NCERT Class 12 Maths Exercise 5.5 Solutions

📘 Concept & Theory

Concept/Theory

›

The given function is a product of three trigonometric functions: \(\cos x\), \(\cos 2x\), and \(\cos 3x\). Direct differentiation by the product rule is possible, but it produces three separate terms and requires repeated application of the product rule.

A more efficient method is logarithmic differentiation. When a function is expressed as a product of several factors, taking logarithms converts multiplication into addition:

\[ \log(abc)=\log a+\log b+\log c \]

We also use the standard derivative

\[ \frac{d}{dx}\log u=\frac{1}{u}\frac{du}{dx} \]

For trigonometric factors, we use

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

\[ \frac{d}{dx}(\cos ax)=-a\sin ax \]

Consequently,

\[ \frac{d}{dx}\log(\cos ax) = \frac{-a\sin ax}{\cos ax} = -a\tan ax \]

Therefore, for a product such as \(\cos x\cos 2x\cos 3x\), logarithmic differentiation provides a compact and systematic route to the derivative.

🗺️ Solution Roadmap

Step-by-step Plan

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Let the given product be \(y\).
Take logarithm on both sides.
Use the logarithm law to split the product into a sum.
Differentiate both sides with respect to \(x\).
Apply the chain rule to \(\cos x\), \(\cos 2x\), and \(\cos 3x\).
Convert the resulting sine-to-cosine ratios into tangent functions.
Multiply by \(y\) to obtain \(\dfrac{dy}{dx}\).
Finally, substitute the original expression for \(y\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Let
\[y=\cos x\cdot\cos 2x\cdot\cos 3x\]
Since the expression contains a product of several factors, logarithmic differentiation is convenient.
Take logarithm on both sides
Taking logarithm on both sides, we get
\[\log y=\log\left(\cos x\cdot\cos 2x\cdot\cos 3x\right)\]
Using
\[\log(abc)=\log a+\log b+\log c\]
we obtain
\[\log y=\log(\cos x)+\log(\cos 2x)+\log(\cos 3x)\]
Differentiate both sides with respect to \(x\)
Differentiating both sides, we get
\[\frac{d}{dx}(\log y)=\frac{d}{dx}\left[\log(\cos x)+\log(\cos 2x)+\log(\cos 3x)\right]\]
Using the chain rule on the left-hand side,
\[ \frac{d}{dx}(\log y) = \frac{1}{y}\frac{dy}{dx} \]
Therefore,
\[ \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\log(\cos x) + \frac{d}{dx}\log(\cos 2x) + \frac{d}{dx}\log(\cos 3x) \]
Differentiate each logarithmic term
For the first term,
\[ \frac{d}{dx}\log(\cos x) = \frac{1}{\cos x}\frac{d}{dx}(\cos x) \]

\[ = \frac{-\sin x}{\cos x} \]

\[ =-\tan x \]
For the second term, we must apply the chain rule because the argument is \(2x\):
\[ \frac{d}{dx}\log(\cos 2x) = \frac{1}{\cos 2x}\frac{d}{dx}(\cos 2x) \]

\[ = \frac{1}{\cos 2x} \left(-\sin 2x\right)(2) \]

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

\[ =-2\tan 2x \]
For the third term, again applying the chain rule,
\[ \frac{d}{dx}\log(\cos 3x) = \frac{1}{\cos 3x}\frac{d}{dx}(\cos 3x) \]

\[ = \frac{1}{\cos 3x} \left(-\sin 3x\right)(3) \]

\[ = -\frac{3\sin 3x}{\cos 3x} \]

\[ =-3\tan 3x \]
Combine the three derivatives
\[ \frac{1}{y}\frac{dy}{dx} = -\tan x-2\tan 2x-3\tan 3x \]
Taking the negative sign common,
\[ \frac{1}{y}\frac{dy}{dx} = -\left(\tan x+2\tan 2x+3\tan 3x\right) \]
Multiply by \(y\)
Multiplying both sides by \(y\),
\[ \frac{dy}{dx} = -y\left(\tan x+2\tan 2x+3\tan 3x\right) \]
Substitute the value of \(y\)
\[y=\cos x\cdot\cos 2x\cdot\cos 3x\]
Therefore,
\[ \boxed{ \frac{dy}{dx} = -\left(\cos x\cdot\cos 2x\cdot\cos 3x\right) \left(\tan x+2\tan 2x+3\tan 3x\right) } \]

🎯 Exam Significance

Exam Significance

›

This problem is important because it tests the application of logarithmic differentiation to a product of composite trigonometric functions. It also checks whether the student can correctly apply the chain rule to expressions such as \(\cos 2x\) and \(\cos 3x\).

It reinforces the logarithmic differentiation technique used for products and complicated functions.
It tests accurate use of the chain rule.
It requires familiarity with
\[ \frac{d}{dx}(\cos ax)=-a\sin ax \]
It provides practice in converting
\[ \frac{\sin ax}{\cos ax} \]
into \(\tan ax\).
It is suitable for step-based differentiation questions where method and intermediate steps contribute to the final answer.

Significance for Competitive Entrance Examinations

For competitive examinations, the main value of this question is the recognition of an efficient differentiation strategy. Logarithmic differentiation can substantially reduce the algebra involved when a function contains several multiplicative factors.

Recognise a product of multiple functions as a candidate for logarithmic differentiation.
Track the coefficients generated by the chain rule: \(1,2,3\) in this problem.
Avoid unnecessary repeated applications of the product rule.
Use
\[ \frac{d}{dx}\ln f(x)=\frac{f'(x)}{f(x)} \]
rapidly in objective-type questions.
The resulting tangent expression may be useful for further simplification in more advanced trigonometric problems.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

Logarithmic differentiation converts a product into a sum.
For
\[ y=\prod_i f_i(x), \]
logarithmic differentiation gives
\[ \frac{y'}{y} = \sum_i\frac{f_i'(x)}{f_i(x)}. \]
The chain rule is essential for composite functions such as \(\cos 2x\) and \(\cos 3x\).
The derivative
\[ \frac{d}{dx}\log(\cos ax)=-a\tan ax \]
is particularly useful in this type of problem.
The coefficients inside the trigonometric arguments must not be lost during differentiation.
The final derivative is
\[ \boxed{ \frac{dy}{dx} = -\cos x\cos 2x\cos 3x \left( \tan x+2\tan 2x+3\tan 3x \right) } \]

📘 Concept & Theory

Concept/Theory

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The given function contains a square root of a quotient involving five linear factors. Direct differentiation would require both the chain rule and the quotient rule, making the calculation unnecessarily lengthy.

Logarithmic differentiation is therefore a convenient method. The important logarithmic identities used are

\[ \log\sqrt{u}=\frac{1}{2}\log u \]

\[ \log\left(\frac{A}{B}\right) = \log A-\log B \]

\[ \log(AB)=\log A+\log B \]

After taking logarithms, every factor can be differentiated separately. For a linear expression \(x-a\),

\[ \frac{d}{dx}\log(x-a) = \frac{1}{x-a} \]

Since the original expression contains a square root, the factor \(\frac{1}{2}\) must be retained throughout the differentiation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the given expression be \(y\).
Take logarithm on both sides.
Use the square-root logarithm rule to introduce the factor \(\frac{1}{2}\).
Expand the logarithm of the quotient into subtraction.
Expand the logarithms of the products into individual terms.
Differentiate both sides with respect to \(x\).
Multiply by \(\frac{y}{2}\) to isolate \(\frac{dy}{dx}\).
Substitute the original value of \(y\).
Write the final derivative in a clear boxed form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Let
\[y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\]
Take logarithm on both sides
Taking logarithm on both sides,
\[\log y=\log\left[\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\right]\]
Using
\[\log\sqrt{u}=\frac{1}{2}\log u\]
we get
\[\log y=\frac{1}{2}\log\left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right]\]
Multiplying both sides by \(2\),
\[2\log y=\log\left[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right]\]
Apply the logarithm rule for a quotient
Using
\[\log\left(\frac{A}{B}\right)=\log A-\log B\]
we obtain
\[2\log y=\log\left[(x-1)(x-2)\right]-\log\left[(x-3)(x-4)(x-5)\right]\]
Expand the logarithms of the products
Using
\[\log(ABC)=\log A+\log B+\log C\]
we get
\[2\log y=\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5)\]
Differentiate both sides
Differentiating with respect to \(x\),
\[\frac{d}{dx}(2\log y)=\frac{d}{dx}\log(x-1)+\frac{d}{dx}\log(x-2)-\frac{d}{dx}\log(x-3)-\frac{d}{dx}\log(x-4)-\frac{d}{dx}\log(x-5)\]
For the left-hand side, applying the chain rule,
\[\frac{d}{dx}(2\log y)=2\frac{1}{y}\frac{dy}{dx}\]
For the individual terms,
\[\frac{d}{dx}\log(x-1)=\frac{1}{x-1}\]

\[\frac{d}{dx}\log(x-2)=\frac{1}{x-2}\]

\[\frac{d}{dx}\log(x-3)=\frac{1}{x-3}\]

\[\frac{d}{dx}\log(x-4)=\frac{1}{x-4}\]

\[\frac{d}{dx}\log(x-5)=\frac{1}{x-5}\]
Therefore,
\[2\frac{1}{y}\frac{dy}{dx}=\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\]
Isolate \(\frac{dy}{dx}\)
Multiplying both sides by \(\frac{y}{2}\),
\[\frac{dy}{dx}=\frac{y}{2}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\]
Substitute the value of \(y\)
Since
\[y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\]
substituting this value gives
\[\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]\]
Hence, the required derivative is
\[\boxed{\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]}\]
Mathematical Note
Because the original function contains a square root, its real domain is restricted to values of \(x\) for which
\[\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\geq 0\]
Also, \(x\neq3,4,5\), because these values make the denominator zero. The logarithmic differentiation is applied on intervals where the function is defined and the logarithmic expressions involved are valid.

🎯 Exam Significance

Exam Significance

›

This problem is a strong application of logarithmic differentiation. It combines a square root, a quotient, multiple factors, and differentiation of logarithmic functions. It is therefore useful for mastering the complete procedure rather than memorising a single derivative.

It demonstrates how logarithmic differentiation simplifies complicated products and quotients.
It tests the correct use of
\[ \log\sqrt{u}=\frac{1}{2}\log u. \]
It reinforces the logarithm rules for products and quotients.
It checks accurate differentiation of linear factors.
It is useful for writing complete, logically connected step-by-step answers in board examinations.

Significance for Competitive Entrance Examinations

In competitive examinations, recognising the appropriate differentiation technique is often as important as carrying out the differentiation itself. The structure

\[ y= \sqrt{\frac{\text{product of factors}} {\text{product of factors}}} \]

strongly suggests logarithmic differentiation.

It reduces a complicated product and quotient to a sum and difference of simple reciprocal terms.
It helps avoid lengthy applications of the product and quotient rules.
The expression
\[ \frac{y'}{y} \]
can often be identified rapidly in objective questions.
It develops pattern recognition for more complicated functions involving powers and multiple factors.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Logarithmic differentiation is particularly effective when several factors occur in a product or quotient.
A square root contributes the factor
\[ \frac{1}{2} \]
when logarithms are taken.
Use
\[ \log\left(\frac{A}{B}\right)=\log A-\log B. \]
Use
\[ \log(ABC)=\log A+\log B+\log C. \]
For every linear factor \(x-a\),
\[ \frac{d}{dx}\log(x-a)=\frac{1}{x-a}. \]
The denominator factors contribute negative terms after logarithmic expansion.
The final result is
\[\boxed{\frac{dy}{dx}=\frac{1}{2}\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}\left[\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5}\right]}\]

📘 Concept & Theory

Concept/Theory

›

The given function has the variable \(x\) appearing in both the base and the exponent:

\[ y=(\log x)^{\cos x} \]

Functions of the form

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

are most conveniently differentiated using logarithmic differentiation.

The key logarithmic identity is

\[ \log(a^b)=b\log a \]

Therefore,

\[ \log y = \cos x\log(\log x) \]

The right-hand side is a product of two functions, \(\cos x\) and \(\log(\log x)\), so the product rule is required.

