=\frac{\left[1+\left(\dfrac{dy}{dx}\right)^2\right]^{3/2}}{\dfrac{d^2y}{dx^2}}\]
Substituting the expressions obtained above,
\[E=\frac{\dfrac{c^3}{|y-b|^3}}{-\dfrac{c^2}{(y-b)^3}}\]
Therefore,
\[E=-c\frac{(y-b)^3}{|y-b|^3}\]
Since
\[\frac{(y-b)^3}{|y-b|^3}=\operatorname{sgn}(y-b),\]
we obtain
\[\boxed{E=-c\,\operatorname{sgn}(y-b)}.\]

🎯 Exam Significance

Exam Significance

›

This problem tests second-order implicit differentiation.
The equation represents a circle with centre $(a,b)$ and radius $c$.
The first derivative is obtained directly by implicit differentiation.
The second derivative requires careful use of the Quotient Rule.
The original equation should be used again to replace
\[ (x-a)^2+(y-b)^2 \]
by $c^2$.
Students should be careful with the distinction between a quantity and its absolute value when fractional powers such as $\frac32$ are involved.

Significance for Competitive Entrance Exams

The problem is closely connected with the radius of curvature of a circle.
The standard formula
\[ \rho= \frac{[1+(y')^2]^{3/2}}{|y''|} \]
immediately suggests that a circle should give its radius.
The parameters $a$ and $b$ represent translations of the circle and therefore should disappear from an intrinsic geometric quantity such as curvature.
Careful handling of absolute values is essential when simplifying fractional powers.
A sign discrepancy can often indicate that a signed curvature-related quantity has been used instead of the radius of curvature.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For the circle,
\[ (x-a)^2+(y-b)^2=c^2, \]
the first derivative is
\[ \frac{dy}{dx}=-\frac{x-a}{y-b}. \]
The second derivative is
\[ \frac{d^2y}{dx^2} = -\frac{c^2}{(y-b)^3}. \]
Also,
\[ \left[ 1+\left(\frac{dy}{dx}\right)^2 \right]^{3/2} = \frac{c^3}{|y-b|^3}. \]
Therefore, the expression exactly as written is
\[ -c\,\operatorname{sgn}(y-b). \]
With the standard absolute value in the denominator, the radius of curvature is
\[ \boxed{c}, \]
independent of $a$ and $b$.

📘 Concept & Theory

Concept/Theory

›

The equation

\[ \cos y=x\cos(a+y) \]

defines $y$ implicitly in terms of $x$. Therefore, we differentiate both sides with respect to $x$, remembering that $y$ is a function of $x$.

The important Chain Rule result is

\[ \frac{d}{dx}\cos(a+y) = -\sin(a+y)\frac{dy}{dx}, \]

because $a$ is constant and

\[ \frac{d}{dx}(a+y)=\frac{dy}{dx}. \]

After implicit differentiation, the main challenge is simplifying the denominator. The original equation itself provides the required substitution.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the given equation implicitly with respect to $x$.
Apply the Product Rule to $x\cos(a+y)$.
Collect all terms containing $\dfrac{dy}{dx}$.
Express $x$ in terms of $y$ using the original equation.
Simplify the denominator using the sine subtraction identity.
Obtain the required result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 26 steps

Given — equation
\[\cos y=x\cos(a+y)\]
Differentiate both sides with respect to $x$
Differentiate the left-hand side: .
\[\frac{d}{dx}(\cos y)=-\sin y\frac{dy}{dx}\]
For the right-hand side, $x\cos(a+y)$ is a product. Therefore, use the Product Rule:
\[\frac{d}{dx}\left[x\cos(a+y)\right]=x\frac{d}{dx}\left[\cos(a+y)\right]+\cos(a+y)\frac{d}{dx}(x)\]
Since
\[\frac{d}{dx}(x)=1\]
and
\[\frac{d}{dx}\left[\cos(a+y)\right]=-\sin(a+y)\frac{dy}{dx},\]
we get
\[\frac{d}{dx}\left[x\cos(a+y)\right]=-x\sin(a+y)\frac{dy}{dx}+\cos(a+y)\]
Therefore,
\[-\sin y\frac{dy}{dx}=\cos(a+y)-x\sin(a+y)\frac{dy}{dx}\]
Collect the terms containing $\dfrac{dy}{dx}$
Add $x\sin(a+y)\dfrac{dy}{dx}$ to both sides:
\[x\sin(a+y)\frac{dy}{dx}-\sin y\frac{dy}{dx}=\cos(a+y)\]
Taking $\dfrac{dy}{dx}$ common,
\[\frac{dy}{dx}\left[x\sin(a+y)-\sin y\right]=\cos(a+y)\]
Hence,
\[\frac{dy}{dx}=\frac{\cos(a+y)}{x\sin(a+y)-\sin y}\]
Simplify the Denominator
We now use the original equation
\[\cos y=x\cos(a+y)\]
Therefore, provided $\cos(a+y)\neq0$,
\[x=\frac{\cos y}{\cos(a+y)}\]
Substitute this into the denominator:
\[\sin(a+y)-\sin y=\frac{\cos y}{\cos(a+y)}\sin(a+y)-\sin y\]
Taking the common denominator $\cos(a+y)$,
\[x\sin(a+y)-\sin y=\frac{\cos y\sin(a+y)-\sin y\cos(a+y)}{\cos(a+y)}\]
Now use the identity
\[\sin A\cos B-\cos A\sin B=\sin(A-B).\]
Taking
\[A=a+y,\quad B=y,\]
we get
\[\sin(a+y)\cos y-\cos(a+y)\sin y=\sin[(a+y)-y]\]
Therefore,
\[\sin(a+y)\cos y-\cos(a+y)\sin y=\sin a\]
Hence,
\[\boxed{x\sin(a+y)-\sin y=\frac{\sin a}{\cos(a+y)}}\]
Substitute into $\dfrac{dy}{dx}$
We already obtained
\[\frac{dy}{dx}=\frac{\cos(a+y)}{x\sin(a+y)-\sin y}\]
Substituting
\[x\sin(a+y)-\sin y=\frac{\sin a}{\cos(a+y)},\]
we get
\[\frac{dy}{dx}=\frac{\cos(a+y)}{\dfrac{\sin a}{\cos(a+y)}}\]
Dividing by a fraction means multiplying by its reciprocal:
\[\frac{dy}{dx}=\cos(a+y)\frac{\cos(a+y)}{\sin a}\]
Therefore,
\[\boxed{\frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a}}\]
Role of the Condition $\cos a\neq\pm1$
The question gives
\[\cos a\neq\pm1.\]
Since $\cos a=\pm1$ would imply
\[\sin a=0\]
the given condition ensures
\[\boxed{\sin a\neq0}\]
This is essential because the final result contains $\sin a$ in the denominator:
\[\frac{dy}{dx}=\frac{\cos^2(a+y)}{\sin a}\]

