NCERT Class 12 Maths Ex 5.2 Solutions

📘 Concept & Theory

Concept/Theory

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This problem illustrates the Chain Rule of Differentiation. The chain rule is used when a function is composed with another function. If

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

then

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

Here, the outer function is \(\sin u\), while the inner function is \(u=x^2+5\). Therefore, we differentiate the outer function first and then multiply by the derivative of the inner function.

🗺️ Solution Roadmap

Step-by-step Plan

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Take the inner expression \(x^2+5\) as a new variable \(u\).
Express the given function as \(y=\sin u\).
Differentiate \(y\) with respect to \(u\).
Differentiate \(u\) with respect to \(x\).
Apply the chain rule \(\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}\).
Substitute \(u=x^2+5\) back into the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Let
\[u=x^2+5\]
Then the given function becomes
\[y=\sin u\]
Using the chain rule,
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
First, differentiate \(y=\sin u\) with respect to \(u\):
\[\frac{du}{dx}=\frac{d}{dx}(x^2+5)\]
Using the sum rule,
\[ \frac{du}{dx} = \frac{d}{dx}(x^2)+\frac{d}{dx}(5). \]
Since
\[\frac{d}{dx}(x^2)=2x\]
and
\[\frac{d}{dx}(5)=0\]
we get
\[\frac{du}{dx}=2x+0=2x\]
Therefore, applying the chain rule,
\[ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} \]

\[ =\cos u\cdot 2x. \]
Now substitute \(u=x^2+5\):
\[\boxed{\frac{dy}{dx}=2x\cos(x^2+5)}\]

💡 Answer

Final Answer

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\[\boxed{2x\cos(x^2+5)}\]

🎯 Exam Significance

Exam Significance

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This is a fundamental application of the chain rule, which is frequently tested in Class 12 Mathematics. The key point is that differentiating \(\sin(x^2+5)\) does not give merely \(\cos(x^2+5)\). The derivative of the inner function \(x^2+5\), namely \(2x\), must also be multiplied.

For board examinations, students should clearly show the identification of the inner function and the application of the chain rule. Writing the intermediate steps helps demonstrate the method and reduces the possibility of losing marks due to an omitted factor.

Significance for Competitive Entrance Examinations

Chain-rule differentiation is a core technique for JEE and other competitive entrance examinations. Problems often involve several nested functions, and the same principle is applied repeatedly. Mastering this simple example provides the foundation for differentiating expressions such as \(\sin(f(x))\), \(e^{f(x)}\), \(\log(f(x))\), and more complicated composite functions.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 8 points

Identify the inner function before differentiating a composite function.
For \(y=f(u)\) and \(u=g(x)\), use the chain rule:
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}.\]
The derivative of \(\sin u\) is \(\cos u\).
The derivative of \(x^2+5\) is \(2x\).
Always multiply by the derivative of the inner function.
The constant \(5\) has derivative \(0\).
For this problem, the required derivative is:
\[\boxed{\frac{d}{dx}\left[\sin(x^2+5)\right]=2x\cos(x^2+5)}\]

📘 Concept & Theory

Concept/Theory

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This problem is an application of the Chain Rule of Differentiation. The given function is a composite function because the function \(\sin x\) is the argument of the outer function \(\cos(\cdot)\).

If

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

then the chain rule states that

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

In this problem, the inner function is \(u=\sin x\), and the outer function is \(y=\cos u\).

🗺️ Solution Roadmap

Step-by-step Plan

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Take \(u=\sin x\) as the inner function.
Rewrite the given expression as \(y=\cos u\).
Find \(\dfrac{dy}{du}\).
Find \(\dfrac{du}{dx}\).
Multiply the two derivatives using the chain rule.
Substitute \(u=\sin x\) into the final expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Let
\[u=\sin x\]
Then the given function becomes
\[y=\cos u\]
Using the chain rule,
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
First, differentiate \(y=\cos u\) with respect to \(u\):
\[\frac{dy}{du}=\frac{d}{du}(\cos u)=-\sin u.\]
Next, differentiate \(u=\sin x\) with respect to \(x\):
\[\frac{du}{dx}=\frac{d}{dx}(\sin x)=\cos x.\]
Therefore, by the chain rule,
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]

\[=(-\sin u)(\cos x).\]
Now substitute \(u=\sin x\):
\[\frac{dy}{dx}=-\sin(\sin x)\cos x.\]
Hence,
\[\boxed{\frac{d}{dx}\left[\cos(\sin x)\right]=-\sin(\sin x)\cos x}\]

💡 Answer

Final Answer

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\[\boxed{-\sin(\sin x)\cos x}\]

🎯 Exam Significance

Exam Significance

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This question tests whether the student can correctly identify and differentiate a composite trigonometric function. It is important to remember that the derivative of \(\cos(\sin x)\) is not simply \(-\sin(\sin x)\). The derivative of the inner function \(\sin x\) must also be multiplied.

In a board examination, clearly writing the substitution \(u=\sin x\), followed by the chain-rule step, makes the method explicit and helps avoid errors involving the inner function.

Significance for Competitive Entrance Examination Aspirants

Nested functions such as \(\cos(\sin x)\) are basic examples of composite functions. Competitive examinations frequently extend this idea to multiple layers of composition. For example, students may encounter functions of the form \(f(g(h(x)))\), where the chain rule has to be applied successively.

A useful mental rule is:

\[ \text{Derivative of outer function}\times\text{Derivative of inner function}. \]

Thus,

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

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

\(\cos(\sin x)\) is a composite function.
For a composite function, apply the chain rule.
If \(y=\cos u\), then \(\dfrac{dy}{du}=-\sin u\).
If \(u=\sin x\), then \(\dfrac{du}{dx}=\cos x\).
The derivatives are multiplied, not added or subtracted.
After differentiation, substitute \(u=\sin x\) back into the expression.
The correct derivative is:

📘 Concept & Theory

Concept/Theory

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This problem is a direct application of the Chain Rule of Differentiation. The expression \(ax+b\) is the inner function of the sine function.