We also use the chain rule:

\[ \frac{d}{dx}\log(\log x) = \frac{1}{\log x}\cdot\frac{1}{x} = \frac{1}{x\log x} \]

Hence, the important idea in this problem is: logarithmic differentiation followed by the product rule and chain rule.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(y=(\log x)^{\cos x}\).
Take logarithm on both sides.
Use \(\log(a^b)=b\log a\) to bring the variable exponent down.
Differentiate both sides with respect to \(x\).
Apply the product rule to \(\cos x\log(\log x)\).
Apply the chain rule to \(\log(\log x)\).
Isolate \(\dfrac{dy}{dx}\).
Substitute the original value of \(y\).
Write the final result in a simplified form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 17 steps

Let
\[y=(\log x)^{\cos x}\]
Take logarithm on both sides
Taking logarithm on both sides,
\[\log y=\log\left[(\log x)^{\cos x}\right]\]
Using the logarithmic identity
\[\log(a^b)=b\log a\]
we obtain
\[\log y=\cos x\log(\log x)\]
Differentiate both sides
Differentiating both sides with respect to \(x\),
\[\frac{d}{dx}(\log y)=\frac{d}{dx}\left[\cos x\log(\log x)\right]\]
On the left-hand side, using the chain rule,
\[\frac{d}{dx}(\log y)=\frac{1}{y}\frac{dy}{dx}\]
On the right-hand side, we have a product of \(\cos x\) and \(\log(\log x)\). Therefore, using the product rule,
\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]
we get
\[\frac{1}{y}\frac{dy}{dx}=\cos x\frac{d}{dx}\left[\log(\log x)\right]+\log(\log x)\frac{d}{dx}(\cos x)\]
Differentiate \(\log(\log x)\)
Apply the chain rule carefully:
\[\frac{d}{dx}\left[\log(\log x)\right]=\frac{1}{\log x}\frac{d}{dx}(\log x)\]
Since
\[\frac{d}{dx}(\log x)=\frac{1}{x}\]
we obtain
\[\frac{d}{dx}\left[\log(\log x)\right]=\frac{1}{\log x}\cdot\frac{1}{x}\]

\[=\frac{1}{x\log x}\]
Differentiate \(\cos x\)
\[\frac{d}{dx}(\cos x)=-\sin x\]
Substituting these two derivatives into the product-rule result,
\[\frac{1}{y}\frac{dy}{dx}=\cos x\left(\frac{1}{x\log x}\right)+\log(\log x)(-\sin x)\]
Therefore,
\[\frac{1}{y}\frac{dy}{dx}=\frac{\cos x}{x\log x}-\sin x\log(\log x)\]
Multiply by \(y\)
Multiplying both sides by \(y\), we get
\[\frac{dy}{dx}=y\left[\frac{\cos x}{x\log x}-\sin x\log(\log x)\right]\]
Substitute the value of \(y\)
Since
\[y=(\log x)^{\cos x}\]
therefore,
\[\boxed{\frac{dy}{dx}=(\log x)^{\cos x}\left[\frac{\cos x}{x\log x}-\sin x\log(\log x)\right]}\]
Hence, the required derivative is
\[\boxed{\frac{d}{dx}\left[(\log x)^{\cos x}\right]=(\log x)^{\cos x}\left[\frac{\cos x}{x\log x}-\sin x\log(\log x)\right]}\]

🎯 Exam Significance

Exam Significance

›

This question is important because it combines three fundamental differentiation techniques: logarithmic differentiation, the product rule, and the chain rule.

It demonstrates how to differentiate a function in which both the base and exponent contain the variable.
It tests the logarithmic identity
\[ \log(a^b)=b\log a. \]
It requires correct differentiation of the nested function \(\log(\log x)\).
It tests the product rule after logarithmic transformation.
Writing all intermediate steps makes the method easy to verify and reduces the possibility of losing a chain-rule factor.

Significance for Competitive Entrance Examinations

For competitive examinations, the key skill is recognising that the standard power rule cannot be applied directly because the exponent \(\cos x\) is variable. Logarithmic differentiation transforms the variable-power expression into a product that can be differentiated systematically.

Recognise expressions of the form
\[ [f(x)]^{g(x)} \]
as candidates for logarithmic differentiation.
Remember that
\[ \frac{d}{dx}\log(\log x) = \frac{1}{x\log x}. \]
Apply the product rule without losing either term.
Keep the original function \(y\) until the final step to avoid unnecessary expansion.
The resulting factored form is often easier to manipulate in subsequent objective-type questions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

For a function of the form
\[ y=[f(x)]^{g(x)}, \]
logarithmic differentiation is generally the appropriate technique.
Use
\[ \log(a^b)=b\log a \]
to bring the variable exponent down.
The expression
\[ \cos x\log(\log x) \]
requires the product rule.
The nested logarithm requires the chain rule:
\[ \frac{d}{dx}\log(\log x) = \frac{1}{x\log x}. \]
The derivative of \(\cos x\) contributes the negative term
\[ -\sin x\log(\log x). \]
The final derivative is
\[ \boxed{ \frac{dy}{dx} = (\log x)^{\cos x} \left[ \frac{\cos x}{x\log x} - \sin x\log(\log x) \right] } \]

📘 Concept & Theory

Concept/Theory

›

The given function is a difference of two functions:

\[ y=x^x-2^{\sin x} \]

The first term \(x^x\) has the variable \(x\) in both the base and exponent. Therefore, logarithmic differentiation is required for \(x^x\).

The second term \(2^{\sin x}\) has a constant base and a variable exponent. It can be differentiated using the standard result

\[ \frac{d}{dx}\left(a^{u}\right) = a^u\log a\frac{du}{dx}. \]

Thus,

\[ \frac{d}{dx}\left(2^{\sin x}\right) = 2^{\sin x}\log 2\cos x. \]

A crucial point in this problem is that logarithmic differentiation cannot be applied to the entire expression \(x^x-2^{\sin x}\) by writing the logarithm of a difference as the difference of logarithms. In general,

\[ \log(A-B)\neq\log A-\log B. \]

Therefore, the two terms must be differentiated separately.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the function as the difference of \(x^x\) and \(2^{\sin x}\).
Differentiate \(x^x\) by logarithmic differentiation.
Take logarithm of \(y_1=x^x\).
Differentiate using the product rule.
Obtain \(\dfrac{d}{dx}(x^x)=x^x(1+\log x)\).
Differentiate \(2^{\sin x}\) using the exponential-function rule and chain rule.
Apply the difference rule to combine the two derivatives.
Write the final result in a simplified form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 24 steps

Let
\[y=x^x-2^{\sin x}\]
We differentiate the two terms separately.
Differentiate \(x^x\)
Since the variable \(x\) occurs in both the base and the exponent, let
\[ y_1=x^x\]
Taking logarithm on both sides,
\[\log y_1=\log(x^x)\]
Using
\[\log(a^b)=b\log a\]
we get
\[\log y_1=x\log x\]
Differentiate both sides
Differentiating with respect to \(x\),
\[\frac{d}{dx}(\log y_1)=\frac{d}{dx}(x\log x)\]
Applying the chain rule on the left-hand side,
\[\frac{1}{y_1}\frac{dy_1}{dx}=\frac{d}{dx}(x\log x)\]
The right-hand side is a product, so applying the product rule,
\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]
we obtain
\[\frac{1}{y_1}\frac{dy_1}{dx}=x\frac{d}{dx}(\log x)+\log x\frac{d}{dx}(x)\]
Since
\[\frac{d}{dx}(\log x)=\frac{1}{x}\]
and
\[\frac{d}{dx}(x)=1,\]
therefore
\[\frac{1}{y_1}\frac{dy_1}{dx}=x\left(\frac{1}{x}\right)+\log x(1)\]

\[\frac{1}{y_1}\frac{dy_1}{dx}=1+\log x\]
Isolate the derivative of \(x^x\)
Multiplying by \(y_1\),
\[\frac{dy_1}{dx}=y_1(1+\log x)\]
Since \(y_1=x^x\),
\[\boxed{\frac{d}{dx}(x^x)=x^x(1+\log x)}\]
Differentiate \(2^{\sin x}\)
Now consider
\[y_2=2^{\sin x}\]
For a constant base \(a\),
\[\frac{d}{dx}(a^{u})=a^u\log a\frac{du}{dx}\]
Here,
\[\a=2,\quad u=\sin x\]
Therefore,
\[\frac{dy_2}{dx}=2^{\sin x}\log 2\frac{d}{dx}(\sin x)\]
Since
\[\frac{d}{dx}(\sin x)=\cos x,\]
we get
\[\boxed{\frac{dy_2}{dx}=2^{\sin x}\log 2\cos x}\]
Apply the difference rule
The original function is
\[y=x^x-2^{\sin x}\]
Therefore,
\[\frac{dy}{dx}=\frac{d}{dx}(x^x)-\frac{d}{dx}(2^{\sin x})\]
Substituting the two derivatives obtained above,
\[\frac{dy}{dx}=x^x(1+\log x)-2^{\sin x}\log 2\cos x\]
Hence, the required derivative is
\[\boxed{\frac{dy}{dx}=x^x(1+\log x)-2^{\sin x}\log 2\cos x}\]

🎯 Exam Significance

Exam Significance

›

This question is particularly useful because it tests whether a student can identify different differentiation techniques within the same expression. The two terms require different methods.

\(x^x\) requires logarithmic differentiation because both its base and exponent contain \(x\).
\(2^{\sin x}\) requires differentiation of an exponential function followed by the chain rule.
The difference rule must be applied correctly at the end.
It tests knowledge of logarithmic identities and helps prevent an important conceptual error involving \(\log(A-B)\).
The step-by-step method provides a clear structure for presenting a complete board-examination solution.

Significance for Competitive Entrance Examinations

For competitive examinations, this problem develops technique selection. Instead of attempting to apply one differentiation formula to the entire expression, each component is identified according to its mathematical structure.

Recognise \(x^x\) as a variable-base, variable-exponent function.
Use
\[ \frac{d}{dx}(x^x)=x^x(1+\log x). \]
Recognise \(2^{\sin x}\) as a composite exponential function.
Use the chain rule to obtain the factor \(\cos x\).
Preserve the minus sign from the original difference.
Avoid the invalid logarithm operation
\[ \log(A-B)=\log A-\log B. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Logarithmic differentiation is useful for functions such as \(x^x\), where both base and exponent are variable.
The important result
\[ \frac{d}{dx}(x^x)=x^x(1+\log x) \]
follows directly from logarithmic differentiation.
For a constant base,
\[ \frac{d}{dx}(a^u) = a^u\log a\frac{du}{dx}. \]
Therefore,
\[ \frac{d}{dx}(2^{\sin x}) = 2^{\sin x}\log 2\cos x. \]
Never use
\[ \log(A-B)=\log A-\log B. \]
Differentiate separate terms separately and then apply the sum or difference rule.

📘 Concept & Theory

Concept/Theory

›

The given function is a product of three algebraic factors, each raised to a different power. Such expressions can certainly be differentiated using the product rule repeatedly, but that approach becomes lengthy and involves differentiating each factor separately.

Logarithmic differentiation provides a much simpler method because logarithms convert products into sums and powers into coefficients.

The logarithmic properties used in this problem are:

\[ \log(ABC)=\log A+\log B+\log C \]

\[ \log(A^n)=n\log A \]

After taking logarithms, differentiate each logarithmic term using

\[ \frac{d}{dx}\log(x+a)=\frac{1}{x+a} \]

Finally, multiply both sides by the original function to obtain the required derivative.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the given product be \(y\).
Take logarithm on both sides.
Use logarithmic identities to convert powers into coefficients.
Convert the product into a sum of logarithms.
Differentiate both sides using the sum rule.
Multiply by \(y\) to isolate \(\dfrac{dy}{dx}\).
Substitute the original expression for \(y\).
Write the final derivative in factored form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Let
\[y=(x+3)^2(x+4)^3(x+5)^4\]
Take logarithm on both sides
Taking logarithm on both sides,
\[\log y=\log\left[(x+3)^2(x+4)^3(x+5)^4\right]\]
Using the product rule of logarithms,
\[\log(ABC)=\log A+\log B+\log C\]
we obtain
\[\log y=\log(x+3)^2+\log(x+4)^3+\log(x+5)^4\]
Now apply the power rule of logarithms,
\[\log(a^n)=n\log a\]
Therefore,
\[\log y=2\log(x+3)+3\log(x+4)+4\log(x+5)\]
Differentiate both sides with respect to \(x\)
Differentiating both sides,
\[\frac{d}{dx}(\log y)=\frac{d}{dx}\left[2\log(x+3)+3\log(x+4)+4\log(x+5)\right]\]
Using the chain rule on the left-hand side,
\[\frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\]
Here we used the standard derivative
\[\frac{d}{dx}\log(x+a)=\frac{1}{x+a}\]
Multiply both sides by \(y\)
Multiplying both sides by \(y\),
\[\frac{dy}{dx}=y\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]\]
Substitute the value of \(y\)
Since
\[y=(x+3)^2(x+4)^3(x+5)^4\]
Therefore,
\[\boxed{\frac{dy}{dx}=(x+3)^2(x+4)^3(x+5)^4\left[\frac{2}{x+3}+\frac{3}{x+4}\frac{4}{x+5}\right]}\]

🎯 Exam Significance

Exam Significance

›

This question is a standard application of logarithmic differentiation for algebraic products. It demonstrates how lengthy product-rule calculations can be simplified into a compact answer.

It reinforces the logarithmic identities for products and powers.
It shows how logarithmic differentiation reduces repeated applications of the product rule.
It develops accuracy in differentiating logarithmic functions.
It is a frequently asked pattern in NCERT and CBSE board examinations.
It encourages writing a clean factored answer instead of expanding polynomial expressions.

Significance for Competitive Entrance Examinations

Competitive examinations often test recognition of efficient differentiation techniques. A product with several powered factors should immediately suggest logarithmic differentiation.