🎯 Exam Significance

Exam Significance

›

This is a standard problem on implicit differentiation.
The Product Rule is required for differentiating $x\cos(a+y)$.
The Chain Rule is required for differentiating $\cos(a+y)$.
The final simplification depends on recognising the sine subtraction identity.
The condition $\cos a\neq\pm1$ should be connected with $\sin a\neq0$.
Writing the denominator simplification separately makes the proof much easier to follow.

Significance for Competitive Entrance Exams

After implicit differentiation, do not stop at
\[ \frac{dy}{dx} = \frac{\cos(a+y)} {x\sin(a+y)-\sin y}. \]
The original equation should be used to eliminate $x$.
The identity
\[ \sin(a+y)\cos y-\cos(a+y)\sin y=\sin a \]
provides the fastest simplification.
When a parameter $a$ appears inside trigonometric arguments, subtraction identities can often eliminate the dependent variable.
The condition $\sin a\neq0$ guarantees that the final denominator does not vanish because of the parameter.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Implicit differentiation gives
\[ \frac{dy}{dx} = \frac{\cos(a+y)} {x\sin(a+y)-\sin y}. \]
The original equation gives
\[ x=\frac{\cos y}{\cos(a+y)}. \]
The denominator simplifies as
\[ x\sin(a+y)-\sin y = \frac{\sin a}{\cos(a+y)}. \]
Therefore,
\[ \boxed{ \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a} }. \]
The condition $\cos a\neq\pm1$ ensures $\sin a\neq0$.

📘 Concept & Theory

Concept/Theory

›

When $x$ and $y$ are both given as functions of a parameter $t$, differentiation is performed parametrically.

The first derivative is

\[ \boxed{ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} } \]

To find the second derivative, we differentiate $\dfrac{dy}{dx}$ with respect to $t$, and then divide by $\dfrac{dx}{dt}$:

\[ \boxed{ \frac{d^2y}{dx^2} = \frac{ \dfrac{d}{dt}\left(\dfrac{dy}{dx}\right) }{ \dfrac{dx}{dt} } } \]

Equivalently, if

\[ \frac{dy}{dx}=\frac{dy/dt}{dx/dt}, \]

then the quotient rule can be used to obtain

\[ \boxed{ \frac{d^2y}{dx^2} = \frac{ \dfrac{d^2y}{dt^2}\dfrac{dx}{dt} - \dfrac{dy}{dt}\dfrac{d^2x}{dt^2} }{ \left(\dfrac{dx}{dt}\right)^3 } } \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate $x$ with respect to $t$.
Differentiate $x$ twice with respect to $t$.
Differentiate $y$ with respect to $t$.
Differentiate $y$ twice with respect to $t$.
Use the parametric second-derivative formula.
Simplify the resulting expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Find $\dfrac{dx}{dt}$
Given —
\[x=a(\cos t+t\sin t)\]
Since $a$ is a constant,
\[\frac{dx}{dt}=a\frac{d}{dt}(\cos t+t\sin t)\]
Using
\[\frac{d}{dt}(\cos t)=-\sin t\]
and and the Product Rule,
\[\frac{d}{dt}(t\sin t)=t\cos t+\sin t\]
Therefore,
\[\begin{aligned}\frac{dx}{dt}&=a[-\sin t+\sin t+t\cos t]\\&=at\cos t\end{aligned}\]
Hence,
\[\boxed{\frac{dx}{dt}=at\cos t}\]
Find $\dfrac{d^2x}{dt^2}$
Differentiate with respect to $t$:
\[\frac{dx}{dt}=at\cos t\]