If

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

then

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

For this problem, we take \(u=ax+b\). The derivative of the outer function \(\sin u\) is \(\cos u\), while the derivative of the inner linear function \(ax+b\) is \(a\).

🗺️ Solution Roadmap

Step-by-step Plan

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Take \(u=ax+b\) as the inner function.
Write the given function as \(y=\sin u\).
Differentiate \(y\) with respect to \(u\).
Differentiate \(u=ax+b\) with respect to \(x\).
Apply the chain rule.
Substitute \(u=ax+b\) back into the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Let
\[u=ax+b\]
Then the given function becomes
\[y=\sin u\]
Using the chain rule,
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
First, differentiate \(y=\sin u\) with respect to \(u\):
\[\frac{dy}{du}=\frac{d}{du}(\sin u)=\cos u\]
Next, differentiate \(u=ax+b\) with respect to \(x\):
\[\frac{du}{dx}=\frac{d}{dx}(ax+b)\]
Using the sum rule,
\[\frac{du}{dx}=\frac{d}{dx}(ax)+\frac{d}{dx}(b)\]
Since \(a\) and \(b\) are constants,
\[\frac{d}{dx}(ax)=a\]
and
\[\frac{d}{dx}(b)=0\]
Therefore,
\[\frac{du}{dx}=a+0=a\]
Now apply the chain rule:
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]

\[=\cos u\cdot a\]

\[=a\cos u\]
Finally, substitute \(u=ax+b\):
\[\boxed{\frac{dy}{dx}=a\cos(ax+b)}\]

💡 Answer

Final Answer

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\[\boxed{a\cos(ax+b)}\]

🎯 Exam Significance

Exam Significance

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This question establishes an important standard result for differentiating a trigonometric function whose argument is a linear expression. It is particularly useful because many board-examination questions require students to differentiate expressions such as \(\sin(ax+b)\), \(\cos(ax+b)\), or \(\tan(ax+b)\).

The essential point is that the derivative of the inner expression \(ax+b\) cannot be omitted. The factor \(a\) appears because

\[ \frac{d}{dx}(ax+b)=a. \]

Showing the chain-rule steps also makes the solution mathematically complete and helps prevent the common error of writing only \(\cos(ax+b)\).

Significance for Competitive Entrance Examination Aspirants

This result is a basic building block for more complicated differentiation problems in JEE and other competitive entrance examinations. Once this pattern is understood, it can be applied immediately to nested and parameterised functions.

For example, the same chain-rule structure gives

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

and

\[ \frac{d}{dx}\tan(ax+b) = a\sec^2(ax+b). \]

Therefore, students should develop the habit of differentiating both the outer function and its inner argument.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

\(\sin(ax+b)\) is a composite function.
Use the chain rule whenever the argument of a function is itself a function of \(x\).
The derivative of \(\sin u\) is \(\cos u\).
The derivative of \(ax+b\) is \(a\).
The constant term \(b\) has derivative \(0\).
The factor \(a\) must be multiplied with the derivative of the outer function.
A useful standard result is:

📘 Concept & Theory

Concept/Theory

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This problem involves a multiple-composition function, so the chain rule must be applied successively. There are three nested functions:

\[ x \longrightarrow \sqrt{x} \longrightarrow \tan(\sqrt{x}) \longrightarrow \sec\left(\tan(\sqrt{x})\right). \]

If

\[ y=f(u), \quad u=g(t), \quad t=h(x), \]

then the chain rule gives

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

The required derivatives are

\[ \frac{d}{du}(\sec u)=\sec u\tan u, \]

\[ \frac{d}{dt}(\tan t)=\sec^2t, \]

and

\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}. \]

🗺️ Solution Roadmap

Step-by-step Plan

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Let \(t=\sqrt{x}\), the innermost function.
Let \(u=\tan t\), the middle function.
Write the given function as \(y=\sec u\).
Differentiate each layer separately: \(\dfrac{dy}{du}\), \(\dfrac{du}{dt}\), and \(\dfrac{dt}{dx}\).
Multiply all three derivatives using the chain rule.
Substitute \(t=\sqrt{x}\) and \(u=\tan(\sqrt{x})\) to express the answer entirely in terms of \(x\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Let
\[t=\sqrt{x}\]
Then
\[u=\tan t\]
Therefore, the given function can be written as
\[y=\sec u\]
Since the function has three layers of composition, apply the chain rule successively
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dt}\cdot\frac{dt}{dx}\]
First, differentiate \(y=\sec u\) with respect to \(u\):
\[\frac{dy}{du}=\frac{d}{du}(\sec u)=\sec u\tan u\]
Next, differentiate \(u=\tan t\) with respect to \(t\):
\[\frac{du}{dt}=\frac{d}{dt}(\tan t)=\sec^2t\]
Now differentiate \(t=\sqrt{x}\) with respect to \(x\). Writing the square root as a power,
\[t=x^{1/2}\]
Therefore, by the power rule,
\[\frac{dt}{dx}=\frac{d}{dx}\left(x^{1/2}\right)=\frac{1}{2}x^{-1/2}\]
Since
\[x^{-1/2}=\frac{1}{\sqrt{x}}\]
we obtain
\[\frac{dt}{dx}=\frac{1}{2\sqrt{x}}\]
Substituting the three derivatives into the chain rule gives
\[\frac{dy}{dx}=\left(\sec u\tan u\right)\left(\sec^2t\right)\left(\frac{1}{2\sqrt{x}}\right)\]
Thus,
\[\frac{dy}{dx}=\frac{\sec u\tan u\sec^2t}{2\sqrt{x}}\]
Now substitute
\[u=\tan t\]
and
\[t=\sqrt{x}\]
First,
\[\sec u=\sec\left(\tan\sqrt{x}\right)\]
and
\[\tan u=\tan\left(\tan\sqrt{x}\right)\]
Also,
\[\sec^2t=\sec^2\sqrt{x}\]
Therefore,
\[ \boxed{ \frac{dy}{dx} = \frac{ \sec\left(\tan\sqrt{x}\right) \tan\left(\tan\sqrt{x}\right) \sec^2\sqrt{x} }{ 2\sqrt{x} } } \]

💡 Answer

Final Answer

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\[ \boxed{ \frac{d}{dx} \left[ \sec\left(\tan\sqrt{x}\right) \right] = \frac{ \sec\left(\tan\sqrt{x}\right) \tan\left(\tan\sqrt{x}\right) \sec^2\sqrt{x} }{ 2\sqrt{x} } } \]

🎯 Exam Significance

Exam Significance

›

This is an important example of successive application of the chain rule. Unlike a simple composite function, this expression contains three nested layers. In a board examination, students should demonstrate each layer clearly to show that no derivative factor has been omitted.