Recognise products of multiple powered factors.
Use logarithmic identities to simplify differentiation.
Avoid repeated product-rule calculations.
Keep the derivative in compact factored form for faster simplification.
Useful for JEE Main, NDA, CUET, and other entrance examinations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Logarithmic differentiation is ideal for products of several powered factors.
Use
\[ \log(a^n)=n\log a. \]
Convert multiplication into addition before differentiation.
Differentiate each logarithmic term separately using the sum rule.
Multiply the result by the original function to obtain the derivative.
The final derivative remains in a compact factored form, making it easier to simplify further if required.

📘 Concept & Theory

Concept/Theory

›

The given expression contains two functions of the form \([f(x)]^{g(x)}\), where both the base and exponent may depend on \(x\). Such expressions are most conveniently differentiated by logarithmic differentiation.

However, there is an important point: because the two functions are connected by addition, we cannot take logarithms of the entire expression and split the result. In general,

\[ \log(A+B)\neq\log A+\log B \]

Therefore, the two terms must be differentiated separately.

We use the following results:

\[ \frac{d}{dx}\left[u(x)\log v(x)\right] = u'(x)\log v(x) + u(x)\frac{v'(x)}{v(x)} \]

\[ \frac{d}{dx}\log x=\frac{1}{x} \]

\[ \frac{d}{dx}\left(x+\frac{1}{x}\right) = 1-\frac{1}{x^2} \]

For a variable-base and variable-exponent function \(z=[f(x)]^{g(x)}\), logarithmic differentiation gives

\[ \frac{z'}{z} = g'(x)\log f(x) + g(x)\frac{f'(x)}{f(x)} \]

This formula allows both terms of the given function to be differentiated systematically.

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the given function into two terms.
Differentiate \(\left(x+\frac{1}{x}\right)^x\) using logarithmic differentiation.
Take logarithm of the first term only.
Differentiate \(x\log\left(x+\frac{1}{x}\right)\) using the product rule and chain rule.
Simplify the derivative of the first term.
Differentiate \(x^{1+\frac{1}{x}}\) using logarithmic differentiation.
Differentiate \(\left(1+\frac{1}{x}\right)\log x\) using the product rule.
Simplify the derivative of the second term.
Add the two derivatives using the sum rule.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 33 steps

Let
\[y_1=\left(x+\frac{1}{x}\right)^x\]
Take logarithm on both sides
\[\log y_1=\log\left[\left(x+\frac{1}{x}\right)^x\right]\]
Using
\[\log(a^b)=b\log a\]
we obtain
\[y_1=x\log\left(x+\frac{1}{x}\right)\]
Differentiate both sides
\[\frac{1}{y_1}\frac{dy_1}{dx}=\frac{d}{dx}\left[x\log\left(x+\frac{1}{x}\right)\right]\]
The right-hand side is a product of \(x\) and \(\log\left(x+\frac{1}{x}\right)\). Hence, the product rule is required.
\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]
Therefore,
\[\frac{1}{y_1}\frac{dy_1}{dx}=\log\left(x+\frac{1}{x}\right)+x\frac{d}{dx}\left[\log\left(x+\frac{1}{x}\right)\right]\]
Differentiate the logarithmic term
Put
\[u=x+\frac{1}{x}\]
Then
\[\frac{du}{dx}=\frac{d}{dx}\left(x+x^{-1}\right)\]

\[\frac{du}{dx}=1-x^{-2}\]

\[=1-\frac{1}{x^2}\]
Using
\[\frac{d}{dx}(\log u)=\frac{1}{u}\frac{du}{dx},\]
we obtain
\[\frac{d}{dx}\left[\log\left(x+\frac{1}{x}\right)\right]=\frac{1}{x+\frac{1}{x}}\left(1-\frac{1}{x^2}\right)\]
Simplify the fraction
First write
\[x+\frac{1}{x}=\frac{x^2+1}{x}\]
Therefore,
\[\frac{1}{x+\frac{1}{x}}=\frac{x}{x^2+1}\]
Also,
\[1-\frac{1}{x^2}=\frac{x^2-1}{x^2}\]
Hence,
\[\frac{d}{dx}\left[\log\left(x+\frac{1}{x}\right)\right]=\frac{x}{x^2+1}\cdot\frac{x^2-1}{x^2}\]

\[=\frac{x^2-1}{x(x^2+1)}\]
Substitute into the derivative
\[\frac{1}{y_1}\frac{dy_1}{dx}=\log\left(x+\frac{1}{x}\right)+x\left[\frac{x^2-1}{x(x^2+1)}\right]\]

\[\frac{1}{y_1}\frac{dy_1}{dx}=\log\left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\]
Thus,
\[\frac{dy_1}{dx}=y_1\left[\log\left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]\]
Since
\[y_1=\left(x+\frac{1}{x}\right)^x,\]
we obtain
\[\boxed{\frac{dy_1}{dx}=\left(x+\frac{1}{x}\right)^x\left[\log\left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]}\]
Differentiating \(x^{1+\frac{1}{x}}\)
Let
\[y_2=x^{1+\frac{1}{x}}\]
Take logarithm on both sides
\[\log y_2=\log\left[x^{1+\frac{1}{x}}\right]\]
Using
\[\log(a^b)=b\log a\]
we get
\[\log y_2=\left(1+\frac{1}{x}\right)\log x\]
Differentiate both sides
\[\frac{1}{y_2}\frac{dy_2}{dx}=\frac{d}{dx}\left[\left(1+\frac{1}{x}\right)\log x\right]\]
Applying the product rule,
\[\frac{1}{y_2}\frac{dy_2}{dx}=\log x\frac{d}{dx}\left(1+\frac{1}{x}\right)+\left(1+\frac{1}{x}\right)\frac{d}{dx}(\log x)\]
Differentiate each factor
\[\frac{d}{dx}\left(1+\frac{1}{x}\right)=0-\frac{1}{x^2}=-\frac{1}{x^2}\]
and
\[\frac{d}{dx}(\log x)=\frac{1}{x}\]
Therefore,
\[\frac{1}{y_2}\frac{dy_2}{dx}=-\frac{\log x}{x^2}+\left(1+\frac{1}{x}\right)\frac{1}{x}\]
Now simplify the second term:
\[\left(1+\frac{1}{x}\right)\frac{1}{x}=\frac{1}{x}+\frac{1}{x^2}=\frac{x+1}{x^2}\]
Hence,
\[\frac{1}{y_2}\frac{dy_2}{dx}=-\frac{\log x}{x^2}+\frac{x+1}{x^2}\]

\[\frac{1}{y_2}\frac{dy_2}{dx}=\frac{x+1-\log x}{x^2}\]
Isolate the derivative
\[\frac{dy_2}{dx}=y_2\frac{x+1-\log x}{x^2}\]
Since
\[y_2=x^{1+\frac{1}{x}},\]
therefore,
\[\boxed{\frac{dy_2}{dx}=x^{1+\frac{1}{x}}\frac{x+1-\log x}{x^2}}\]
Combine the Two Derivatives
The original function is
\[y=y_1+y_2\]
Therefore, by the sum rule,
\[\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}\]
Substituting the results obtained above,
\[\frac{dy}{dx}=\left(x+\frac{1}{x}\right)^x\left[\log\left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]+x^{1+\frac{1}{x}}\frac{x+1-\log x}{x^2}\]
Hence, the required derivative is
\[\boxed{\frac{dy}{dx}=\left(x+\frac{1}{x}\right)^x\left[\log\left(x+\frac{1}{x}\right)+\frac{x^2-1}{x^2+1}\right]+\frac{x^{1+\frac{1}{x}}}{x^2}\left(x+1-\log x\right)}\]

🎯 Exam Significance

Exam Significance

›

This problem is valuable because it combines logarithmic differentiation with the product rule and chain rule in two different variable-power functions.

It tests recognition of functions in which the exponent is variable.
It reinforces the correct use of logarithmic differentiation.
It requires careful differentiation of
\[ \log\left(x+\frac{1}{x}\right). \]
It tests the product rule in expressions such as
\[ x\log\left(x+\frac{1}{x}\right). \]
It tests the sum rule because the original function consists of two added terms.

Significance for Competitive Entrance Examinations

For competitive examinations, the main skill tested here is choosing the correct differentiation strategy for each component. The problem also provides practice in simplifying logarithmic derivatives efficiently.

Identify variable-base and variable-exponent functions.
Apply logarithmic differentiation independently to each term.
Use the product rule efficiently after taking logarithms.
Apply the chain rule to nested functions.
Preserve the sum between independent terms.
Avoid the invalid identity
\[ \log(A+B)=\log A+\log B. \]

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

A sum of complicated functions should be differentiated term by term.
Do not apply logarithmic differentiation to a sum unless the logarithm can legitimately be simplified.
For
\[ y=[f(x)]^{g(x)}, \]
logarithmic differentiation is highly effective.
The derivative
\[ \frac{d}{dx} \left[ x\log\left(x+\frac{1}{x}\right) \right] = \log\left(x+\frac{1}{x}\right) + \frac{x^2-1}{x^2+1} \]
is obtained using both the product rule and chain rule.
For
\[ x^{1+\frac{1}{x}}, \]
logarithmic differentiation gives the factor
\[ \frac{x+1-\log x}{x^2}. \]
Since the original terms are added, their derivatives must also be added.

📘 Concept & Theory

Concept/Theory

›

The given function is the sum of two variable-power functions:

\[ y=(\log x)^x+x^{\log x} \]

In the first term, the base is \(\log x\) and the exponent is \(x\). In the second term, the base is \(x\) and the exponent is \(\log x\). In both cases, the exponent is variable, so logarithmic differentiation is the natural method.

However, because the two terms are connected by addition, we must differentiate them separately. We cannot write

\[ \log(A+B)=\log A+\log B. \]

This identity is not valid.

For a variable-power function

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

logarithmic differentiation gives

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

The first term requires the derivative of \(\log(\log x)\), while the second term has a useful simplification:

\[ x^{\log x}=e^{(\log x)^2}. \]

Nevertheless, logarithmic differentiation can be used directly for both terms.

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the two terms of the given sum.
Differentiate \((\log x)^x\) using logarithmic differentiation.
Take logarithm of the first term.
Differentiate \(x\log(\log x)\) using the product rule and chain rule.
Differentiate \(x^{\log x}\) using logarithmic differentiation.
Differentiate \((\log x)^2\) using the chain rule.
Multiply each logarithmic derivative by its corresponding original function.
Add the two derivatives using the sum rule.
Present the final result in a factored and simplified form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 28 steps

Differentiating \((\log x)^x\)
Let
\[y_1=(\log x)^x\]
Take logarithm on both sides
\[\log y_1=\log\left[(\log x)^x\right]\]
Using
\[\log(a^b)=b\log a\]
we get
\[\log y_1=x\log(\log x)\]
Differentiate both sides
\[\frac{1}{y_1}\frac{dy_1}{dx}=\frac{d}{dx}\left[x\log(\log x)\right]\]
The right-hand side is a product of \(x\) and \(\log(\log x)\). Therefore, applying the product rule,
\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx},\]
we obtain
\[\frac{1}{y_1}\frac{dy_1}{dx}=x\frac{d}{dx}\left[\log(\log x)\right]+\log(\log x)\frac{d}{dx}(x)\]
Since
\[\frac{d}{dx}(x)=1,\]
we have
\[\frac{1}{y_1}\frac{dy_1}{dx}=x\frac{d}{dx}\left[\log(\log x)\right]+\log(\log x)\]
Differentiate \(\log(\log x)\)
Applying the chain rule,
\[\frac{d}{dx}\left[\log(\log x)\right]=\frac{1}{\log x}\frac{d}{dx}(\log x)\]
Since
\[\frac{d}{dx}(\log x)=\frac{1}{x}\]
therefore,
\[\frac{d}{dx}\left[\log(\log x)\right]=\frac{1}{x\log x}\]
Substitute into the derivative
\[\frac{1}{y_1}\frac{dy_1}{dx}=x\left(\frac{1}{x\log x}\right)+\log(\log x)\]

\[\frac{1}{y_1}\frac{dy_1}{dx}=\frac{1}{\log x}+\log(\log x)\]
Multiplying by \(y_1\),
\[\frac{dy_1}{dx}=y_1\left[\frac{1}{\log x}+\log(\log x)\right]\]
Since
\[y_1=(\log x)^x\]
we obtain
\[\boxed{\frac{dy_1}{dx}=(\log x)^x\left[\frac{1}{\log x}+\log(\log x)\right]}\]
Differentiating \(x^{\log x}\)
Let
\[y_2=x^{\log x}\]
Take logarithm on both sides
\[\log y_2=\log\left(x^{\log x}\right)\]
Using
\[\log(a^b)=b\log a\]
we obtain
\[\log y_2=(\log x)(\log x)\]

\[\log y_2=(\log x)^2\]
Differentiate both sides
\[\frac{1}{y_2}\frac{dy_2}{dx}=\frac{d}{dx}\left[(\log x)^2\right]\]
Applying the chain rule,
\[\frac{d}{dx}\left[(\log x)^2\right]=2\log x\frac{d}{dx}(\log x)\]

\[=2\log x\left(\frac{1}{x}\right)\]

\[=\frac{2\log x}{x}\]
Hence,
\[\frac{1}{y_2}\frac{dy_2}{dx}=\frac{2\log x}{x}\]
Isolate the derivative of \(x^{\log x}\)
Multiplying by \(y_2\),
\[\frac{dy_2}{dx}=y_2\frac{2\log x}{x}\]
Since
\[y_2=x^{\log x}\]
we obtain
\[\boxed{\frac{dy_2}{dx}=x^{\log x}\frac{2\log x}{x}}\]
Combine the Two Derivatives
The original function is
\[y=y_1+y_2\]
Therefore, by the sum rule,
\[\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}\]
Substituting the derivatives obtained above,
\[\frac{dy}{dx}=(\log x)^x\left[\frac{1}{\log x}+\log(\log x)\right]+x^{\log x}\frac{2\log x}{x}\]
Therefore, the required derivative is
\[\boxed{\frac{dy}{dx}=(\log x)^x\left[\frac{1}{\log x}+\log(\log x)\right]+\frac{2x^{\log x}\log x}{x}}\]

🎯 Exam Significance

Exam Significance

›

This question is important because it tests logarithmic differentiation in two different variable-power structures within the same expression.