\[\frac{d^2x}{dt^2}=a\frac{d}{dt}(t\cos t)\]
Using the Product Rule,
\[\frac{d}{dt}(t\cos t)=\cos t-t\sin t\]
Therefore,
\[\boxed{\frac{d^2x}{dt^2}=a(\cos t-t\sin t)}\]
Find $\dfrac{dy}{dt}$
Given —
\[y=a(\sin t-t\cos t)\]
Differentiate with respect to $t$:
\[\frac{dy}{dt}=a\frac{d}{dt}(\sin t-t\cos t)\]
Now,
\[\frac{d}{dt}(\sin t)=\cos t\]
Using the Product Rule,
\[\frac{d}{dt}(t\cos t)=\cos t-t\sin t\]
Hence,
\[\begin{aligned}\frac{dy}{dt}&=a[\cos t-(\cos t-t\sin t)]\\ &=a[\cos t-\cos t+t\sin t]\\ &=at\sin t\end{aligned}\]
Thus,
\[\boxed{\frac{dy}{dt}=at\sin t}\]
Find $\dfrac{d^2y}{dt^2}$
Differentiate
\[\frac{dy}{dt}=at\sin t\]
with respect to $t$:
\[\frac{d^2y}{dt^2}=a\frac{d}{dt}(t\sin t)\]
Using the Product Rule,
\[\frac{d}{dt}(t\sin t)=\sin t+t\cos t\]
Therefore,
\[\boxed{\frac{d^2y}{dt^2}=a(\sin t+t\cos t)}\]
Apply the Formula for the Second Parametric Derivative
The required second derivative is
\[\frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}\]
First,
\[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\]
Using
\[\frac{dy}{dt}=at\sin t\]
and
\[\frac{dx}{dt}=at\cos t,\]
we get
\[\begin{aligned}\frac{dy}{dx}&=\frac{at\sin t}{at\cos t}\\ &=\tan t\end{aligned}\]
Now differentiate with respect to $t$:
\[\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dt}(\tan t)=\sec^2t\]
Therefore,
\[\frac{d^2y}{dx^2}=\frac{\sec^2t}{at\cos t}\]
Since
\[\sec^2t=\frac{1}{\cos^2t},\]
we obtain
\[\boxed{\frac{d^2y}{dx^2}=\frac{1}{at\cos^3t}}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the second derivative of a function given parametrically.
The Product Rule is required repeatedly.
A clean solution should first calculate $dx/dt$ and $dy/dt$.
The safest route here is to simplify
\[ \frac{dy}{dx}=\tan t \]
before calculating the second derivative.
Writing the correct second-derivative formula is essential for avoiding a major conceptual error.

Significance for Competitive Entrance Exams

The fastest approach is to obtain
\[ \frac{dy}{dx}=\tan t \]
and then differentiate with respect to $t$.
Recognising cancellations in parametric expressions can significantly reduce calculation.
Always distinguish
\[ \frac{d^2y}{dt^2} \]
from
\[ \frac{d^2y}{dx^2}. \]
They are different derivatives.
The factor $a$ cancels from the first derivative but remains in the second derivative.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For parametric equations,
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. \]
For the given equations,
\[ \frac{dx}{dt}=at\cos t, \qquad \frac{dy}{dt}=at\sin t. \]
Therefore,
\[ \frac{dy}{dx}=\tan t. \]
Hence,
\[ \frac{d}{dt}\left(\frac{dy}{dx}\right)=\sec^2t. \]
Finally,
\[ \frac{d^2y}{dx^2} = \frac{\sec^2t}{at\cos t} = \boxed{\frac{1}{at\cos^3t}}. \]

📘 Concept & Theory

Concept/Theory

›

The absolute value function is defined by

\[|x|=\begin{cases}x, & x\geq0,\\ -x, & x<0.\end{cases}\]

Therefore,

\[|x|^3=\begin{cases}x^3, & x\geq0,\\ (-x)^3, & x<0.\end{cases}\]

Hence,

\[\boxed{f(x)=\begin{cases}-x^3, & x<0,\\x^3, & x\geq0.\end{cases}}\]

Since the question specifically asks us to prove that the second derivative exists for every real number, the important point is $x=0$. For $x\neq0$, ordinary differentiation of the appropriate polynomial is straightforward.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write $f(x)=|x|^3$ as a piecewise function.
Differentiate once to obtain $f'(x)$.
Check carefully that $f'(0)$ exists.
Differentiate $f'(x)$ for $x\neq0$.
Check separately whether $f''(0)$ exists.
Combine the results into a single expression for $f''(x)$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 44 steps