The most common error is to differentiate only the outer function and forget one or more inner derivatives. Here, all three factors are essential:

\[ \sec u\tan u, \qquad \sec^2t, \qquad \frac{1}{2\sqrt{x}}. \]

Their product gives the complete derivative.

Significance for Competitive Entrance Examination Aspirants

Competitive entrance examinations frequently use nested functions to test whether students can apply the chain rule efficiently. This example is particularly useful because it combines three standard derivatives:

\[ \frac{d}{dx}(\sec x)=\sec x\tan x, \]

\[ \frac{d}{dx}(\tan x)=\sec^2x, \]

\[ \frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}. \]

Recognising these derivative patterns allows a student to solve complicated-looking composite functions quickly and accurately.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 8 points

For multiple nested functions, apply the chain rule successively.
Identify the innermost, middle, and outermost functions before differentiating.
For \(y=\sec u\), \(\dfrac{dy}{du}=\sec u\tan u\).
For \(u=\tan t\), \(\dfrac{du}{dt}=\sec^2t\).
For \(t=\sqrt{x}\), \(\dfrac{dt}{dx}=\dfrac{1}{2\sqrt{x}}\).
All derivative factors must be multiplied together.
Temporary variables must be substituted back before presenting the final answer.
The final derivative is:
\[ \boxed{ \frac{ \sec\left(\tan\sqrt{x}\right) \tan\left(\tan\sqrt{x}\right) \sec^2\sqrt{x} }{ 2\sqrt{x} } } \]

📘 Concept & Theory

Concept/Theory

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This problem combines two important rules of differentiation: the Quotient Rule and the Chain Rule. The numerator and denominator are both composite trigonometric functions.

For two differentiable functions \(f(x)\) and \(g(x)\), the quotient rule is

\[ \frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}, \qquad g(x)\neq 0. \]

Also, for a composite function \(F(u)\), where \(u=u(x)\), the chain rule gives

\[ \frac{d}{dx}F(u) = \frac{dF}{du}\cdot\frac{du}{dx}. \]

Here, we take

\[ u=ax+b, \qquad v=cx+d. \]

Then

\[ y=\frac{\sin u}{\cos v}. \]

While applying the quotient rule, the derivatives of \(\sin u\) and \(\cos v\) must themselves be found using the chain rule.

🗺️ Solution Roadmap

Step-by-step Plan

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Take \(u=ax+b\) and \(v=cx+d\).
Rewrite the function as \(y=\dfrac{\sin u}{\cos v}\).
Apply the quotient rule.
Use the chain rule to differentiate \(\sin u\) with respect to \(x\).
Use the chain rule to differentiate \(\cos v\) with respect to \(x\).
Substitute these derivatives into the quotient-rule formula.
Replace \(u\) and \(v\) by \(ax+b\) and \(cx+d\).
Simplify the result using \(\sec\theta=\dfrac{1}{\cos\theta}\) and \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 24 steps

Let
\[u=ax+b\]
and
\[v=cx+d\]
Therefore,
\[y=\frac{\sin u}{\cos v}.\]
Using the quotient rule,
\[ \frac{dy}{dx} = \frac{ \cos v\cdot\dfrac{d}{dx}(\sin u) - \sin u\cdot\dfrac{d}{dx}(\cos v) }{ (\cos v)^2 }. \]
We now differentiate the numerator function \(\sin u\) using the chain rule:
\[\frac{d}{dx}(\sin u)=\frac{d}{du}(\sin u)\cdot\frac{du}{dx}\]
Since
\[\frac{d}{du}(\sin u)=\cos u\]
and
\[\frac{du}{dx}=\frac{d}{dx}(ax+b)=a\]
we obtain
\[\frac{d}{dx}(\sin u)=a\cos u\]
Similarly, differentiate the denominator function \(\cos v\):
\[\frac{d}{dx}(\cos v)=\frac{d}{dv}(\cos v)\cdot\frac{dv}{dx}\]
Now,
\[\frac{d}{dv}(\cos v)=-\sin v\]
and
\[\frac{dv}{dx}=\frac{d}{dx}(cx+d)=c\]
Therefore,
\[\frac{d}{dx}(\cos v)=-c\sin v\]
Substituting these derivatives into the quotient-rule formula,
\[ \frac{dy}{dx} = \frac{ \cos v(a\cos u) - \sin u(-c\sin v) }{ (\cos v)^2 }. \]
Because the second term contains two negative signs, it becomes positive:
\[ \frac{dy}{dx} = \frac{ a\cos u\cos v + c\sin u\sin v }{ \cos^2v }. \]
Now substitute
\[u=ax+b\]
and
\[v=cx+d\]
Hence,
\[ \frac{dy}{dx} = \frac{ a\cos(ax+b)\cos(cx+d) + c\sin(ax+b)\sin(cx+d) }{ \cos^2(cx+d) }. \]
This is already a correct form of the derivative.
Simplification of the Result
To obtain an alternative form, split the fraction into two terms:
\[\frac{dy}{dx}=\frac{a\cos(ax+b)\cos(cx+d)}{\cos^2(cx+d)}+\frac{c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}\]
For the first term, cancel one factor of \(\cos(cx+d)\):
\[\frac{a\cos(ax+b)\cos(cx+d)}{\cos^2(cx+d)}=\frac{a\cos(ax+b)}{\cos(cx+d)}\]
Using
\[\sec\theta=\frac{1}{\cos\theta}\]
we get
\[\frac{a\cos(ax+b)}{\cos(cx+d)}=a\cos(ax+b)\sec(cx+d).\]
For the second term, write
\[\frac{c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}\]
as
\[c\sin(ax+b)\left(\frac{\sin(cx+d)}{\cos(cx+d)}\right)\left(\frac{1}{\cos(cx+d)}\right)\]
Using
\[\tan\theta=\frac{\sin\theta}{\cos\theta}\]
and
\[\sec\theta=\frac{1}{\cos\theta}\]
we obtain
\[c\sin(ax+b)\tan(cx+d)\sec(cx+d)\]
Therefore,
\[\boxed{\frac{dy}{dx}=a\cos(ax+b)\sec(cx+d)+c\sin(ax+b)\tan(cx+d)\sec(cx+d)}\]
Equivalently, taking \(\sec(cx+d)\) as a common factor,
\[\boxed{\frac{dy}{dx}=\sec(cx+d)\left[a\cos(ax+b)+c\sin(ax+b)\tan(cx+d)\right]}\]