It reinforces the method for differentiating
\[ (\log x)^x. \]
It reinforces the method for differentiating
\[ x^{\log x}. \]
It tests the chain rule through
\[ \log(\log x) \]
and
\[ (\log x)^2. \]
It tests correct application of the sum rule.
It highlights an important logarithm property: logarithms cannot be distributed across addition.

Significance for Competitive Entrance Examinations

For competitive examinations, this problem develops rapid recognition of variable-base and variable-exponent functions. It also demonstrates how apparently complicated powers can be converted into simpler logarithmic expressions.

Recognise expressions of the form
\[ [f(x)]^{g(x)}. \]
Use logarithmic differentiation instead of attempting to apply the ordinary power rule.
Recognise the useful transformation
\[ x^{\log x}=e^{(\log x)^2}. \]
Apply the chain rule accurately.
Keep separate terms separate when a function contains a sum.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Logarithmic differentiation is suitable for functions with variable bases and variable exponents.
For
\[ (\log x)^x, \]
the logarithmic derivative is
\[ \frac{1}{\log x}+\log(\log x). \]
For
\[ x^{\log x}, \]
the logarithmic derivative is
\[ \frac{2\log x}{x}. \]
The identity
\[ \log(A+B)=\log A+\log B \]
is not valid.
The expression
\[ \log x\cdot\log x \]
equals
\[ (\log x)^2. \]
The final derivative is
\[ \boxed{ \frac{dy}{dx} = (\log x)^x \left[ \frac{1}{\log x} + \log(\log x) \right] + \frac{2x^{\log x}\log x}{x} } \]

📘 Concept & Theory

Concept/Theory

›

The given function is the sum of two functions:

\[ y=(\sin x)^x+\sin^{-1}\sqrt{x} \]

The first term, \((\sin x)^x\), has both a variable base and a variable exponent. Therefore, it is differentiated using logarithmic differentiation.

The second term, \(\sin^{-1}\sqrt{x}\), is an inverse-trigonometric composite function. It must be differentiated using the chain rule.

The relevant standard results are

\[ \frac{d}{dx}(\sin x)^x = (\sin x)^x \left[ \log(\sin x)+x\cot x \right] \]

and

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

For \(u=\sqrt{x}\),

\[ \frac{du}{dx} = \frac{1}{2\sqrt{x}}. \]

Therefore, the second derivative contains both factors \(\frac{1}{\sqrt{1-x}}\) and \(\frac{1}{2\sqrt{x}}\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the two terms because the function is a sum.
Differentiate \((\sin x)^x\) by logarithmic differentiation.
Take logarithm and use the power rule of logarithms.
Differentiate \(x\log(\sin x)\) using the product rule.
Use the chain rule to differentiate \(\log(\sin x)\).
Differentiate \(\sin^{-1}\sqrt{x}\) using the inverse-sine derivative and chain rule.
Substitute \(u=\sqrt{x}\) carefully into the inverse-trigonometric derivative formula.
Add the two derivatives using the sum rule.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 32 steps

Differentiating \((\sin x)^x\)
Let
\[y_1=(\sin x)^x\]
Take logarithm on both sides
\[\log y_1=\log\left[(\sin x)^x\right]\]
Using
\[\log(a^b)=b\log a\]
we obtain
\[\log y_1=x\log(\sin x)\]
Differentiate both sides
\[\frac{1}{y_1}\frac{dy_1}{dx}=\frac{d}{dx}\left[x\log(\sin x)\right]\]
The right-hand side is a product of \(x\) and \(\log(\sin x)\). Hence, applying the product rule,
\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx},\]
we get
\[\frac{1}{y_1}\frac{dy_1}{dx}=\log(\sin x)\frac{d}{dx}(x)+x\frac{d}{dx}\left[\log(\sin x)\right]\]
Since
\[\frac{d}{dx}(x)=1\]
therefore,
\[\frac{1}{y_1}\frac{dy_1}{dx}=\log(\sin x)+x\frac{d}{dx}\left[\log(\sin x)\right]\]
Differentiate \(\log(\sin x)\)
Using the chain rule,
\[\frac{d}{dx}\log(\sin x)=\frac{1}{\sin x}\frac{d}{dx}(\sin x)\]

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

\[=\cot x\]
Substitute the derivative
\[\frac{1}{y_1}\frac{dy_1}{dx}=\log(\sin x)+x\cot x\]
Multiplying both sides by \(y_1\),
\[\frac{dy_1}{dx}=y_1\left[\log(\sin x)+x\cot x\right]\]
Since
\[y_1=(\sin x)^x,\]
we obtain
\[\boxed{\frac{dy_1}{dx}=(\sin x)^x\left[\log(\sin x)+x\cot x\right]}\]
Differentiating \(\sin^{-1}\sqrt{x}\)
Let
\[y_2=\sin^{-1}\sqrt{x}\]
Introduce the inner function
Put
\[u=\sqrt{x}\]
Then
\[y_2=\sin^{-1}u\]
Differentiate using the inverse-sine formula
We know that
\[\frac{d}{du}(\sin^{-1}u)=\frac{1}{\sqrt{1-u^2}}\]
Therefore, by the chain rule,
\[\frac{dy_2}{dx}=\frac{1}{\sqrt{1-u^2}}\frac{du}{dx}\]
Differentiate \(u=\sqrt{x}\)
\[u=x^{1/2}\]
Hence,
\[\frac{du}{dx}=\frac{1}{2}x^{-1/2}\]

\[=\frac{1}{2\sqrt{x}}\]
Substitute \(u=\sqrt{x}\), Since
\[ u=\sqrt{x},\]
we have
\[u^2=(\sqrt{x})^2=x\]
Therefore,
\[\sqrt{1-u^2}=\sqrt{1-x}\]
Thus,
\[\frac{dy_2}{dx}=\frac{1}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}}\]

\[\boxed{\frac{dy_2}{dx}=\frac{1}{2\sqrt{x}\sqrt{1-x}}}\]
Combine the Two Derivatives
The original function is
\[y=y_1+y_2\]
Therefore, using the sum rule,
\[\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}\]
Substituting the two derivatives,
\[\frac{dy}{dx}=(\sin x)^x\left[\log(\sin x)+x\cot x\right]+\frac{1}{2\sqrt{x}\sqrt{1-x}}\]
Hence, the required derivative is
\[\boxed{\frac{dy}{dx}=(\sin x)^x\left[\log(\sin x)+x\cot x\right]+\frac{1}{2\sqrt{x}\sqrt{1-x}}}\]
Domain Note
For the real-valued inverse sine term,
\[\sin^{-1}\sqrt{x}\]
we require
\[0\leq x\leq1.\]
The derivative
\[ \frac{1}{2\sqrt{x}\sqrt{1-x}} \]
is defined for
\[ 0 < x < 1. \]

🎯 Exam Significance

Exam Significance

›

This question combines logarithmic differentiation with inverse trigonometric differentiation and the chain rule. It is therefore useful for testing whether the student can switch between differentiation techniques within a single question.

It reinforces logarithmic differentiation of
\[ (\sin x)^x. \]
It tests the product rule after taking logarithms.
It tests the standard derivative of \(\sin^{-1}u\).
It checks correct application of the chain rule to \(\sqrt{x}\).
It reinforces the sum rule when combining the two derivatives.

Significance for Competitive Entrance Examinations

For competitive examinations, the main value of this problem lies in quickly identifying the correct technique for each term.

Recognise \((\sin x)^x\) as a variable-base, variable-exponent function.
Recognise \(\sin^{-1}\sqrt{x}\) as a composite inverse-trigonometric function.
Apply the chain rule without dropping
\[ \frac{1}{2\sqrt{x}}. \]
Keep the two terms separate until their individual derivatives have been obtained.
Use the compact logarithmic-derivative form to reduce calculation time.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Use logarithmic differentiation for functions such as
\[ (\sin x)^x. \]
The logarithmic derivative of \((\sin x)^x\) is
\[ \log(\sin x)+x\cot x. \]
For
\[ y=\sin^{-1}u, \]
use
\[ \frac{dy}{dx} = \frac{u'}{\sqrt{1-u^2}}. \]
When \(u=\sqrt{x}\),
\[ u'=\frac{1}{2\sqrt{x}}. \]
Therefore,
\[ \frac{d}{dx}\sin^{-1}\sqrt{x} = \frac{1}{2\sqrt{x}\sqrt{1-x}}. \]
The derivative of the sum is the sum of the individual derivatives.

📘 Concept & Theory

Concept/Theory

›

The given function is a sum of two variable-base, variable-exponent functions. Such expressions are most conveniently differentiated using 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 expression requires both the product rule and the chain rule.

Since the original expression is a sum, the two terms must be differentiated separately. We cannot take the logarithm of the complete sum and split it, because in general

\[ \log(A+B)\neq\log A+\log B. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the given function into two terms.
Use logarithmic differentiation for \(x^{\sin x}\).
Use logarithmic differentiation for \((\sin x)^{\cos x}\).
Apply the product rule after taking logarithms.
Apply the chain rule to \(\log(\sin x)\).
Add the two derivatives.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Let
\[y=x^{\sin x}+(\sin x)^{\cos x}.\]
Since the expression is a sum, differentiate the two terms separately.
First Term: Differentiation of \(x^{\sin x}\)
Let
\[y_1=x^{\sin x}.\]
Taking logarithm on both sides,
\[\log y_1=\sin x\log x.\]
Differentiate both sides using the product rule:
\[ \frac{1}{y_1}\frac{dy_1}{dx} = \cos x\log x + \sin x\frac{1}{x}. \]
Therefore,
\[ \frac{1}{y_1}\frac{dy_1}{dx} = \cos x\log x+\frac{\sin x}{x}. \]
Multiplying by \(y_1=x^{\sin x}\),
\[ \boxed{ \frac{dy_1}{dx} = x^{\sin x} \left[ \cos x\log x+\frac{\sin x}{x} \right] }. \]
Second Term: Differentiation of \((\sin x)^{\cos x}\)
Let
\[y_2=(\sin x)^{\cos x}.\]
Taking logarithm on both sides,
\[\log y_2=\cos x\log(\sin x).\]
Differentiate both sides using the product rule:
\[\frac{1}{y_2}\frac{dy_2}{dx}=(-\sin x)\log(\sin x)+\cos x\frac{d}{dx}\left[\log(\sin x)\right].\]
Using the chain rule,
\[\frac{d}{dx}\left[\log(\sin x)\right]=\frac{\cos x}{\sin x}=\cot x.\]
Hence,
\[ \frac{1}{y_2}\frac{dy_2}{dx} = -\sin x\log(\sin x) + \cos x\cot x. \]
Multiplying by \(y_2=(\sin x)^{\cos x}\),
\[ \boxed{ \frac{dy_2}{dx} = (\sin x)^{\cos x} \left[ \cos x\cot x - \sin x\log(\sin x) \right] }\]
.
Combining Both Derivatives
Since
\[ y=y_1+y_2, \]
we have
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx}. \]
Therefore,
\[ \boxed{ \frac{dy}{dx} = x^{\sin x} \left[ \cos x\log x+\frac{\sin x}{x} \right] + (\sin x)^{\cos x} \left[ \cos x\cot x - \sin x\log(\sin x) \right] } \]

🎯 Exam Significance

Exam Significance

›

This problem is an important application of logarithmic differentiation because both the base and exponent are functions of \(x\). It tests the combined use of logarithms, product rule, and chain rule.

For board examinations, students should write the logarithmic differentiation of each term separately and clearly show the multiplication by the original function after differentiation.