Express $f(x)$ in Piecewise Form
Given —
\[f(x)=|x|^3\]
For $x>0$,
\[|x|=x\]
so
\[f(x)=x^3\]
For $x < 0$,
\[|x|=-x\]
and therefore
\[f(x)=(-x)^3=-x^3\]
Thus,
\[ \boxed{ f(x)= \begin{cases} -x^3, & x < 0,\\ x^3, & x\geq0. \end{cases} } \]
Find $f'(x)$ for $x\neq0$
For $x < 0$,
\[f(x)=-x^3\]
Therefore,
\[f'(x)=-3x^2\]
For $x>0$,
\[f(x)=x^3\]
so
\[f'(x)=3x^2\]
Hence, for $x\neq0$,
\[ f'(x)= \begin{cases} -3x^2, & x < 0,\\ 3x^2, & x>0. \end{cases} \]
This can also be written compactly as
\[\boxed{f'(x)=3x|x|}\]
Verify that $f'(0)$ Exists
Because the question concerns the existence of the second derivative at every real number, we must first establish $f'(0)$.
Using the definition of derivative,
\[f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h}.\]
Since
\[f(h)=|h|^3\]
and
\[f(0)=0\]
we get
\[f'(0)=\lim_{h\to0}\frac{|h|^3}{h}\]
Now,
\[\left|\frac{|h|^3}{h}\right|=\frac{|h|^3}{|h|}=|h|^2\]
Therefore,
\[\lim_{h\to0}\left|\frac{|h|^3}{h}\right|=\lim_{h\to0}|h|^2=0\]
Hence,
\[\boxed{f'(0)=0}\]
Thus the complete first derivative is
\[\boxed{f'(x)=\begin{cases}-3x^2, & x < 0,\\0, & x=0,\\3x^2, & x>0\end{cases}}\]
Equivalently,
\[\boxed{f'(x)=3x|x|}\]
Find $f''(x)$ for $x\neq0$
For $x < 0$,
\[f'(x)=-3x^2\]
Therefore,
\[f''(x)=-6x\]
For $x>0$,
\[f'(x)=3x^2\]
Therefore,
\[f''(x)=6x\]
Thus, for $x\neq0$,
\[ f''(x)= \begin{cases} -6x, & x < 0,\\ 6x, & x>0. \end{cases} \]
This can be written as
\[\boxed{f''(x)=6|x|,\quad x\neq0}\]
Verify that $f''(0)$ Exists
This is the most important step. We cannot simply substitute $x=0$ into the expressions obtained for $x < 0$ and $x>0$ without checking the derivative at the point where the absolute value changes form.
By definition,
\[f''(0)=\lim_{h\to0}\frac{f'(0+h)-f'(0)}{h}\]
We already know that
\[f'(0)=0\]
Therefore,
\[f''(0)=\lim_{h\to0}\frac{f'(h)}{h}\]
Since
\[f'(h)=3h|h|,\]
we have
\[f''(0)=\lim_{h\to0}\frac{3h|h|}{h}\]
For $h\neq0$,
\[\frac{3h|h|}{h}=3|h|\]
Hence,
\[f''(0)=\lim_{h\to0}3|h|=0\]
Therefore,
\[\boxed{f''(0)=0}\]
Combine the Results
For $x < 0$,
\[f''(x)=-6x=6|x|\]
For $x=0$,
\[f''(0)=0=6|0|\]
For $x>0$,
\[f''(x)=6x=6|x|\]
Therefore, for every real $x$,
\[\boxed{f''(x)=6|x|}\]
Final Answer
\[\boxed{f''(x)=6|x|,\quad x\in\mathbb{R}}\]
Hence, $f''(x)$ exists for all real values of $x$
Key Observation at $x=0$
Although $f(x)=|x|^3$ contains an absolute value, the power $3$ makes the function sufficiently smooth at $x=0$.
The first derivative is
\[f'(x)=3x|x|\]
which is continuous at $x=0$, and its derivative exists there as well:
\[f''(0)=0.\]
Thus, the absolute value does not prevent the existence of the second derivative in this example.

🎯 Exam Significance

Exam Significance

›

This problem tests the concept of differentiability at a point where a function changes its algebraic form.
The point $x=0$ must be checked separately.
Using the definition of derivative at $x=0$ gives a rigorous proof that $f'(0)$ and $f''(0)$ exist.
A complete board answer should not simply differentiate $x^3$ and $-x^3$; it should verify the derivative at the joining point.
The final answer can be expressed compactly as
\[ f''(x)=6|x|. \]

Significance for Competitive Entrance Exams

Piecewise representation is often the quickest way to handle absolute-value functions.
At a point where the definition changes, always check the derivative from the definition if existence is being asked.
The compact identity
\[ f'(x)=3x|x| \]
makes the second-derivative calculation particularly efficient.
For $f''(0)$, the limit
\[ \lim_{h\to0}3|h|=0 \]
immediately establishes existence.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

\[ |x|^3= \begin{cases} -x^3,&x<0,\\ x^3,&x\geq0. \end{cases} \]
\[ f'(x)=3x|x|. \]
\[ f'(0)=0. \]
\[ f''(0)=0. \]
For $x\neq0$,
\[ f''(x)=6|x|. \]
Therefore, for every $x\in\mathbb{R}$,
\[ \boxed{f''(x)=6|x|}. \]

📘 Concept & Theory

Concept/Theory

›

We are given the sine addition formula

\[ \boxed{ \sin(A+B)=\sin A\cos B+\cos A\sin B }. \]

To derive the corresponding addition formula for cosine, differentiate both sides with respect to $A$, while treating $B$ as a constant.

Since $B$ is constant with respect to $A$,

\[ \frac{dB}{dA}=0. \]

We shall use

\[ \frac{d}{dA}(\sin A)=\cos A \]

and

\[ \frac{d}{dA}(\cos A)=-\sin A. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Start with the sine addition formula.
Differentiate both sides with respect to $A$.
Apply the Chain Rule to $\sin(A+B)$.
Differentiate the right-hand side using the Product Rule.
Use the fact that $B$ is constant.
Simplify to obtain the cosine addition formula.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Given — Identity
\[\sin(A+B)=\sin A\cos B+\cos A\sin B\]
Differentiate the Left-Hand Side
Differentiate both sides with respect to $A$:
\[\frac{d}{dA}\left[\sin(A+B)\right]=\frac{d}{dA}\left[\sin A\cos B+\cos A\sin B\right]\]
Using the Chain Rule,
\[\frac{d}{dA}\sin(A+B)=\cos(A+B)\frac{d}{dA}(A+B)\]
Since $B$ is constant,
\[\frac{d}{dA}(A+B)=1\]
Therefore,
\[\boxed{\frac{d}{dA}\sin(A+B)=\cos(A+B)}\]
Differentiate the Right-Hand Side
Consider the first term:
\[\frac{d}{dA}(\sin A\cos B)\]
Since $\cos B$ is constant with respect to $A$,
\[\frac{d}{dA}(\sin A\cos B)=\cos B\frac{d}{dA}(\sin A).\]
Hence,
\[\frac{d}{dA}(\sin A\cos B)=\cos A\cos B\]
Now consider the second term:
\[\frac{d}{dA}(\cos A\sin B)\]
Since $\sin B$ is constant with respect to $A$,
\[\frac{d}{dA}(\cos A\sin B)=\sin B\frac{d}{dA}(\cos A)\]
Since
\[\frac{d}{dA}(\cos A)=-\sin A\]
,
we obtain
\[\frac{d}{dA}(\cos A\sin B)=-\sin A\sin B\]
Combine the Results
Therefore, differentiating the complete right-hand side gives
\[\cos A\cos B-\sin A\sin B\]
Thus,
\[\cos(A+B)=\cos A\cos B-\sin A\sin B\]
Final Answer
\[\boxed{\cos(A+B)=\cos A\cos B-\sin A\sin B}\]