💡 Answer

Final Answer

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The derivative may be written in quotient form as

\[\boxed{\frac{d}{dx}\left[\frac{\sin(ax+b)}{\cos(cx+d)}\right]=\frac{a\cos(ax+b)\cos(cx+d)+c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}}\]

or, equivalently, in trigonometric-function form as

\[\boxed{\frac{d}{dx}\left[\frac{\sin(ax+b)}{\cos(cx+d)}\right]=\sec(cx+d)\left[a\cos(ax+b)+c\sin(ax+b)\tan(cx+d)\right]}\]

🎯 Exam Significance

Exam Significance

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This question is important because it tests three skills simultaneously: the quotient rule, the chain rule, and the simplification of trigonometric expressions. In board examinations, writing the quotient-rule formula first and then differentiating the numerator and denominator separately provides a systematic and reliable solution.

Special attention should be paid to the sign of the derivative of cosine:

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

Since the quotient rule itself contains a subtraction, differentiating the cosine denominator creates two negative signs, which produce the positive term in the numerator.

Significance for Competitive Entrance Examination Aspirants

For competitive examinations, this problem demonstrates the value of choosing an efficient representation before differentiating. Although the quotient rule works directly, rewriting

\[ \frac{1}{\cos(cx+d)} = \sec(cx+d) \]

converts the function into

\[ y=\sin(ax+b)\sec(cx+d), \]

which can then be differentiated using the product rule and chain rule. Recognising such transformations can reduce algebraic complexity and lower the risk of sign errors in time-bound examinations.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

The problem requires the quotient rule together with the chain rule.
The derivative of \(\sin(ax+b)\) is \(a\cos(ax+b)\).
The derivative of \(\cos(cx+d)\) is \(-c\sin(cx+d)\).
Be particularly careful with the two negative signs produced while differentiating the denominator.
The parameters \(a\), \(b\), \(c\), and \(d\) are treated as constants.
The expression can alternatively be rewritten as \(\sin(ax+b)\sec(cx+d)\) and differentiated using the product rule.
Both forms of the final answer are mathematically equivalent.

📘 Concept & Theory

Concept/Theory

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This problem involves the Product Rule together with the Chain Rule. The given expression is the product of two functions:

\[ y=\cos(x^3)\cdot\sin(x^5). \]

For two differentiable functions \(f(x)\) and \(g(x)\), the product rule states that

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

Both factors are also composite functions. Therefore, the chain rule is required when differentiating \(\cos(x^3)\) and \(\sin(x^5)\).

The relevant chain-rule results are

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

and

\[ \frac{d}{dx}\sin(x^5) = \cos(x^5)\cdot 5x^4. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the two factors \(f(x)=\cos(x^3)\) and \(g(x)=\sin(x^5)\).
Apply the product rule.
Differentiate \(\sin(x^5)\) using the chain rule.
Differentiate \(\cos(x^3)\) using the chain rule.
Substitute both derivatives into the product-rule formula.
Simplify the resulting expression.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 18 steps

Let
\[y=\cos(x^3)\cdot\sin(x^5)\]
We have a product of two functions. Therefore, using the product rule,
\[ \frac{dy}{dx} = \cos(x^3)\frac{d}{dx}\left[\sin(x^5)\right] + \sin(x^5)\frac{d}{dx}\left[\cos(x^3)\right]. \]
Differentiate \(\sin(x^5)\)
Let
\[u=x^5\]
Then
\[\sin(x^5)=\sin u\]
Using the chain rule,
\[ \frac{d}{dx}(\sin u) = \frac{d}{du}(\sin u)\cdot\frac{du}{dx}. \]
Now,
\[\frac{d}{du}(\sin u)=\cos u\]
and
\[\frac{du}{dx}=\frac{d}{dx}(x^5)=5x^4\]
Therefore,
\[\frac{d}{dx}\left[\sin(x^5)\right]=\cos(x^5)\cdot 5x^4\]
Hence,
\[\boxed{\frac{d}{dx}\left[\sin(x^5)\right]=5x^4\cos(x^5)}\]
Differentiate \(\cos(x^3)\)
Let
\[v=x^3\]
Then
\[\cos(x^3)=\cos v\]
Again, using the chain rule,
\[\frac{d}{dx}(\cos v)=\frac{d}{dv}(\cos v)\cdot\frac{dv}{dx}\]
Now,
\[\frac{d}{dv}(\cos v)=-\sin v\]
and
\[\frac{dv}{dx}=\frac{d}{dx}(x^3)=3x^2\]
Therefore,
\[\frac{d}{dx}\left[\cos(x^3)\right]=-\sin(x^3)\cdot3x^2\]
Hence,
\[\boxed{\frac{d}{dx}\left[\cos(x^3)\right]=-3x^2\sin(x^3)}\]
Apply the Product Rule
We previously obtained
\[\frac{dy}{dx}=\cos(x^3)\frac{d}{dx}\left[\sin(x^5)\right]+\sin(x^5)\frac{d}{dx}\left[\cos(x^3)\right]\]
Substituting the derivatives found above,
\[\frac{dy}{dx}=\cos(x^3)\left[5x^4\cos(x^5)\right]+\sin(x^5)\left[-3x^2\sin(x^3)\right]\]
Therefore,
\[\frac{dy}{dx}=5x^4\cos(x^3)\cos(x^5)-3x^2\sin(x^3)\sin(x^5)\]
Hence, the required derivative is
\[\boxed{\frac{dy}{dx}=5x^4\cos(x^3)\cos(x^5)-3x^2\sin(x^3)\sin(x^5)}\]