Significance for Competitive Entrance Examinations

Expressions of the form

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

occur frequently in differential calculus. The standard identity

\[ \frac{d}{dx}\left([f(x)]^{g(x)}\right) = [f(x)]^{g(x)} \left[ g'(x)\log f(x) + g(x)\frac{f'(x)}{f(x)} \right] \]

is useful for quickly differentiating such functions.

This question combines two such expressions, making it a useful practice problem for applying the technique accurately under examination time constraints.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Variable-base and variable-exponent functions are efficiently handled using logarithmic differentiation.
A sum of such functions must be differentiated term-by-term.
For
\[ y=[f(x)]^{g(x)}, \]
logarithmic differentiation requires both the product rule and chain rule.
The identity
\[ \frac{d}{dx}\log(\sin x)=\cot x \]
is an important chain-rule application.
Careful preservation of the original function after logarithmic differentiation is essential.

📘 Concept & Theory

Concept/Theory

›

The given function is the sum of two functions:

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

The first term \(x^{x\cos x}\) has a variable base and a variable exponent, so logarithmic differentiation is the appropriate method.

The second term is a quotient of two functions, so it is differentiated using the quotient rule.

The two important formulas are

\[ \frac{d}{dx}(f^g) = f^g \left[ g'log f+g\frac{f'}{f} \right] \]

and

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu'-uv'}{v^2}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the two terms using the sum rule.
Apply logarithmic differentiation to \(u=x^{x\cos x}\).
Take logarithm of both sides correctly.
Differentiate \(x\cos x\) using the product rule.
Differentiate \(\log x\) and apply the product rule to \(x\cos x\log x\).
Differentiate the rational function using the quotient rule.
Combine the two derivatives.
Present the result in a compact and mathematically correct form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 22 steps

Differentiating \(x^{x\cos x}\)
\[u=x^{x\cos x}\]
Take logarithm on both sides
\[\log u=\log\left(x^{x\cos x}\right)\]
Using
\[\log(a^b)=b\log a,\]
we get
\[\log u=x\cos x\log x\]
Differentiate both sides
\[\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\left(x\cos x\log x\right)\]
This is a product of three factors: \(x\), \(\cos x\), and \(\log x\). We can first write
\[x\cos x\log x=(x\cos x)\log x\]
Applying the product rule,
\[\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(x\cos x)\log x+x\cos x\frac{d}{dx}(\log x)\]
Differentiate \(x\cos x\)
Using the product rule,
\[\frac{d}{dx}(x\cos x)=x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(x)\]

\[=x(-\sin x)+\cos x\]

\[=\cos x-x\sin x\]
Also,
\[\frac{d}{dx}(\log x)=\frac{1}{x}\]
Substitute these derivatives
\[\frac{1}{u}\frac{du}{dx}=(\cos x-x\sin x)\log x+x\cos x\left(\frac{1}{x}\right)\]

\[=(\cos x-x\sin x)\log x+\cos x\]
Therefore,
\[\frac{du}{dx}=u\left[(\cos x-x\sin x)\log x+\cos x\right]\]
Since
\[u=x^{x\cos x},\]
we obtain
\[\boxed{\frac{du}{dx}=x^{x\cos x}\left[(\cos x-x\sin x)\log x+\cos x\right]}\]
Differentiating the Rational Function
Let
\[v=\frac{x^2+1}{x^2-1}\]
Using the quotient rule,
\[\frac{dv}{dx}=\frac{(x^2-1)\frac{d}{dx}(x^2+1)-(x^2+1)\frac{d}{dx}(x^2-1)}{(x^2-1)^2}\]
Now,
\[\frac{d}{dx}(x^2+1)=2x\]
and
\[\frac{d}{dx}(x^2-1)=2x\]
Therefore,
\[\frac{dv}{dx}=\frac{(x^2-1)(2x)-(x^2+1)(2x)}{(x^2-1)^2}\]

\[=\frac{2x\left[(x^2-1)-(x^2+1)\right]}{(x^2-1)^2}\]

\[=\frac{2x(x^2-1-x^2-1)}{(x^2-1)^2}\]

\[=\frac{2x(-2)}{(x^2-1)^2}\]

\[\boxed{\frac{dv}{dx}=-\frac{4x}{(x^2-1)^2}}\]
Combine the Results
The original function is
\[y=u+v\]
Therefore, by the sum rule,
\[\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\]
Substituting the derivatives obtained above,
\[\frac{dy}{dx}=x^{x\cos x}\left[(\cos x-x\sin x)\log x+\cos x\right]-\frac{4x}{(x^2-1)^2}\]
Hence, the required derivative is
\[\boxed{\frac{dy}{dx}=x^{x\cos x}\left[\cos x+(\cos x-x\sin x)\log x\right]-\frac{4x}{(x^2-1)^2}}\]

🎯 Exam Significance

Exam Significance

›

This question is particularly useful because it combines logarithmic differentiation, the product rule, and the quotient rule in a single problem.

It tests differentiation of a function with both variable base and variable exponent.
It requires the product rule for \(x\cos x\).
It requires careful differentiation of \(x\cos x\log x\).
It reinforces the quotient rule for rational functions.
It tests the correct use of the sum rule when combining the two parts.

Significance for Competitive Entrance Examinations

For competitive examinations, the key skill is recognising the structure of the function immediately. The expression \(x^{x\cos x}\) should trigger logarithmic differentiation, while the rational term can be differentiated independently.

Variable base and exponent:
\[ x^{x\cos x} \]
→ logarithmic differentiation.
Product inside the exponent:
\[ x\cos x \]
→ product rule.
Product
\[ (x\cos x)\log x \]
→ product rule again.
Rational function
\[ \frac{x^2+1}{x^2-1} \]
→ quotient rule.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For
\[ u=x^{x\cos x}, \]
take logarithms only once:
\[ \log u=x\cos x\log x. \]
The derivative of \(x\cos x\) is
\[ \cos x-x\sin x. \]
Hence,
\[ \frac1u\frac{du}{dx} = (\cos x-x\sin x)\log x+\cos x. \]
The rational term differentiates to
\[ -\frac{4x}{(x^2-1)^2}. \]
Logarithmic differentiation should not be extended unnecessarily beyond the equation
\[ \log u=x\cos x\log x. \]

📘 Concept & Theory

Concept/Theory

›

The given function is a sum of two functions:

\[ y=(x\cos x)^x+(x\sin x)^{1/x} \]

Both terms have variable bases and variable exponents. Therefore, logarithmic differentiation is the most suitable method for both terms.

For the first term, the exponent is \(x\), while for the second term the exponent is \(\frac{1}{x}\). Careful use of the product rule is required after taking logarithms.

The key formula is

\[ \frac{d}{dx}[f(x)]^{g(x)} = [f(x)]^{g(x)} \left[ g'(x)\log f(x) + g(x)\frac{f'(x)}{f(x)} \right]. \]

We shall nevertheless derive both terms step by step rather than applying the formula directly.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate \((x\cos x)^x\) using logarithmic differentiation.
Expand \(\log(x\cos x)\) into \(\log x+\log(\cos x)\).
Use the product rule to differentiate \(x[\log x+\log(\cos x)]\).
Differentiate \((x\sin x)^{1/x}\) using logarithmic differentiation.
Write \(\frac1x[\log x+\log(\sin x)]\) as a sum of two products.
Apply the product rule carefully to both products.
Combine the two derivatives using the sum rule.
Check that the original factors \(x\cos x\) and \(x\sin x\) have not been lost.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 33 steps

Differentiating \((x\cos x)^x\)
Let
\[y_1=(x\cos x)^x\]
Take logarithm on both sides
\[\log y_1=\log\left[(x\cos x)^x\right]\]
Using
\[\log(a^b)=b\log a\]
we obtain
\[\log y_1=x\log(x\cos x)\]
Expand the logarithm
Using
\[ \log(ab)=\log a+\log b\]
we get
\[\log y_1=x[\log x+\log(\cos x)]\]
Differentiate both sides
\[ \frac{1}{y_1}\frac{dy_1}{dx} = \frac{d}{dx} \left[ x\log x+x\log(\cos x) \right] \]
Applying the product rule to the first product,
\[ \frac{d}{dx}(x\log x) = x\frac{d}{dx}(\log x) + \log x\frac{d}{dx}(x) \]

\[ = x\left(\frac1x\right)+\log x \]

\[ =1+\log x \]
For the second product,
\[ \frac{d}{dx}[x\log(\cos x)] = x\frac{d}{dx}[\log(\cos x)] + \log(\cos x)\frac{d}{dx}(x) \]
Now,
\[\frac{d}{dx}[\log(\cos x)]=\frac{1}{\cos x}(-\sin x)=-\tan x\]
Therefore,
\[\frac{d}{dx}[x\log(\cos x)]=-x\tan x+\log(\cos x)\]
Combine the derivatives
\[\frac{1}{y_1}\frac{dy_1}{dx}=1+\log x+\log(\cos x)-x\tan x\]
Multiplying by \(y_1\),
\[ \frac{dy_1}{dx} = y_1 \left[ 1+\log x+\log(\cos x)-x\tan x \right] \]
Since
\[y_1=(x\cos x)^x,\]
we obtain
\[\boxed{\frac{dy_1}{dx}=(x\cos x)^x\left[1+\log x+\log(\cos x)-x\tan x\right]}\]
Differentiating \((x\sin x)^{1/x}\)
Let
\[y_2=(x\sin x)^{1/x}\]
Take logarithm on both sides
\[\log y_2=\log\left[(x\sin x)^{1/x}\right]\]
Using the logarithm power rule,
\[\log y_2=\frac1x\log(x\sin x)\]
Expanding the logarithm of the product,
\[\log y_2=\frac1x[\log x+\log(\sin x)]\]
Rewrite for easier differentiation
\[\log y_2=\frac{\log x}{x}+\frac{\log(\sin x)}{x}\]
Differentiate the first term
\[\frac{d}{dx}\left(\frac{\log x}{x}\right)=\frac{d}{dx}\left(x^{-1}\log x\right)\]
Using the product rule,
\[=x^{-1}\frac1x+\log x(-x^{-2})\]

\[=\frac1{x^2}-\frac{\log x}{x^2}\]

\[=\frac{1-\log x}{x^2}\]
Differentiate the second term
\[\frac{d}{dx}\left[\frac{\log(\sin x)}{x}\right]=\frac{d}{dx}\left[x^{-1}\log(\sin x)\right]\]
Applying the product rule,
\[=x^{-1}\frac{d}{dx}[\log(\sin x)]+\log(\sin x)\frac{d}{dx}(x^{-1})\]
Now,
\[\frac{d}{dx}[\log(\sin x)]=\frac{\cos x}{\sin x}=\cot x\]
and
\[\frac{d}{dx}(x^{-1})=-x^{-2}\]
Therefore,
\[\frac{d}{dx}\left[\frac{\log(\sin x)}{x}\right]=\frac{\cot x}{x}-\frac{\log(\sin x)}{x^2}\]
Combine the derivatives
\[\frac1{y_2}\frac{dy_2}{dx}=\frac{1-\log x}{x^2}+\frac{\cot x}{x}-\frac{\log(\sin x)}{x^2}\]
Taking \(1/x^2\) common,
\[\frac1{y_2}\frac{dy_2}{dx}=\frac{1-\log x+x\cot x-\log(\sin x)}{x^2}\]
Hence,
\[\frac{dy_2}{dx}=\frac{y_2}{x^2}\left[1+x\cot x-\log x-\log(\sin x)\right]\]
Since
\[y_2=(x\sin x)^{1/x}\]
we obtain
\[\boxed{\frac{dy_2}{dx}=\frac{(x\sin x)^{1/x}}{x^2}\left[1+x\cot x-\log x-\log(\sin x)\right]}\]
Combining Both Derivatives
The original function is
\[y=y_1+y_2\]
Therefore, using the sum rule,
\[\frac{dy}{dx}=\frac{dy_1}{dx}+\frac{dy_2}{dx}\]
Substituting the two results,
\[\frac{dy}{dx}=(x\cos x)^x\left[1+\log x+\log(\cos x)-x\tan x\right]+\frac{(x\sin x)^{1/x}}{x^2}\left[1+x\cot x-\log x-\log(\sin x)\right]\]
Hence, the required derivative is
\[\boxed{\frac{dy}{dx}=(x\cos x)^x\left[1+\log x+\log(\cos x)-x\tan x\right]+\frac{(x\sin x)^{1/x}}{x^2}\left[1+x\cot x-\log x-\log(\sin x)\right]}\]

🎯 Exam Significance

Exam Significance

›

This problem is valuable for board examinations because it combines logarithmic differentiation with repeated applications of the product rule.

It tests differentiation of variable-base and variable-exponent functions.
It requires correct expansion of logarithms.
It tests the product rule for \(x\cos x\) and \(x\sin x\).
It tests differentiation of \(1/x\).
It reinforces the importance of retaining the complete multiplier \(y\) after logarithmic differentiation.

Significance for Competitive Entrance Examinations

The question is useful for developing speed and accuracy with composite differentiation structures. The two terms look similar, but their exponents are different, so their logarithmic derivatives have different structures.