🎯 Exam Significance

Exam Significance

›

This problem demonstrates how one trigonometric identity can be derived from another using differentiation.
The Chain Rule is essential on the left-hand side.
The Product Rule is involved when differentiating products such as $\sin A\cos B$.
It is important to explicitly state that $B$ is treated as a constant while differentiating with respect to $A$.
The final result is one of the fundamental trigonometric addition formulae.

Significance for Competitive Entrance Exams

The derivation reinforces the relationship between differentiation and trigonometric identities.
Always identify the differentiation variable before differentiating an expression containing more than one symbol.
Constants such as $\sin B$ and $\cos B$ can be taken outside differentiation with respect to $A$.
The signs in derivatives of sine and cosine are crucial:
\[ (\sin A)'=\cos A, \qquad (\cos A)'=-\sin A. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Start with
\[ \sin(A+B)=\sin A\cos B+\cos A\sin B. \]
Differentiate with respect to $A$, treating $B$ as constant.
The left-hand side becomes
\[ \cos(A+B). \]
The right-hand side becomes
\[ \cos A\cos B-\sin A\sin B. \]
Therefore,
\[ \boxed{\cos(A+B)=\cos A\cos B-\sin A\sin B}. \]

📘 Concept & Theory

Concept/Theory

›

Recall that:

A function is continuous at $x=a$ if
\[ \lim_{x\to a}f(x)=f(a). \]
A function is differentiable at $x=a$ if its derivative exists there, i.e. its left-hand and right-hand derivatives are equal.

The absolute value function $|x|$ is continuous everywhere but is not differentiable at $x=0$. Similarly, $|x-1|$ is continuous everywhere but is not differentiable at $x=1$.

Therefore, the sum

\[ f(x)=|x|+|x-1| \]

provides a natural example having exactly two points of non-differentiability.

🗺️ Solution Roadmap

Step-by-step Plan

›

Construct an appropriate function.
Express it in piecewise form.
Show that it is continuous at $x=0$ and $x=1$.
Find its derivative in each interval.
Check the left-hand and right-hand derivatives at $x=0$.
Check the left-hand and right-hand derivatives at $x=1$.
Show that the derivative exists everywhere else.
Conclude that the function is continuous everywhere but not differentiable exactly at two points.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 40 steps

Define the Function
Consider
\[f(x)=|x|+|x-1|\]
The points at which the absolute-value expressions change their form are $x=0$ and $x=1$
We therefore consider three intervals:
\[x < 0,\quad 0\leq x < 1,\quad x\geq1\]
Piecewise Representation
For $x < 0$,
\[|x|=-x\]
and
\[|x-1|=-(x-1)=1-x\]
Hence,
\[f(x)=-x+(1-x)=1-2x\]
For $0\leq x<1$,
\[|x|=x\]
and
\[|x-1|=1-x\]
Therefore,
\[f(x)=x+(1-x)=1\]
For $x\geq1$,
\[|x|=x\]
and
\[|x-1|=x-1\]
Thus,
\[f(x)=x+(x-1)=2x-1\]
Therefore, the function can be written as
\[\boxed{f(x)=\begin{cases}1-2x, & x<0,\\[4pt] 1, & 0\leq x<1,\\[4pt] 2x-1, & x\geq1\end{cases}}\]
Show that $f(x)$ is Continuous Everywhere
Each of the three expressions
\[1-2x,\quad 1,\quad 2x-1\]
is continuous on its respective interval.
Therefore, we only need to check the joining points $x=0$ and $x=1$.
Continuity at $x=0$
The left-hand limit is
\[\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(1-2x)=1\]
The right-hand limit is
\[\lim_{x\to0^+}f(x)=\lim_{x\to0^+}1=1\]
Also,
\[f(0)=|0|+|0-1|=0+1=1\]
Hence,
\[\lim_{x\to0}f(x)=f(0)=1\]
Therefore, $f$ is continuous at $x=0$.
Continuity at $x=1$
The left-hand limit is
\[\lim_{x\to1^-}f(x)=\lim_{x\to1^-}1=1\]
The right-hand limit is
\[\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(2x-1)=1\]
Also,
\[f(1)=|1|+|1-1|=1+0=1\]
Therefore,
\[\lim_{x\to1}f(x)=f(1)=1\]
Thus, $f$ is continuous at $x=1$.
Consequently,
\[\boxed{f(x)\text{ is continuous for every }x\in\mathbb R}\]
Examine Differentiability
Differentiate the piecewise function for points away from $0$ and $1$.
For $x < 0$,
\[f(x)=1-2x\]
so
\[f'(x)=-2\]
For $0 < x < 1$,
\[f(x)=1\]
so
\[f'(x)=0\]
For $x>1$,
\[f(x)=2x-1\]
so
\[f'(x)=2\]
Thus, for $x\neq0,1$,
\[\boxed{f'(x)=\begin{cases}-2, & x < 0,\\[4pt] 0, & 0 < x < 1,\\[4pt] 2, & x>1.\end{cases}}\]
Check Differentiability at $x=0$
The left-hand derivative at $x=0$ is
\[f'_-(0)=-2\]
The right-hand derivative at $x=0$ is
\[f'_+(0)=0\]
Since
\[f'_-(0)\neq f'_+(0)\]
the derivative does not exist at $x=0$.
Check Differentiability at $x=1$
The left-hand derivative at $x=1$ is
\[f'_-(1)=0\]
The right-hand derivative at $x=1$ is
\[f'_+(1)=2\]
Since
\[f'_-(1)\neq f'_+(1)\]
the derivative does not exist at $x=1$.
Show that These Are the Only Two Points
For every $x<0$, $f(x)=1-2x$, which is differentiable.
\[f'(x)=-2,\qquad x < 0\]
For every $0 < x < 1$, $f(x)=1$, which is differentiable.
\[f'(x)=0,\qquad 0 < x < 1.\]
For every $x >1$, $f(x)=2x-1$, which is differentiable.
\[f'(x)=2,\qquad x>1.\]
Hence, there are no other points where differentiability fails.
Therefore, the function is:
- continuous for every $x\in\mathbb R$,
- not differentiable at $x=0$,
- not differentiable at $x=1$,
- differentiable at every other real number.