💡 Answer

Final Answer

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\[ \boxed{ \frac{d}{dx} \left[ \cos(x^3)\sin(x^5) \right] = 5x^4\cos(x^3)\cos(x^5) - 3x^2\sin(x^3)\sin(x^5) } \]

🎯 Exam Significance

Exam Significance

›

This problem is particularly useful for competitive examinations because it combines multiple differentiation rules in a compact expression. Students should quickly identify the structure:

\[ \underbrace{\cos(x^3)}_{\text{first factor}} \cdot \underbrace{\sin(x^5)}_{\text{second factor}}. \]

The derivative of each factor contains an additional power of \(x\) from the chain rule:

\[ \cos(x^3)\longrightarrow -3x^2\sin(x^3) \]

and

\[ \sin(x^5)\longrightarrow 5x^4\cos(x^5). \]

Recognising these patterns quickly is valuable in time-bound entrance examinations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Use the Product Rule when two functions are multiplied.
Use the Chain Rule when the argument of a function is itself a function of \(x\).
For \(\sin(x^5)\), the derivative is \(5x^4\cos(x^5)\).
For \(\cos(x^3)\), the derivative is \(-3x^2\sin(x^3)\).
Do not introduce a square on \(\sin(x^5)\); the original function contains only \(\sin(x^5)\).
When applying the product rule, differentiate both factors and preserve the other factor in each term.
The final derivative is:

📘 Concept & Theory

Concept/Theory

›

This problem requires the Chain Rule to be applied through two nested layers:

\[x\longrightarrow x^2\longrightarrow\cot(x^2)\longrightarrow\sqrt{\cot(x^2)}\]

The outer function is a square-root function, while the inner function is \(\cot(x^2)\). The derivative of the inner cotangent function itself requires another application of the chain rule.

The relevant derivative formulas are

\[\frac{d}{du}(\sqrt{u})=\frac{1}{2\sqrt{u}},\]

\[\frac{d}{dv}(\cot v)=-\operatorname{cosec}^2v,\]

and

\[ \frac{d}{dx}(x^2)=2x. \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Let \(u=\cot(x^2)\).
Rewrite the function as \(y=2\sqrt{u}\).
Differentiate \(y\) with respect to \(u\).
Differentiate \(u=\cot(x^2)\) with respect to \(x\), applying the chain rule again.
Multiply the derivatives.
Substitute \(u=\cot(x^2)\) back into the answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 18 steps

Let
\[u=\cot(x^2)\]
Then the given function becomes
\[y=2\sqrt{u}\]
Using the chain rule,
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
Differentiate \(y=2\sqrt{u}\)
Write the square root in index form:
\[y=2u^{1/2}\]
Therefore, using the power rule,
\[\frac{dy}{du}=2\cdot\frac{1}{2}u^{-1/2}\]
Hence,
\[\frac{dy}{du}=u^{-1/2}\]
Since
\[u^{-1/2}=\frac{1}{\sqrt{u}}\]
we obtain
\[\boxed{\frac{dy}{du}=\frac{1}{\sqrt{u}}}\]
Differentiate \(u=\cot(x^2)\)
Let
\[v=x^2\]
Then
\[u=\cot v\]
Using the chain rule,
\[\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}\]
Now,
\[\frac{du}{dv}=\frac{d}{dv}(\cot v)=-\operatorname{cosec}^2v\]
and
\[\frac{dv}{dx}=\frac{d}{dx}(x^2)=2x\]
Therefore,
\[\frac{du}{dx}=-\operatorname{cosec}^2v\cdot2x\]
Substituting \(v=x^2\),
\[\boxed{\frac{du}{dx}=-2x\operatorname{cosec}^2(x^2)}\]
Apply the Chain Rule
We have
\[\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\]
Substituting the two derivatives,
\[\frac{dy}{dx}=\frac{1}{\sqrt{u}}\left[-2x\operatorname{cosec}^2(x^2)\right]\]
Thus,
\[\frac{dy}{dx}=-\frac{2x\operatorname{cosec}^2(x^2)}{\sqrt{u}}\]
Finally, since
\[u=\cot(x^2)\]
we obtain
\[\boxed{\frac{dy}{dx}=-\frac{2x\operatorname{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}}\]

💡 Answer

Final Answer

›

\[\boxed{\frac{d}{dx}\left[2\sqrt{\cot(x^2)}\right]=-\frac{2x\operatorname{cosec}^2(x^2)}{\sqrt{\cot(x^2)}}}\]

🎯 Exam Significance

Exam Significance

›

This question is a useful example of successive application of the chain rule. Students must differentiate both the square-root function and the cotangent function correctly.

The most important point is that the derivative of the inner expression \(\cot(x^2)\) contains two factors:

\[\frac{d}{dx}\cot(x^2)=\frac{d}{d(x^2)}(\cot(x^2))\cdot\frac{d}{dx}(x^2)\]

Therefore,

\[\frac{d}{dx}\cot(x^2)=-\operatorname{cosec}^2(x^2)\cdot2x\]

Writing this intermediate step explicitly is particularly helpful in board examinations because it shows that the chain rule has been applied completely.