\((x\cos x)^x\) has exponent \(x\), giving a \(+1\) contribution from differentiating \(x\).
\((x\sin x)^{1/x}\) has exponent \(1/x\), whose derivative is \(-1/x^2\).
The factor \(1\) in the second answer arises from
\[ \frac{d}{dx}\left(\frac{\log x}{x}\right). \]
Keeping the logarithmic derivative in factored form reduces algebraic errors.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For a variable-base, variable-exponent function, logarithmic differentiation is highly effective.
For
\[ (x\cos x)^x, \]
the logarithmic derivative is
\[ 1+\log x+\log(\cos x)-x\tan x. \]
For
\[ (x\sin x)^{1/x}, \]
the logarithmic derivative is
\[ \frac{ 1+x\cot x-\log x-\log(\sin x) }{x^2}. \]
The constant \(1\) in the second result is essential and comes from differentiating \(\log x/x\).
Always multiply the logarithmic derivative by the original function before writing the final derivative.

📘 Concept & Theory

Concept/Theory

›

This is an implicit differentiation problem. Both \(x\) and \(y\) occur as variables, and \(y\) is an implicit function of \(x\).

The expressions \(x^y\) and \(y^x\) have both variable bases and variable exponents. Therefore, logarithmic differentiation is required for each term.

For the first term, let

\[ u=x^y \]

Then

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

For the second term, let

\[ v=y^x \]

Then

\[ \log v=x\log y. \]

Since \(y\) is a function of \(x\), whenever we differentiate an expression containing \(y\), the factor \(\dfrac{dy}{dx}\) must be included.

🗺️ Solution Roadmap

Step-by-step Plan

›

Start with \(x^y+y^x=1\).
Differentiate \(x^y\) using logarithmic differentiation.
Differentiate \(y^x\) using logarithmic differentiation.
Differentiate both sides of the original implicit equation.
Collect all terms containing \(\dfrac{dy}{dx}\) on one side.
Move the remaining terms to the other side.
Solve algebraically for \(\dfrac{dy}{dx}\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 25 steps

Given — Equation
\[x^y+y^x=1\]
Differentiate \(x^y\)
Let
\[u=x^y\]
Taking logarithm on both sides,
\[\log u=\log(x^y)\]

\[\log u=y\log x\]
Differentiate both sides with respect to \(x\):
\[\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(y\log x)\]
Using the product rule,
\[\frac{1}{u}\frac{du}{dx}=y\frac{d}{dx}(\log x)+\log x\frac{dy}{dx}\]

\[=\frac{y}{x}+\log x\frac{dy}{dx}\]
Multiplying by \(u\),
\[\frac{du}{dx}=u\left[\frac{y}{x}+\log x\frac{dy}{dx}\right]\]
Since \(u=x^y\),
\[\boxed{\frac{d}{dx}(x^y)=x^y\left[\frac{y}{x}+\log x\frac{dy}{dx}\right]}\]
Differentiate \(y^x
Let
\[v=y^x\]
Taking logarithm on both sides,
\[\log v=\log(y^x)\]

\[\log v=x\log y\]
Differentiate both sides with respect to \(x\):
\[\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}(x\log y)\]
Using the product rule,
\[\frac{1}{v}\frac{dv}{dx}=\log y\frac{d}{dx}(x)+x\frac{d}{dx}(\log y)\]

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

\[=\log y+\frac{x}{y}\frac{dy}{dx}\]
Multiplying by \(v\),
\[\frac{dv}{dx}=v\left[\log y+\frac{x}{y}\frac{dy}{dx}\right]\]
Since \(v=y^x\),
\[\boxed{\frac{d}{dx}(y^x)=y^x\left[\log y+\frac{x}{y}\frac{dy}{dx}\right]}\]
Differentiate the complete equation
Starting with
\[x^y+y^x=1\]
we obtain
\[x^y\left[\frac{y}{x}+\log x\frac{dy}{dx}\right]+y^x\left[\log y+\frac{x}{y}\frac{dy}{dx}\right]=0\]
Expanding the brackets,
\[\frac{x^y y}{x}+x^y\log x\frac{dy}{dx}+y^x\log y+\frac{xy^x}{y}\frac{dy}{dx}=0\]
Collect the \(\dfrac{dy}{dx}\) terms
\[x^y\log x\frac{dy}{dx}+\frac{xy^x}{y}\frac{dy}{dx}=-\frac{x^y y}{x}-y^x\log y\]
Taking \(\dfrac{dy}{dx}\) common,
\[\frac{dy}{dx}\left[x^y\log x+\frac{xy^x}{y}\right]=-\left[\frac{x^y y}{x}+y^x\log y\right]\]
Solve for \(\dfrac{dy}{dx}\)
\[\frac{dy}{dx}=-\frac{\dfrac{x^y y}{x}+y^x\log y}{x^y\log x+\dfrac{xy^x}{y}}\]
Therefore,
\[\boxed{\frac{dy}{dx}=-\frac{\dfrac{x^y y}{x}+y^x\log y}{x^y\log x+\dfrac{xy^x}{y}}}\]
Multiplying numerator and denominator by \(xy\), we obtain
\[\frac{dy}{dx}=-\frac{xy\left(\dfrac{x^y y}{x}+y^x\log y\right)}{xy\left(x^y\log x+\dfrac{xy^x}{y}\right)}\]
The numerator becomes
\[xy\left(\frac{x^y y}{x}\right)=x^y y^2\]
and
\[xy(y^x\log y)=xy^{x+1}\log y.\]
The denominator becomes
\[xy(x^y\log x)=x^{y+1}y\log x\]
and
\[xy\left(\frac{xy^x}{y}\right)=x^2y^x.\]
Hence,
\[\boxed{\frac{dy}{dx}=-\frac{x^y y^2+xy^{x+1}\log y}{x^{y+1}y\log x+x^2y^x}}\]
Equivalently, moving the negative sign into the numerator,
\[\boxed{\frac{dy}{dx}=\frac{-x^y y^2-xy^{x+1}\log y}{x^{y+1}y\log x+x^2y^x}}\]

🎯 Exam Significance

Exam Significance

›

This is an important implicit-differentiation problem because both terms contain variable bases and variable exponents.

It tests logarithmic differentiation of \(x^y\).
It tests logarithmic differentiation of \(y^x\).
It requires the product rule in both logarithmic equations.
It tests correct treatment of \(y\) as a function of \(x\).
It requires careful collection and simplification of \(\dfrac{dy}{dx}\) terms.

Significance for Competitive Entrance Examinations

The problem is useful for developing fluency with implicit differentiation involving variable powers. The main challenge is algebraic accuracy after the differentiation has been completed.

Recognise \(x^y\) and \(y^x\) as logarithmic-differentiation forms immediately.
Remember that
\[ \frac{d}{dx}(y)=\frac{dy}{dx}. \]
For \(y^x\), remember that both the exponent \(x\) and the base \(y\) vary with \(x\).
Keep the negative sign outside the numerator until the final simplification to avoid sign errors.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For
\[ x^y, \]
logarithmic differentiation gives
\[ \frac{d}{dx}(x^y) = x^y \left[ \frac{y}{x}+\log x\frac{dy}{dx} \right]. \]
For
\[ y^x, \]
logarithmic differentiation gives
\[ \frac{d}{dx}(y^x) = y^x \left[ \log y+\frac{x}{y}\frac{dy}{dx} \right]. \]
Since the equation is implicit, every differentiation involving \(y\) must account for \(\dfrac{dy}{dx}\).
After differentiating, collect all \(\dfrac{dy}{dx}\) terms before solving.
The negative sign applies to the complete numerator when solving the final linear equation in \(\dfrac{dy}{dx}\).

📘 Concept & Theory

Concept/Theory

›

The equation

\[ y^x=x^y \]

is an implicit equation because \(y\) is a function of \(x\). Both sides contain variable bases and variable exponents. Therefore, logarithmic differentiation is required.

We can differentiate both sides separately and then equate the resulting derivatives. The important point is that when differentiating \(y^x\), both the exponent \(x\) and the base \(y\) must be treated as variable quantities.

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate \(y^x\) using logarithmic differentiation.
Differentiate \(x^y\) using logarithmic differentiation.
Equate the derivatives because \(y^x=x^y\).
Expand both sides carefully.
Collect the \(\dfrac{dy}{dx}\) terms on one side.
Solve for \(\dfrac{dy}{dx}\).
Use the original relation \(y^x=x^y\) if a further simplification is desired.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Differentiating \(y^x\)
Let
\[u=y^x\]
Taking logarithm on both sides,
\[\log u=\log(y^x)\]

\[\log u=x\log y\]
Differentiate both sides with respect to \(x\):
\[\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}(x\log y)\]
Using the product rule,
\[\frac{1}{u}\frac{du}{dx}=\log y\frac{d}{dx}(x)+x\frac{d}{dx}(\log y)\]

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

\[=\log y+\frac{x}{y}\frac{dy}{dx}\]
Multiplying by \(u\),
\[\frac{du}{dx}=u\left[\log y+\frac{x}{y}\frac{dy}{dx}\right]\]
Since \(u=y^x\),
\[\boxed{\frac{d}{dx}(y^x)=y^x\left[\log y+\frac{x}{y}\frac{dy}{dx}\right]}\]
Differentiating \(x^y\)
Let
\[v=x^y\]
Taking logarithm on both sides,
\[\log v=\log(x^y)\]

\[\log v=y\log x\]
Differentiate both sides with respect to \(x\):
\[\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}(y\log x)\]
Using the product rule,
\[\frac{1}{v}\frac{dv}{dx}=\log x\frac{dy}{dx}+y\frac{d}{dx}(\log x)\]

\[=\log x\frac{dy}{dx}+\frac{y}{x}\]
Multiplying by \(v\),
\[\frac{dv}{dx}=v\left[\log x\frac{dy}{dx}+\frac{y}{x}\right]\]
Since \(v=x^y\),
\[\boxed{\frac{d}{dx}(x^y)=x^y\left[\log x\frac{dy}{dx}+\frac{y}{x}\right]}\]
Equate the Derivatives
Since the original equation is
\[y^x=x^y\]
their derivatives are equal:
\[y^x\left[\log y+\frac{x}{y}\frac{dy}{dx}\right]=x^y\left[\log x\frac{dy}{dx}+\frac{y}{x}\right]\]
Expand both sides
\[y^x\log y+\frac{xy^x}{y}\frac{dy}{dx}=x^y\log x\frac{dy}{dx}+\frac{x^y y}{x}\]
Since
\[\frac{xy^x}{y}=xy^{x-1}\]
and
\[\frac{x^y y}{x}=x^{y-1}y,\]
we get
\[y^x\log y+xy^{x-1}\frac{dy}{dx}=x^y\log x\frac{dy}{dx}+x^{y-1}y\]
Collect the \(\dfrac{dy}{dx}\) terms
\[xy^{x-1}\frac{dy}{dx}-x^y\log x\frac{dy}{dx}=x^{y-1}y-y^x\log y\]
Taking \(\dfrac{dy}{dx}\) common,
\[\frac{dy}{dx}\left[xy^{x-1}-x^y\log x\right]=x^{y-1}y-y^x\log y\]
Solve for \(\dfrac{dy}{dx}\)
\[\boxed{\frac{dy}{dx}=\frac{x^{y-1}y-y^x\log y}{xy^{x-1}-x^y\log x}}\]
Use the relation \(y^x=x^y\)
The result can also be simplified by using the original equation. Since
\[y^x=x^y\]
we may write
\[x^{y-1}y=\frac{y}{x}x^y\]
and
\[xy^{x-1}=\frac{x}{y}y^x=\frac{x}{y}x^y.\]
Therefore,
\[\frac{dy}{dx}=\frac{x^y\left(\frac{y}{x}-\log y\right)}{x^y\left(\frac{x}{y}-\log x\right)}\]

\[\boxed{\frac{dy}{dx}=\frac{\dfrac{y}{x}-\log y}{\dfrac{x}{y}-\log x}}\]
This is a more compact form of the answer.

🎯 Exam Significance

Exam Significance

›

This is an important implicit differentiation problem because both sides contain variable-base and variable-exponent functions.

It tests logarithmic differentiation of \(y^x\).
It tests logarithmic differentiation of \(x^y\).
It requires the product rule after taking logarithms.
It tests the correct treatment of \(y\) as a function of \(x\).
It requires careful algebraic collection of \(\dfrac{dy}{dx}\).

Significance for Competitive Entrance Examinations

This problem is useful for recognising symmetry between \(x^y\) and \(y^x\). The compact form obtained after using \(y^x=x^y\) is especially useful for algebraic simplification.