💡 Answer

Final Answer

›

Yes, such a function exists. One example is

\[ \boxed{f(x)=|x|+|x-1|}. \]

It is continuous everywhere, but it is not differentiable exactly at

\[ \boxed{x=0\quad\text{and}\quad x=1}. \]

Thus, the required function exists.

🎯 Exam Significance

Exam Significance

›

This problem tests the distinction between continuity and differentiability.
A function can be continuous at a point even when it is not differentiable there.
For piecewise or absolute-value functions, the joining points must be checked separately.
To prove non-differentiability, compare the left-hand and right-hand derivatives.
The construction $f(x)=|x-a|+|x-b|$ provides a general method for creating functions that are continuous everywhere but non-differentiable at two prescribed points $a$ and $b$, provided $a\neq b$.

Significance for Competitive Entrance Exams

Absolute-value functions are frequently used to test differentiability at corner points.
The fastest way to locate possible non-differentiability points is to identify where the arguments of absolute-value terms become zero.
For
\[ |x|+|x-1|, \]
the only candidate points are $x=0$ and $x=1$.
Once these points are checked, differentiability elsewhere follows because the function is linear on each open interval.

Common Mistakes

Confusing continuity with differentiability: A function may be continuous but not differentiable.
Checking only the function values: Equality of the function's left and right limits proves continuity, not differentiability.
Failing to check one-sided derivatives: At $x=0$ and $x=1$, the left and right derivatives must be compared.
Claiming there are infinitely many non-differentiable points: The function is linear on each of the three intervals and hence differentiable there.
Forgetting the exact points: The two points are
\[ \boxed{0\text{ and }1}. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Continuity does not imply differentiability.
The function
\[ \boxed{f(x)=|x|+|x-1|} \]
is continuous everywhere.
Its piecewise form is
\[ f(x)= \begin{cases} 1-2x,&x<0,\\ 1,&0\leq x<1,\\ 2x-1,&x\geq1. \end{cases} \]
At $x=0$,
\[ f'_-(0)=-2,\qquad f'_+(0)=0. \]
Hence $f'(0)$ does not exist.
At $x=1$,
\[ f'_-(1)=0,\qquad f'_+(1)=2. \]
Hence $f'(1)$ does not exist.
At every other point, $f$ is differentiable.
Therefore,
\[ \boxed{\text{Yes, such a function exists.}} \]

Q21

NUMERIC3 marks

if \[ y= \begin{vmatrix} f(x) & g(x) & h(x)\\ l & m & n\\ a & b & c \end{vmatrix}, \] prove that \[ \frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{vmatrix} \]

📘 Concept & Theory

Concept/Theory

›

The determinant is a function of $x$ because only its first row contains functions of $x$:

\[ f(x),\qquad g(x),\qquad h(x). \]

The remaining entries $l,m,n,a,b,c$ are constants. Therefore, we can expand the determinant along the first row. The resulting expression is a linear combination of $f(x),g(x),h(x)$ with constant coefficients.

Differentiating this expanded expression will therefore differentiate only $f(x),g(x),h(x)$, while the cofactors remain constant.

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand the determinant along its first row.
Identify the constant cofactors of $f(x),g(x),h(x)$.
Differentiate the expanded expression with respect to $x$.
Recognise the resulting expression as the determinant obtained by replacing $f,g,h$ with $f',g',h'$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[ y= \begin{vmatrix} f(x) & g(x) & h(x)\\ l & m & n\\ a & b & c \end{vmatrix} \]
Expanding along the first row,
\[ \begin{aligned} y &= f(x) \begin{vmatrix} m & n\\ b & c \end{vmatrix} - g(x) \begin{vmatrix} l & n\\ a & c \end{vmatrix} + h(x) \begin{vmatrix} l & m\\ a & b \end{vmatrix}. \end{aligned} \]
Now evaluate the $2\times2$ determinants:
\[ \begin{vmatrix} m & n\\ b & c \end{vmatrix} = mc-nb, \]

\[ \begin{vmatrix} l & n\\ a & c \end{vmatrix} = lc-na, \]