Significance for Competitive Entrance Examination Aspirants

This example trains students to recognise nested functions quickly. The expression contains three functional layers:

\[x^2,\quad\cot(x^2),\quad 2\sqrt{\cot(x^2)}\]

The derivative is obtained by multiplying the derivative contributed by each layer:

\[\frac{1}{\sqrt{\cot(x^2)}}\cdot\left[-\operatorname{cosec}^2(x^2)\right]\cdot2x\]

This layered approach is particularly useful in JEE and other competitive examinations, where complicated composite functions are often designed to test whether any chain-rule factor has been missed.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Identify every layer of a composite function before differentiating.
The derivative of \(2\sqrt{u}\) is \(\dfrac{1}{\sqrt{u}}\).
The derivative of \(\cot u\) is \(-\operatorname{cosec}^2u\).
The derivative of \(x^2\) is \(2x\).
The argument \(x^2\) must be retained correctly inside \(\operatorname{cosec}^2(x^2)\).
When several functions are nested, multiply the derivative of each successive layer.

📘 Concept & Theory

Concept/Theory

›

This problem is a straightforward application of the Chain Rule of Differentiation. The expression \(\cos\sqrt{x}\) is a composite function because the argument of the cosine function is \(\sqrt{x}\).

If

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

then the chain rule states that

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

Here,

\[ t=\sqrt{x}. \]

Therefore, we first differentiate \(\cos t\) with respect to \(t\), and then multiply by the derivative of \(\sqrt{x}\) with respect to \(x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Take \(t=\sqrt{x}\) as the inner function.
Rewrite the given function as \(y=\cos t\).
Find \(\dfrac{dy}{dt}\).
Find \(\dfrac{dt}{dx}\).
Apply the chain rule.
Substitute \(t=\sqrt{x}\) into the final result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Let
\[t=\sqrt{x}\]
Then the given function becomes
\[y=\cos t\]
Using the chain rule,
\[\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}\]
Differentiate \(y=\cos t\)
Differentiating with respect to \(t\),
\[\frac{dy}{dt}=\frac{d}{dt}(\cos t)=-\sin t\]
Differentiate \(t=\sqrt{x}\)
Write the square root in index form:
\[t=x^{1/2}\]
Using the power rule,
\[\frac{dt}{dx}=\frac{d}{dx}\left(x^{1/2}\right)=\frac{1}{2}x^{-1/2}\]
Since
\[x^{-1/2}=\frac{1}{\sqrt{x}}\]
we get
\[\frac{dt}{dx}=\frac{1}{2\sqrt{x}}\]
Apply the Chain Rule
Substituting the two derivatives into the chain-rule formula,
\[\frac{dy}{dx}=(-\sin t)\left(\frac{1}{2\sqrt{x}}\right)\]
Therefore,
\[\frac{dy}{dx}=-\frac{\sin t}{2\sqrt{x}}\]
Finally, substituting \(t=\sqrt{x}\),
\[\boxed{\frac{dy}{dx}=-\frac{\sin\sqrt{x}}{2\sqrt{x}}}\]

💡 Answer

Final Answer

›

\[\boxed{\frac{d}{dx}\left(\cos\sqrt{x}\right)=-\frac{\sin\sqrt{x}}{2\sqrt{x}}}\]

🎯 Exam Significance

Exam Significance

›

This is a basic but important chain-rule problem. It demonstrates that when a function such as \(\cos x\) contains a non-trivial expression \(\sqrt{x}\) as its argument, the derivative of that inner expression must also be included.

A common mistake is to write

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

which omits the derivative of \(\sqrt{x}\). The correct application is

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

Writing the chain-rule step explicitly is therefore useful in board examinations.

Significance for Competitive Entrance Examination Aspirants

This problem develops the ability to recognise a composite function immediately. In competitive examinations, the inner function may be an algebraic, trigonometric, exponential, or logarithmic expression. The same principle applies in every case.

Here the differentiation can be viewed as three simple steps:

\[\sqrt{x}\longrightarrow\cos(\sqrt{x})\longrightarrow -\sin(\sqrt{x})\]

with the derivative of the inner function providing the additional factor

\[ \frac{1}{2\sqrt{x}}. \]

Recognising this pattern quickly helps reduce calculation time and prevents omission of chain-rule factors.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

\(\cos\sqrt{x}\) is a composite function.
Use the chain rule for composite functions.
The derivative of \(\cos t\) is \(-\sin t\).
The derivative of \(\sqrt{x}\) is \(\dfrac{1}{2\sqrt{x}}\).
The derivative of the inner function must be multiplied by the derivative of the outer function.
Writing \(\sqrt{x}=x^{1/2}\) makes its differentiation using the power rule explicit.
The final derivative is:

📘 Concept & Theory

Concept/Theory

›

A function \(f(x)\) is differentiable at \(x=a\) if and only if its left-hand derivative and right-hand derivative at \(x=a\) exist and are equal.

The left-hand derivative is

\[f'_-(a)=\lim_{h\to0^-}\frac{f(a+h)-f(a)}{h}\]

and the right-hand derivative is

\[f'_+(a)=\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}\]

Therefore, to prove that \(f(x)\) is not differentiable at \(x=1\), it is sufficient to show that

\[f'_-(1)\neq f'_+(1)\]

For the modulus function, it is also useful to remember that

\[|x-1|=\begin{cases}1-x, & x<1,\\x-1, & x\geq1\end{cases} \]

Thus, the function has different algebraic forms on the two sides of \(x=1\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Find \(f(1)\).
Express \(f(x)=|x-1|\) in piecewise form on either side of \(x=1\).
Calculate the left-hand derivative at \(x=1\).
Calculate the right-hand derivative at \(x=1\).
Compare the two derivatives.
Since they are unequal, conclude that \(f\) is not differentiable at \(x=1\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 28 steps