Recognise both \(x^y\) and \(y^x\) as logarithmic-differentiation forms.
Remember that the exponent and base may both depend on \(x\).
Collect \(\dfrac{dy}{dx}\) terms systematically rather than simplifying prematurely.
Use the original relation after differentiation whenever it simplifies the result.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For
\[ y^x, \]
logarithmic differentiation gives
\[ \frac{d}{dx}(y^x) = y^x \left[ \log y+\frac{x}{y}\frac{dy}{dx} \right]. \]
For
\[ x^y, \]
logarithmic differentiation gives
\[ \frac{d}{dx}(x^y) = x^y \left[ \log x\frac{dy}{dx}+\frac{y}{x} \right]. \]
The relation \(y^x=x^y\) must be preserved throughout the implicit differentiation.
After differentiation, collect all \(\dfrac{dy}{dx}\) terms before solving.
The final result can be written compactly as
\[ \frac{dy}{dx} = \frac{\dfrac{y}{x}-\log y} {\dfrac{x}{y}-\log x}. \]

📘 Concept & Theory

Concept/Theory

›

The given equation is an implicit relation between \(x\) and \(y\):

\[ (\cos x)^y=(\cos y)^x \]

Both sides contain a variable base and a variable exponent. Therefore, logarithmic differentiation is required.

Since \(y\) depends on \(x\), differentiating either side requires the chain rule and product rule. In particular,

\[ \frac{d}{dx}(\log(\cos y)) = -\tan y\frac{dy}{dx}. \]

After differentiating both sides, we collect the terms containing \(\dfrac{dy}{dx}\) and solve for the required derivative.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(u=(\cos x)^y\) and differentiate it logarithmically.
Let \(v=(\cos y)^x\) and differentiate it logarithmically.
Remember that \(y\) is a function of \(x\).
Equate the derivatives because the two original expressions are equal.
Collect all \(\dfrac{dy}{dx}\) terms on the left.
Move the remaining terms to the right.
Solve for \(\dfrac{dy}{dx}\).
Use the original relation to obtain a more compact equivalent form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 24 steps

Differentiating \((\cos x)^y\)
Let
\[u=(\cos x)^y\]
Take logarithm on both sides
\[\log u=\log\left[(\cos x)^y\right]\]

\[\log u=y\log(\cos x)\]
Differentiate both sides
\[\frac{1}{u}\frac{du}{dx}=\frac{d}{dx}\left[y\log(\cos x)\right]\]
Using the product rule,
\[\frac{1}{u}\frac{du}{dx}=\log(\cos x)\frac{dy}{dx}+y\frac{d}{dx}[\log(\cos x)]\]
Now, by the chain rule,
\[\frac{d}{dx}[\log(\cos x)]=\frac{1}{\cos x}(-\sin x)=-\tan x\]
Therefore,
\[\frac{1}{u}\frac{du}{dx}=\log(\cos x)\frac{dy}{dx}-y\tan x\]
Multiplying by \(u\),
\[\frac{du}{dx}=u\left[\log(\cos x)\frac{dy}{dx}-y\tan x\right]\]
Since \(u=(\cos x)^y\),
\[\boxed{\frac{du}{dx}=(\cos x)^y\left[\log(\cos x)\frac{dy}{dx}-y\tan x\right]}\]
Differentiating \((\cos y)^x\)
Let
\[v=(\cos y)^x\]
Take logarithm on both sides
\[\log v=\log\left[(\cos y)^x\right]\]

\[\log v=x\log(\cos y)\]
Differentiate both sides
\[\frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}\left[x\log(\cos y)\right]\]
Using the product rule,
\[\frac{1}{v}\frac{dv}{dx}=\log(\cos y)\frac{d}{dx}(x)+x\frac{d}{dx}[\log(\cos y)]\]

\[=\log(\cos y)+x\frac{d}{dx}[\log(\cos y)]\]
Now apply the chain rule:
\[\frac{d}{dx}[\log(\cos y)]=\frac{1}{\cos y}\left(-\sin y\right)\frac{dy}{dx}\]

\[=-\tan y\frac{dy}{dx}\]
Therefore,
\[\frac{1}{v}\frac{dv}{dx}=\log(\cos y)-x\tan y\frac{dy}{dx}\]
Multiplying by \(v\),
\[\frac{dv}{dx}=v\left[\log(\cos y)-x\tan y\frac{dy}{dx}\right]\]
Since \(v=(\cos y)^x\),
\[\boxed{\frac{dv}{dx}=(\cos y)^x\left[\log(\cos y)-x\tan y\frac{dy}{dx}\right]}\]
Equating the Derivatives
Given —
\[(\cos x)^y=(\cos y)^x,\]
therefore,
\[(\cos x)^y\left[\log(\cos x)\frac{dy}{dx}-y\tan x\right]=(\cos y)^x\left[\log(\cos y)-x\tan y\frac{dy}{dx}\right]\]
Expand both sides
\[(\cos x)^y\log(\cos x)\frac{dy}{dx}-y(\cos x)^y\tan x\]

\[=(\cos y)^x\log(\cos y)-x(\cos y)^x\tan y\frac{dy}{dx}\]
Collect the \(\dfrac{dy}{dx}\) terms
Move the second derivative term to the left:
\[(\cos x)^y\log(\cos x)\frac{dy}{dx}+x(\cos y)^x\tan y\frac{dy}{dx}\]

\[=(\cos y)^x\log(\cos y)+y(\cos x)^y\tan x\]
Taking \(\dfrac{dy}{dx}\) common,
\[\frac{dy}{dx}\left[(\cos x)^y\log(\cos x)+x(\cos y)^x\tan y\right]\]

\[=(\cos y)^x\log(\cos y)+y(\cos x)^y\tan x\]
Solve for \(\dfrac{dy}{dx}\)
\[\boxed{\frac{dy}{dx}=\frac{(\cos y)^x\log(\cos y)+y(\cos x)^y\tan x}{(\cos x)^y\log(\cos x)+x(\cos y)^x\tan y}}\]
Simplify using the original equation
Since
\[(\cos x)^y=(\cos y)^x\]
let their common value be \(K\). Then
\[\frac{dy}{dx}=\frac{K\log(\cos y)+yK\tan x}{K\log(\cos x)+xK\tan y}\]
Canceling the common non-zero factor \(K\),
\[\boxed{\frac{dy}{dx}=\frac{\log(\cos y)+y\tan x}{\log(\cos x)+x\tan y}}\]

🎯 Exam Significance

Exam Significance

›

This problem combines implicit differentiation, logarithmic differentiation, product rule, and chain rule. It is therefore a useful higher-order practice problem for Continuity and Differentiability.

It tests variable-base and variable-exponent differentiation.
It reinforces the derivative of \(\log(\cos x)\).
It requires the chain rule when differentiating \(\log(\cos y)\).
It tests systematic collection of \(\dfrac{dy}{dx}\) terms.
It demonstrates how the original relation can simplify the final answer.

Significance for Competitive Entrance Examinations

The symmetry of the equation makes this a useful problem for developing algebraic efficiency. Instead of carrying large powers throughout the calculation, the relation can be used at the end to cancel the common factor.

Recognise logarithmic differentiation immediately.
Remember
\[ \frac{d}{dx}\log(\cos y) = -\tan y\frac{dy}{dx}. \]
Collect all \(\dfrac{dy}{dx}\) terms before dividing.
Use the original equality to simplify the final expression.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

For a variable-base, variable-exponent function, logarithmic differentiation is the appropriate technique.
The derivative
\[ \frac{d}{dx}\log(\cos x)=-\tan x \]
is frequently useful.
When the argument contains \(y\),
\[ \frac{d}{dx}\log(\cos y) = -\tan y\frac{dy}{dx}. \]
After differentiating an implicit equation, collect all \(\dfrac{dy}{dx}\) terms on one side.
The relation
\[ (\cos x)^y=(\cos y)^x \]
allows the common factor to be canceled from the final expression.

📘 Concept & Theory

Concept/Theory

›

The given equation is an implicit relation between \(x\) and \(y\):

\[ xy=e^{x-y}. \]

Since \(y\) is a function of \(x\), we use implicit differentiation.

The left-hand side \(xy\) is a product, so the product rule is required:

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

The right-hand side is an exponential composite function. Therefore, the chain rule is required:

\[ \frac{d}{dx}\left(e^{x-y}\right) = e^{x-y}\frac{d}{dx}(x-y). \]

Because \(y\) depends on \(x\),

\[ \frac{d}{dx}(x-y) = 1-\frac{dy}{dx}. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate \(xy\) using the product rule.
Differentiate \(e^{x-y}\) using the chain rule.
Remember that \(\dfrac{d}{dx}(y)=\dfrac{dy}{dx}\).
Equate the two derivatives.
Collect all \(\dfrac{dy}{dx}\) terms on one side.
Solve for \(\dfrac{dy}{dx}\).
Optionally use the original equation to obtain an alternative form.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 20 steps

Given — Equation
\[xy=e^{x-y}\]
Differentiate the left-hand side
Using the product rule,
\[\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\frac{dx}{dx}\]
Since
\[\frac{dx}{dx}=1\]
we get
\[\boxed{\frac{d}{dx}(xy)=x\frac{dy}{dx}+y}\]
Differentiate the right-hand side
\[\frac{d}{dx}\left(e^{x-y}\right)\]
Using the chain rule,
\[=e^{x-y}\frac{d}{dx}(x-y)\]
Since
\[\frac{d}{dx}(x)=1\]
and
\[\frac{d}{dx}(y)=\frac{dy}{dx}\]
therefore,
\[\frac{d}{dx}(x-y)=1-\frac{dy}{dx}\]
Hence,
\[\frac{d}{dx}\left(e^{x-y}\right)=e^{x-y}\left(1-\frac{dy}{dx}\right)\]
Exoanding
\[=e^{x-y}-e^{x-y}\frac{dy}{dx}.\]
Thus,
\[\boxed{\frac{d}{dx}\left(e^{x-y}\right)=e^{x-y}-e^{x-y}\frac{dy}{dx}}\]
Equate both derivatives
\[x\frac{dy}{dx}+y=e^{x-y}-e^{x-y}\frac{dy}{dx}\]
Collect the \(\dfrac{dy}{dx}\) terms. Move
\[-e^{x-y}\frac{dy}{dx}\]
to the left-hand side:
\[x\frac{dy}{dx}+e^{x-y}\frac{dy}{dx}=e^{x-y}-y\]
Taking \(\dfrac{dy}{dx}\) common,
\[\frac{dy}{dx}\left(x+e^{x-y}\right)=e^{x-y}-y\]
Solve for \(\dfrac{dy}{dx}\)
\[\boxed{\frac{dy}{dx}=\frac{e^{x-y}-y}{x+e^{x-y}}}\]
Alternative form using the original equation
From the original equation,
\[xy=e^{x-y}\]
Substituting this into the result gives
\[\frac{dy}{dx}=\frac{xy-y}{x+xy}\]

\[=\frac{y(x-1)}{x(1+y)}.\]
Therefore, an equivalent form is
\[\boxed{\frac{dy}{dx}=\frac{y(x-1)}{x(1+y)}}\]
The form
\[ \frac{e^{x-y}-y}{x+e^{x-y}} \]
follows directly from differentiation, while the latter form uses the given equation to simplify the result.

🎯 Exam Significance

Exam Significance

›

This is a fundamental implicit-differentiation problem that combines two highly important differentiation rules.

It tests the product rule through the term \(xy\).
It tests the chain rule through \(e^{x-y}\).
It reinforces that \(y\) must be treated as a function of \(x\).
It tests correct collection of \(\dfrac{dy}{dx}\) terms.
It provides practice in simplifying an answer using the original equation.

Significance for Competitive Entrance Examinations

The problem is useful for developing speed in implicit differentiation. The structure is straightforward once the product and chain rules are recognised.

Product \(xy\) immediately suggests the product rule.
Exponential composite function \(e^{x-y}\) immediately suggests the chain rule.
The term \(-\dfrac{dy}{dx}\) in the exponent is a common source of sign errors.
Using \(e^{x-y}=xy\) can shorten the final expression.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Use the product rule for
\[ xy. \]
Use the chain rule for
\[ e^{x-y}. \]
Since \(y\) depends on \(x\),
\[ \frac{d}{dx}(y)=\frac{dy}{dx}. \]
Always collect the \(\dfrac{dy}{dx}\) terms before solving.
The original relation
\[ xy=e^{x-y} \]
can be used after differentiation to obtain a simpler equivalent result.

📘 Concept & Theory

Concept/Theory

›

The given function is a product of four factors, each involving a different power of \(x\). Direct differentiation using the product rule would require repeated application of the product rule.