\[ \begin{vmatrix} l & m\\ a & b \end{vmatrix} = lb-ma. \]
Therefore,
\[ \boxed{ y=f(x)(mc-nb)-g(x)(lc-na)+h(x)(lb-ma) } \]
Differentiate with Respect to $x$
Differentiate both sides:
\[ \frac{dy}{dx} = \frac{d}{dx} \left[ f(x)(mc-nb)-g(x)(lc-na)+h(x)(lb-ma) \right]. \]
Since $l,m,n,a,b,c$ are constants, the quantities
\[ mc-nb,\quad lc-na,\quad lb-ma \]
are also constants.
Therefore,
\[\frac{dy}{dx}=f'(x)(mc-nb)-g'(x)(lc-na)+h'(x)(lb-ma)\]
Recognise the Determinant Form
Now consider the determinant
\[ \begin{vmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{vmatrix}. \]
Expanding this determinant along the first row gives
\[\begin{aligned}&\begin{vmatrix}f'(x) & g'(x) & h'(x)\\ l & m & n\\a & b & c\end{vmatrix}\\ &=f'(x)\begin{vmatrix}m & n\\b & c\end{vmatrix}-g'(x)\begin{vmatrix}l & n\\a & c\end{vmatrix}+h'(x)\begin{vmatrix}l & m\\a & b\end{vmatrix}\end{aligned}\]
Substituting the $2\times2$ determinants,
\[\begin{aligned}&\begin{vmatrix}f'(x) & g'(x) & h'(x)\\l & m & n\\a & b & c\end{vmatrix}\\ &=f'(x)(mc-nb)-g'(x)(lc-na)+h'(x)(lb-ma)\end{aligned}\]
But this is exactly the expression obtained for $\dfrac{dy}{dx}$.
Hence,
\[ \boxed{ \frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{vmatrix} }. \]
Conclusion
Therefore, the derivative of the determinant can be obtained by differentiating the entries of the row that contains functions of $x$, provided the other rows consist entirely of constants.
\[ \boxed{ \frac{d}{dx} \begin{vmatrix} f(x) & g(x) & h(x)\\ l & m & n\\ a & b & c \end{vmatrix} = \begin{vmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{vmatrix} }. \]

🎯 Exam Significance

Exam Significance

›

This question tests differentiation combined with determinant expansion.
Expanding along the first row makes the differentiation straightforward.
The cofactors of the first-row elements are constants because the second and third rows contain only constants.
Remember the determinant expansion signs:
\[ +,\quad-,\quad+. \]
A clear proof should show both the expansion and the reconstruction of the differentiated determinant.

Significance for Competitive Entrance Exams

The key idea is linearity: the determinant is linear in each row when the other rows are fixed.
Since only one row contains functions of $x$, differentiation acts on that row while the other rows remain unchanged.
The determinant form can therefore be recognised without fully expanding it once the underlying property is understood.
This concept can save substantial calculation time in determinant-based differentiation problems.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Expand the determinant along the row containing the variable functions.
The cofactors remain constant because the other rows contain constants.
Differentiate only $f(x),g(x),h(x)$.
Reconstruct the resulting expression as a determinant.
The final result is
\[ \boxed{ \frac{dy}{dx} = \begin{vmatrix} f'(x) & g'(x) & h'(x)\\ l & m & n\\ a & b & c \end{vmatrix} }. \]

📘 Concept & Theory

Concept/Theory

›

The given function is a composite exponential function:

\[ y=e^{a\cos^{-1}x}. \]

To find its first and second derivatives, we need the Chain Rule and the derivative

\[ \frac{d}{dx}(\cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}, \qquad -1 < x < 1. \]

Since the required result contains both $y'$ and $y''$, we first calculate these derivatives and then substitute them into the expression

\[ (1-x^2)y''-xy'-a^2y. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate $y=e^{a\cos^{-1}x}$ using the Chain Rule.
Differentiate $y'$ again to obtain $y''$.
Multiply $y''$ by $1-x^2$.
Subtract $xy'$ and $a^2y$.
Show that all terms cancel and the result is $0$.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 29 steps