Given —
\[f(x)=|x-1|\]
We have to determine whether \(f\) is differentiable at \(x=1\).
Find \(f(1)\)
Substituting \(x=1\),
\[f(1)=|1-1|=0\]
Express the Function in Piecewise Form
Recall that
\[|a|=\begin{cases}-a, & a<0,\\a, & a\geq0\end{cases}\]
Here,
\[a=x-1\]
When \(x<1\), we have \(x-1<0\). Therefore,
\[|x-1|=-(x-1)=1-x\]
When \(x\geq1\), we have \(x-1\geq0\). Therefore,
\[|x-1|=x-1\]
Hence,
\[f(x)=\begin{cases}1-x, & x<1,\\x-1, & x\geq1\end{cases}\]
Find the Left-Hand Derivative at \(x=1\)
For \(x<1\),
\[f(x)=1-x\]
Therefore, the derivative from the left is
\[f'_-(1)=\frac{d}{dx}(1-x)\]
Hence,
\[f'_-(1)=-1\]
Alternatively, using the definition of the left-hand derivative,
\[f'_-(1)=\lim_{h\to0^-}\frac{f(1+h)-f(1)}{h}\]
Since \(h<0\), we have \(1+h<1\). Thus,
\[f(1+h)=|(1+h)-1|=|h|=-h\]
Also,
\[f(1)=0\]
Therefore,
\[f'_-(1)=\lim_{h\to0^-}\frac{-h-0}{h}\]

\[=\lim_{h\to0^-}(-1)\]

\[=-1\]
Thus,
\[\boxed{f'_-(1)=-1}\]
Find the Right-Hand Derivative at \(x=1\)
For \(x>1\),
\[f(x)=x-1\]
Therefore, the derivative from the right is
\[f'_+(1)=\frac{d}{dx}(x-1)\]
Hence,
\[f'_+(1)=1\]
Alternatively, using the definition of the right-hand derivative,
\[f'_+(1)=\lim_{h\to0^+}\frac{f(1+h)-f(1)}{h}\]
Since \(h>0\), we have \(1+h>1\). Thus,
\[f(1+h)=|(1+h)-1|=|h|=h\]
Also,
\[f(1)=0\]
Therefore,
\[f'_+(1)=\lim_{h\to0^+}\frac{h-0}{h}\]

\[=\lim_{h\to0^+}(1)\]

\[=1\]
Thus,
\[\boxed{f'_+(1)=1}\]
Compare the Left-Hand and Right-Hand Derivatives
We have obtained
\[f'_-(1)=-1\]
and
\[f'_+(1)=1\]
Therefore,
\[f'_-(1)\neq f'_+(1)\]
Hence, the left-hand derivative and right-hand derivative are not equal at \(x=1\).
Therefore, the derivative \(f'(1)\) does not exist.
\[\boxed{\text{Hence, }f(x)=|x-1|\text{ is not differentiable at }x=1.}\]

🎯 Exam Significance

Exam Significance

›

This is a standard conceptual problem from Continuity and Differentiability. It tests whether students understand that differentiability at a point requires equality of the left-hand and right-hand derivatives.

For a board examination, the safest approach is to calculate both one-sided derivatives explicitly:

\[ f'_-(1)=-1, \qquad f'_+(1)=1. \]

Since these are unequal, the function is not differentiable at \(x=1\).

Students should also remember that the function is continuous at \(x=1\). Therefore, this example clearly demonstrates that continuity alone is not sufficient for differentiability.

Significance for Competitive Entrance Examination Aspirants

This example is important for understanding corner points and one-sided derivatives. Competitive examinations frequently use modulus functions to test differentiability at points where the expression changes its algebraic form.

A quick way to recognise the issue is that

\[ |x-1| \]

changes from \(1-x\) for \(x<1\) to \(x-1\) for \(x>1\). Consequently, the slopes on the two sides are different:

\[ -1\quad\text{and}\quad+1. \]

This immediately indicates a sharp corner at \(x=1\), so the derivative does not exist there.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

For differentiability at \(x=a\), the left-hand and right-hand derivatives must be equal.
The modulus function must often be expressed in piecewise form before analysing differentiability.
For \(x<1\), \(|x-1|=1-x\).
For \(x>1\), \(|x-1|=x-1\).
The left-hand derivative at \(x=1\) is \(-1\).
The right-hand derivative at \(x=1\) is \(1\).
Since \(-1\neq1\), \(f'(1)\) does not exist.
The function is continuous at \(x=1\), but not differentiable there.
Geometrically, the failure of differentiability corresponds to the sharp corner of the V-shaped graph.

📘 Concept & Theory

Concept/Theory

›

The greatest integer function \([x]\), also written as \(\lfloor x\rfloor\), represents the greatest integer less than or equal to \(x\).

On the interval \(0 < x < 3\), its value changes at the integers \(1\) and \(2\). Consequently, the function can be represented as

\[f(x)=\begin{cases}0, & 0 < x < 1,\\1, & 1\leq x<2,\\2, & 2\leq x<3\end{cases}\]

A fundamental result from continuity and differentiability is

\[ \boxed{\text{If a function is differentiable at a point, then it must be continuous there.}} \]

Thus, showing that \(f\) is discontinuous at \(x=1\) and \(x=2\) is already sufficient to prove that it is not differentiable at those points. We will also verify the result directly using the left-hand and right-hand derivatives.

🗺️ Solution Roadmap

Step-by-step Plan

›

First, identify the values of the greatest integer function on the three subintervals. Next, analyse the behaviour of the function immediately to the left and right of \(x=1\), and then repeat the same procedure at \(x=2\). Finally, compare the one-sided derivatives and use the continuity criterion as an additional verification.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 31 steps