Since the function is a product of positive factors on an appropriate interval, logarithmic differentiation provides a much more efficient approach. If

\[ y=u_1u_2u_3u_4, \]

then

\[ \log y=\log u_1+\log u_2+\log u_3+\log u_4. \]

Differentiating gives

\[ \frac{1}{y}\frac{dy}{dx} = \frac{u_1'}{u_1} + \frac{u_2'}{u_2} + \frac{u_3'}{u_3} + \frac{u_4'}{u_4}. \]

Here, logarithmic differentiation converts the complicated product into a sum of four simple rational expressions.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let the given product be \(y\).
Take logarithm on both sides.
Use the logarithm product rule to separate the four factors.
Differentiate using the chain rule.
Multiply by \(y\) to obtain \(dy/dx=f'(x)\).
Substitute \(x=1\) to calculate \(f'(1)\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Let
\[y=(1+x)(1+x^2)(1+x^4)(1+x^8)\]
Taking logarithm on both sides,
\[\log y=\log\left[(1+x)(1+x^2)(1+x^4)(1+x^8)\right]\]
Using
\[\log(ab)=\log a+\log b,\]
we get
\[\log y=\log(1+x)+\log(1+x^2)+\log(1+x^4)+\log(1+x^8)\]
Differentiate both sides with respect to \(x\):
\[\frac{1}{y}\frac{dy}{dx}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\]
Multiplying both sides by \(y\),
\[\frac{dy}{dx}=y\left[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right]\]
Substituting the value of \(y\),
\[\boxed{f'(x)=(1+x)(1+x^2)(1+x^4)(1+x^8)\left[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+\frac{8x^7}{1+x^8}\right]}\]
Finding \(f'(1)\)
Put \(x=1\) in the derivative:
\[f'(1)=(1+1)(1+1^2)(1+1^4)(1+1^8)\left[\frac{1}{1+1}+\frac{2(1)}{1+1^2}+\frac{4(1)^3}{1+1^4}+\frac{8(1)^7}{1+1^8}\right]\]
Simplifying the product,
\[(1+1)(1+1^2)(1+1^4)(1+1^8)=2\cdot2\cdot2\cdot2=16\]
Now simplify the bracket:
\[\frac12+\frac22+\frac42+\frac82=\frac{1+2+4+8}{2}\]

\[=\frac{15}{2}\]
Therefore,
\[f'(1)=16\left(\frac{15}{2}\right)\]

\[f'(1)=8(15)\]

\[\boxed{f'(1)=120}\]

🎯 Exam Significance

Exam Significance

›

This problem is important because it tests the application of logarithmic differentiation to a product involving powers of \(x\). It also tests whether the student can correctly apply the chain rule after taking logarithms.

In board examinations, the important scoring steps are:

Taking logarithm correctly.
Splitting the logarithm of the product.
Differentiating every logarithmic term correctly.
Multiplying by \(y\).
Substituting \(x=1\) carefully.

Significance for Competitive Entrance Examinations

For competitive examinations, this problem illustrates how logarithmic differentiation can reduce a lengthy product-rule calculation to a short sum. The same technique is particularly useful for expressions containing several factors raised to variable powers.

A useful pattern to remember is

\[ y=\prod_{k=1}^{n}u_k(x) \quad\Longrightarrow\quad \frac{y'}{y} = \sum_{k=1}^{n}\frac{u_k'(x)}{u_k(x)}. \]

Recognising this pattern can save substantial calculation time in time-constrained examinations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Logarithmic differentiation is highly effective for products containing several factors.
Always apply the chain rule when differentiating logarithmic expressions.
For a product \(y=\prod u_i\),
\[ \frac{y'}{y}=\sum\frac{u_i'}{u_i}. \]
The derivative of the given function is
\[ f'(x) = (1+x)(1+x^2)(1+x^4)(1+x^8) \left[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} \right]. \]
At \(x=1\),
\[ f'(1)=120. \]

📘 Concept & Theory

Concept/Theory

›

This question demonstrates that the derivative of the same function can be obtained using different differentiation techniques. The three methods are:

Product Rule:
\[ \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}. \]
Expansion: Expand the product first and then differentiate the resulting polynomial term-by-term.
Logarithmic Differentiation: Take logarithms of both sides, convert the product into a sum, differentiate, and finally multiply by \(y\).

The final derivative obtained by all three methods must be identical because all three methods differentiate the same function.

🗺️ Solution Roadmap

Step-by-step Plan

›

Define the two quadratic factors as \(u\) and \(v\).
Apply the product rule directly.
Expand the product and differentiate the resulting polynomial.
For logarithmic differentiation, take \(\log\) on both sides.
Differentiate the resulting sum of logarithms.
Multiply by \(y\) and simplify.
Verify that all three methods give the same result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Method I: Using the Product Rule
Let
\[u=x^2-5x+8\]

\[v=x^2+7x+9\]
Therefore,
\[y=uv\]
Differentiate \(u\):
\[\frac{du}{dx}=2x-5\]
Differentiate \(v\):
\[\frac{dv}{dx}=2x+7\]
Using the product rule,
\[\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\]
Substituting the values of \(u,v,u'\) and \(v'\),
\[\frac{dy}{dx}=(x^2-5x+8)(2x+7)+(x^2+7x+9)(2x-5)\]
Expand the first product:
\[(x^2-5x+8)(2x+7)\]

\[=2x^3+7x^2-10x^2-35x+16x+56\]

\[=2x^3-3x^2-19x+56\]
Expand the second product:
\[(x^2+7x+9)(2x-5)\]

\[=2x^3-5x^2+14x^2-35x+18x-45\]

\[=2x^3+9x^2-17x-45\]
Therefore,
\[\frac{dy}{dx}=(2x^3-3x^2-19x+56)+(2x^3+9x^2-17x-45)\]

\[\boxed{\frac{dy}{dx}=4x^3+6x^2-36x+11}\]
Method II: Expanding the Product First
We have
\[y=(x^2-5x+8)(x^2+7x+9)\]
Multiply each term of the first polynomial by the second polynomial:
\[y=x^2(x^2+7x+9)-5x(x^2+7x+9)+8(x^2+7x+9)\]
Expanding each part,
\[y=(x^4+7x^3+9x^2)+(-5x^3-35x^2-45x)+(8x^2+56x+72)\]
Collecting like terms,
\[y=x^4+(7-5)x^3+(9-35+8)x^2+(-45+56)x+72\]

\[y=x^4+2x^3-18x^2+11x+72\]
Differentiate term-by-term:
\[\frac{dy}{dx}=4x^3+6x^2-36x+11\]
Hence,
\[\boxed{\frac{dy}{dx}=4x^3+6x^2-36x+11}\]
Method III: Using Logarithmic Differentiation
Let
\[y=(x^2-5x+8)(x^2+7x+9)\]
Taking logarithm on both sides,
\[\log y=\log(x^2-5x+8)+\log(x^2+7x+9)\]
Differentiate both sides:
\[\frac{1}{y}\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{2x+7}{x^2+7x+9}\]
Multiplying by \(y\),
\[\frac{dy}{dx}=(x^2-5x+8)(x^2+7x+9)\left[\frac{2x-5}{x^2-5x+8}+\frac{2x+7}{x^2+7x+9}\right]\]
Cancel the corresponding factors:
\[\frac{dy}{dx}=(2x-5)(x^2+7x+9)+(2x+7)(x^2-5x+8)\]
This is exactly the product-rule result.
\[\boxed{\frac{dy}{dx}=4x^3+6x^2-36x+11}\]

🎯 Exam Significance

Exam Significance

›

This question is particularly useful because it asks students to compare three standard differentiation techniques. It strengthens conceptual understanding rather than relying on a single formula.

For board examinations, students should be able to identify when the product rule is more convenient and when logarithmic differentiation provides a cleaner approach.

Significance for Competitive Entrance Examinations

Competitive examinations frequently test the ability to recognise the shortest valid differentiation route. For a product of two simple polynomials, the product rule or expansion may be faster, whereas for complicated products and variable powers, logarithmic differentiation can be substantially more efficient.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

The product rule is directly applicable to the original expression.
Expansion converts the function into a polynomial that can be differentiated term-by-term.
Logarithmic differentiation converts the product into a sum.
All three methods produce the same derivative.

Q18

NUMERIC3 marks

If \(u\), \(v\) and \(w\) are functions of \(x\), show that \[\frac{d}{dx}(u\cdot v\cdot w)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}\]

Prove the result in the following two ways:

By repeated application of the product rule.
By logarithmic differentiation.

📘 Concept & Theory

Concept/Theory

›

The product rule for two differentiable functions is

\[ \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}. \]

When three functions are multiplied together, the product rule can be applied repeatedly. The resulting derivative contains three terms, with one factor differentiated in each term while the other two factors remain unchanged.

Alternatively, logarithmic differentiation converts the product into a sum:

\[ \log(uvw)=\log u+\log v+\log w. \]

This provides a short and elegant derivation of the same formula.

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(y=uvw\).
For the first method, regard \(uv\) as one function and apply the product rule to \((uv)w\).
Apply the product rule again to \(uv\).
For the second method, take logarithms of \(y=uvw\).
Split the logarithm of the product into a sum.
Differentiate both sides.
Multiply by \(y=uvw\) and simplify.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 18 steps

Method I: Repeated Application of Product Rule
Let
\[y=uvw\]
Group the first two factors together:
\[y=(uv)w\]
Applying the product rule to \((uv)w\),
\[\frac{dy}{dx}=\frac{d}{dx}(uv)\cdot w+(uv)\frac{dw}{dx}\]
We now differentiate \(uv\) using the product rule:
\[\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}\]
Substituting this into the previous equation,
\[\frac{dy}{dx}=\left(u\frac{dv}{dx}+v\frac{du}{dx}\right)w+uv\frac{dw}{dx}\]
Distributing \(w\),
\[\frac{dy}{dx}=u\frac{dv}{dx}w+v\frac{du}{dx}w+uv\frac{dw}{dx}\]
Rearranging the first two terms,
\[\frac{dy}{dx}=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}\]
Therefore,
\[\boxed{\frac{d}{dx}(uvw)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}}\]
Method II: Logarithmic Differentiation
Let
\[y=uvw\]
Taking logarithm on both sides,
\[\log y=\log(uvw).\]
Using
\[\log(uvw)=\log u+\log v+\log w,\]
we obtain
\[\log y=\log u+\log v+\log w\]
Differentiate both sides with respect to \(x\):
\[\frac{1}{y}\frac{dy}{dx}=\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\]
Multiply both sides by \(y\):
\[\frac{dy}{dx}=y\left[\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\right]\]
Since \(y=uvw\),
\[\frac{dy}{dx}=uvw\left[\frac{1}{u}\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}\frac{dw}{dx}\right]\]
Distribute \(uvw\) term-by-term:
\[\frac{dy}{dx}=uvw\left(\frac{1}{u}\frac{du}{dx}\right)+uvw\left(\frac{1}{v}\frac{dv}{dx}\right)+uvw\left(\frac{1}{w}\frac{dw}{dx}\right)\]
Simplifying each term,
\[uvw\left(\frac{1}{u}\frac{du}{dx}\right)=\frac{du}{dx}vw,\]

\[uvw\left(\frac{1}{v}\frac{dv}{dx}\right)=u\frac{dv}{dx}w,\]
and
\[uvw\left(\frac{1}{w}\frac{dw}{dx}\right)=uv\frac{dw}{dx}\]
Hence,
\[\boxed{\frac{d}{dx}(uvw)=\frac{du}{dx}vw+u\frac{dv}{dx}w+uv\frac{dw}{dx}}\]
Thus, the required result is proved by both methods.

🎯 Exam Significance

Exam Significance

›

This question is important because it tests both the conceptual and procedural understanding of the product rule. It also demonstrates how logarithmic differentiation can be used to prove a differentiation identity.

For a board examination, students should clearly show the intermediate step

\[ y=(uv)w \]

before applying the product rule twice. This makes the derivation easy to follow and avoids missing any term.

Significance for Competitive Entrance Examinations

The result is a fundamental extension of the product rule and is useful whenever several functions are multiplied together.

The logarithmic form is especially useful for a product of many functions:

\[ y=u_1u_2\cdots u_n \]

gives

\[ \frac{1}{y}\frac{dy}{dx} = \frac{u_1'}{u_1} + \frac{u_2'}{u_2} +\cdots+ \frac{u_n'}{u_n}. \]

Recognising this pattern can considerably reduce the amount of algebra required in competitive calculus problems.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

The product rule for three functions can be obtained by applying the two-function product rule twice.
The three-function product rule is
\[ (uvw)'=u'vw+uv'w+uvw'. \]
Logarithmic differentiation provides a concise alternative proof.
In every term, exactly one factor is differentiated and the remaining two factors are left unchanged.
Both methods lead to the same result, confirming the identity.

Continuity and Differentiability

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

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CONTINUITY AND DIFFERENTIABILITY — Learning Resources

Frequently Asked Questions

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Continuity and Differentiability

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

Recent posts

CONTINUITY AND DIFFERENTIABILITY — Learning Resources

Frequently Asked Questions

Q 01 What is covered in NCERT Class 12 Mathematics Exercise 5.5?

Q 02 How do you differentiate a product of three functions?

Q 03 How can the product rule for three functions be proved?

Q 04 How is logarithmic differentiation used in Exercise 5.5?

Q 05 What are the three methods used in Question 17?

Q 06 Why is logarithmic differentiation useful for products?

Q 07 What is the product rule for two functions?

Q 08 What is the product rule for three functions?

Q 09 What is the answer to Question 17 of Exercise 5.5?

Q 10 What is the main exam benefit of Exercise 5.5?

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