Given —
\[y=e^{a\cos^{-1}x}\]
Find $\dfrac{dy}{dx}$
Differentiate both sides with respect to $x$.
\[\frac{dy}{dx}=e^{a\cos^{-1}x}\frac{d}{dx}(a\cos^{-1}x)\]
Since $a$ is a constant,
\[\frac{d}{dx}(a\cos^{-1}x)=a\frac{d}{dx}(\cos^{-1}x)\]
Using
\[\frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}},\]
we obtain
\[\frac{dy}{dx}=-\frac{ae^{a\cos^{-1}x}}{\sqrt{1-x^2}}\]
Since
\[e^{a\cos^{-1}x}=y,\]
we can write
\[\boxed{\frac{dy}{dx}=-\frac{ay}{\sqrt{1-x^2}}}\]
Find $\dfrac{d^2y}{dx^2}$
From the previous step,
\[\frac{dy}{dx}=-ay(1-x^2)^{-1/2}\]
Differentiate both sides with respect to $x$:
\[\frac{d^2y}{dx^2}=-a\frac{d}{dx}\left[y(1-x^2)^{-1/2}\right]\]
Apply the Product Rule:
\[\frac{d^2y}{dx^2}=-a\left[\frac{dy}{dx}(1-x^2)^{-1/2}+y\frac{d}{dx}(1-x^2)^{-1/2}\right]\]
Now calculate the derivative of the second factor:
\[\frac{d}{dx}(1-x^2)^{-1/2}=-\frac12(1-x^2)^{-3/2}(-2x)\]
Therefore,
\[\frac{d}{dx}(1-x^2)^{-1/2}=x(1-x^2)^{-3/2}\]
Hence,
\[\frac{d^2y}{dx^2}=-a\left[\frac{dy}{dx}(1-x^2)^{-1/2}+xy(1-x^2)^{-3/2}\right]\]
Substitute the First Derivative
We know that
\[\frac{dy}{dx}=-ay(1-x^2)^{-1/2}\]
Substitute this into the expression for $y''$:
\[ \begin{aligned} \frac{d^2y}{dx^2} &= -a \left[ -ay(1-x^2)^{-1/2}(1-x^2)^{-1/2} + xy(1-x^2)^{-3/2} \right]. \end{aligned} \]
Since
\[(1-x^2)^{-1/2}(1-x^2)^{-1/2}=(1-x^2)^{-1},\]
we get
\[\begin{aligned}\frac{d^2y}{dx^2}&=-a\left[-\frac{ay}{1-x^2}+\frac{xy}{(1-x^2)^{3/2}}\right]\\ &=\frac{a^2y}{1-x^2}-\frac{axy}{(1-x^2)^{3/2}}\end{aligned}\]
Therefore,
\[\boxed{\frac{d^2y}{dx^2}=\frac{a^2y}{1-x^2}-\frac{axy}{(1-x^2)^{3/2}}}\]
Substitute into the Required Expression
We need to prove
\[(1-x^2)y''-xy'-a^2y=0\]
Consider the left-hand side:
\[L=(1-x^2)y''-xy'-a^2y\]
Substitute the expression for $y''$:
\[\begin{aligned}(1-x^2)y''&=(1-x^2)\left[\frac{a^2y}{1-x^2}-\frac{axy}{(1-x^2)^{3/2}}\right]\\ &=a^2y-\frac{axy}{\sqrt{1-x^2}}\end{aligned}\]
Also, from the first derivative,
\[y'=-\frac{ay}{\sqrt{1-x^2}}\]
Therefore,
\[-xy'=-x\left(-\frac{ay}{\sqrt{1-x^2}}\right)=\frac{axy}{\sqrt{1-x^2}}\]
Hence,
\[\begin{aligned}L&=\left(a^2y-\frac{axy}{\sqrt{1-x^2}}\right)+\frac{axy}{\sqrt{1-x^2}}-a^2y\\&=0\end{aligned}\]
Thus,
\[\boxed{(1-x^2)\frac{d^2y}{dx^2}-x\frac{dy}{dx}-a^2y=0}\]
Important Domain Note
The given domain is
\[-1\leq x\leq1\]
The derivative of $\cos^{-1}x$ is finite only for
\[-1 < x < 1.\]
Thus, the differential equation is understood on the open interval
\[-1 < x < 1\]
The endpoint values $x=\pm1$ belong to the domain of the original function, but the ordinary derivative formula for $\cos^{-1}x$ is not finite there.

🎯 Exam Significance

Exam Significance

›

This problem tests repeated differentiation of a composite exponential function.
The Chain Rule is essential while finding $y'$.
The Product Rule is required while finding $y''$.
Expressing the derivatives in terms of $y$ makes the final substitution much easier.
The cancellation of the terms involving
\[ \frac{axy}{\sqrt{1-x^2}} \]
is the key step in proving the differential equation.

Significance for Competitive Entrance Exams

The substitution
\[ y=e^{a\cos^{-1}x} \]
immediately gives
\[ y'=-\frac{ay}{\sqrt{1-x^2}}. \]
Keeping the derivative in terms of $y$ avoids repeatedly writing the exponential expression.
When a question asks to prove a differential equation, calculate only the derivatives necessary to substitute into the target expression.
Careful handling of the sign in
\[ \frac{d}{dx}(\cos^{-1}x) = -\frac1{\sqrt{1-x^2}} \]
is essential.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Given
\[ y=e^{a\cos^{-1}x}, \]
the first derivative is
\[ \boxed{ y'=-\frac{ay}{\sqrt{1-x^2}} }. \]
The second derivative is
\[ \boxed{ y''= \frac{a^2y}{1-x^2} - \frac{axy}{(1-x^2)^{3/2}} }. \]
Substitution gives complete cancellation:
\[ (1-x^2)y''-xy'-a^2y=0. \]
Therefore,
\[ \boxed{ (1-x^2)\frac{d^2y}{dx^2} -x\frac{dy}{dx} -a^2y=0 }. \]

Continuity and Differentiability

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Domain Consideration

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Continuity and Differentiability

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Significance for Competitive Entrance Examinations

Useful Identities

Significance for Competitive Entrance Examinations

Domain Consideration

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

Common Mistakes

Significance for Competitive Entrance Exams

Significance for Competitive Entrance Exams

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CONTINUITY AND DIFFERENTIABILITY — Learning Resources

Frequently Asked Questions

Q 01 What is covered in the NCERT Class 12 Mathematics Chapter 5 Miscellaneous Exercise?

Q 02 Are these NCERT Chapter 5 Miscellaneous Exercise solutions useful for CBSE board exams?

Q 03 Which differentiation techniques are used in this Miscellaneous Exercise?

Q 04 How do I find the second derivative of a parametric function?

Q 05 Does a continuous function always have to be differentiable?

Q 06 How can I check whether a function is differentiable at a point?

Q 07 What is the importance of implicit differentiation in Chapter 5?

Q 08 How can I prepare Chapter 5 Continuity and Differentiability effectively?

Q 09 Are these Miscellaneous Exercise solutions useful for competitive entrance exams?

Q 10 What are the key topics to revise before attempting the Chapter 5 Miscellaneous Exercise?

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