Write the Greatest Integer Function in Piecewise Form
For \(0 < x < 1\), the greatest integer less than or equal to \(x\) is \(0\). Hence,
\[f(x)=0,\qquad 0 < x < 1\]
For \(1\leq x < 2\), the greatest integer is \(1\), so
\[f(x)=1,\qquad 1\leq x < 2\]
For \(2\leq x < 3\), the greatest integer is \(2\), giving
\[f(x)=2,\quad 2\leq x < 3\]
Therefore,
\[\boxed{f(x)=\begin{cases}0, & 0 < x < 1,\\1, & 1\leq x < 2,\\2, & 2\leq x < 3\end{cases}}\]
Part I: Differentiability at \(x=1\)
Determine \(f(1)\)
Since \([1]=1\),
\[f(1)=[1]=1\]
Calculate the Left-Hand Derivative at \(x=1\)
The left-hand derivative, if finite, is given by
\[f'_-(1)=\lim_{h\to0^-}\frac{f(1+h)-f(1)}{h}\]
When \(h\to0^-\), we have \(h<0\), and hence
\[0 < 1+h < 1\]
From the piecewise definition,
\[f(1+h)=[1+h]=0\]
Since \(f(1)=1\), the difference quotient becomes
\[\frac{f(1+h)-f(1)}{h}=\frac{0-1}{h}=-\frac{1}{h}\]
Consequently,
\[f'_-(1)=\lim_{h\to0^-}\left(-\frac{1}{h}\right)\]
As \(h\) approaches \(0\) through negative values, \(-1/h\) becomes arbitrarily large and positive. Therefore, the left-hand derivative does not exist as a finite real number:
\[\boxed{f'_-(1)\text{ does not exist as a finite number}}\]
Calculate the Right-Hand Derivative at \(x=1\)
Now consider \(h\to0^+\). Then \(h>0\), so for sufficiently small \(h\),
\[1 < 1+h < 2\]
Hence,
\[f(1+h)=[1+h]=1\]
Therefore,
\[\frac{f(1+h)-f(1)}{h}=\frac{1-1}{h}=0\]
Taking the right-hand limit gives
\[f'_+(1)=\lim_{h\to0^+}0=0\]
Thus,
\[\boxed{f'_+(1)=0}\]
Conclusion at \(x=1\)
At \(x=1\), the left-hand derivative is not finite, whereas the right-hand derivative equals \(0\). Hence there cannot be a common derivative at this point.
\[\boxed{\therefore f(x)=[x]\text{ is not differentiable at }x=1}\]
Part II: Differentiability at \(x=2\)
Determine \(f(2)\)
Since \([2]=2\),
\[f(2)=[2]=2\]
Calculate the Left-Hand Derivative at \(x=2\)
The left-hand derivative is given by
\[f'_-(2)=\lim_{h\to0^-}\frac{f(2+h)-f(2)}{h}\]
For \(h<0\) sufficiently close to \(0\),
\[1 < 2+h < 2\]
Therefore,
\[f(2+h)=[2+h]=1\]
Using \(f(2)=2\), we obtain
\[\frac{f(2+h)-f(2)}{h}=\frac{1-2}{h}=-\frac{1}{h}\]
Hence,
\[f'_-(2)=\lim_{h\to0^-}\left(-\frac{1}{h}\right)\]
Again, as \(h\to0^-\),
\[-\frac{1}{h}\to+\infty\]
Therefore, the left-hand derivative does not exist as a finite real number:
\[\boxed{f'_-(2)\text{ does not exist as a finite number}.}\]
Calculate the Right-Hand Derivative at \(x=2\)
For the right-hand derivative, let \(h\to0^+\). Then
\[2 < 2+h < 3\]
Consequently,
\[f(2+h)=[2+h]=2\]
Since \(f(2)=2\),
\[\frac{f(2+h)-f(2)}{h}=\frac{2-2}{h}=0\]
Thus,
\[f'_+(2)=\lim_{h\to0^+}0=0\]
Hence,
\[\boxed{f'_+(2)=0}\]
Conclusion at \(x=2\)
The left-hand derivative is not finite, while the right-hand derivative is \(0\). Therefore, \(f'(2)\) does not exist.
\[\boxed{\therefore f(x)=[x]\text{ is not differentiable at }x=2}\]

🎯 Exam Significance

Exam Significance

›

This question is a useful test of the connection between continuity and differentiability. The key theorem is that every differentiable function is necessarily continuous, although the converse is not true.

For a board examination, the most efficient proof is to demonstrate that the greatest integer function is discontinuous at the specified points. A more detailed solution can calculate the one-sided derivatives directly.

At \(x=1\), the relevant values are

\[ f'_+(1)=0, \qquad f'_-(1)\text{ is not finite}, \]

and at \(x=2\),

\[ f'_+(2)=0, \qquad f'_-(2)\text{ is not finite}. \]

These calculations make the failure of differentiability explicit.

Significance for Competitive Entrance Examination Aspirants

The greatest integer function is particularly important in problems involving discontinuity, one-sided limits, and differentiability. Between consecutive integers, \([x]\) remains constant, but at every integer it jumps to a new value.

For the present domain \(0

\[ x=1 \]

and

\[ x=2. \]

Recognising this step-function structure can often provide a much faster route to the conclusion than carrying out lengthy differentiation calculations.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The greatest integer function gives the greatest integer less than or equal to \(x\).
For \(0\[ f(x)= \begin{cases} 0, & 0
The function has jump discontinuities at \(x=1\) and \(x=2\).
At \(x=1\), the right-hand derivative is \(0\), while the left-hand difference quotient tends to \(+\infty\).
At \(x=2\), the right-hand derivative is \(0\), while the left-hand difference quotient tends to \(+\infty\).
Since the required one-sided derivatives do not give a common finite value, the function is not differentiable at either point.
The quickest conceptual proof uses the theorem that differentiability implies continuity.

Continuity and Differentiability

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

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 Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Significance for Competitive Entrance Examination Aspirants

Chapter Complete!

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CONTINUITY AND DIFFERENTIABILITY — Learning Resources

Frequently Asked Questions

Q 01 What is covered in NCERT Class 12 Maths Exercise 5.2?

Q 02 What is the Chain Rule of Differentiation?

Q 03 How do you differentiate sin(ax+b)?

Q 04 How do you differentiate a function containing several nested functions?

Q 05 How do you differentiate a product of two functions?

Q 06 How do you differentiate a quotient of two functions?

Q 07 Why is the function |x-1| not differentiable at x=1?

Q 08 Why is the greatest integer function [x] not differentiable at x=1 and x=2?

Q 09 Is every continuous function differentiable?

Q 10 Why are NCERT Exercise 5.2 solutions important for JEE preparation?

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