Differentiation Using the Chain Rule | NCERT Class 12 Mathematics

Quick Facts

• Exercise 5.2 focuses mainly on differentiation of composite functions and differentiability.
• The Chain Rule is used when one function is inside another function.
• If $y=f(u)$ and $u=g(x)$, then:

$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$$

• Standard derivatives frequently used in this exercise:

$$\frac{d}{dx}(\sin x)=\cos x$$$$\frac{d}{dx}(\cos x)=-\sin x$$$$\frac{d}{dx}(\tan x)=\sec^2x$$$$\frac{d}{dx}(\cot x)=-\csc^2x$$$$\frac{d}{dx}(\sec x)=\sec x\tan x$$$$\frac{d}{dx}(x^n)=nx^{n-1}$$

• A function can be continuous but not differentiable at a point, as illustrated by modulus and greatest integer functions in this exercise.

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Story

Imagine a function as a machine. Sometimes $x$ goes directly into the machine:$$y=\sin x$$

But sometimes $x$ first enters one machine and its output enters another:$$x\rightarrow x^2+5\rightarrow \sin(x^2+5)$$

Here, the function is composite. To differentiate it, we need to follow the same journey in reverse.

For example,$$y=\sin(x^2+5)$$

First differentiate the outside function:$$\cos(x^2+5)$$

Then multiply by the derivative of the inside function:$$2x$$

Therefore,$$\boxed{\frac{dy}{dx}=2x\cos(x^2+5)}$$

This step-by-step process is the Chain Rule, which the textbook introduces for composite functions.

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Key Terms

Term	Meaning
Composite Function	A function formed by applying one function to another
Chain Rule	Rule used to differentiate composite functions
Inner Function	The function inside another function
Outer Function	The function acting on the inner function
Differentiability	Existence of a finite derivative at a point
Left-hand Derivative	Derivative obtained by approaching a point from the left
Right-hand Derivative	Derivative obtained by approaching a point from the right
Greatest Integer Function	$[x]$, the greatest integer less than or equal to $x$
Modulus Function	(

The textbook states that a function is differentiable at a point when the relevant left- and right-hand derivative limits are finite and equal.

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Questions – Detailed Solutions

The PDF contains 10 questions in Exercise 5.2: Questions 1–8 ask for differentiation, while Questions 9–10 ask for proofs of non-differentiability.

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Question 1

Differentiate:

$$y=\sin(x^2+5)$$

Solution

Let$$u=x^2+5$$

Then,$$y=\sin u$$

Using the Chain Rule,$$\frac{dy}{dx}=\cos u\frac{du}{dx}$$

Now,$$\frac{du}{dx}=2x$$

Therefore,$$\boxed{\frac{dy}{dx}=2x\cos(x^2+5)}$$

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Question 2

Differentiate:

$$y=\cos(\sin x)$$

Let$$u=\sin x$$

Then$$y=\cos u$$

Therefore,$$\frac{dy}{dx}=-\sin u\frac{du}{dx}$$

Since$$\frac{du}{dx}=\cos x$$

we get$$\boxed{\frac{dy}{dx}=-\cos x\sin(\sin x)}$$

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Question 3

Differentiate:

$$y=\sin(ax+b)$$

Let$$u=ax+b$$

Then$$y=\sin u$$

Using the Chain Rule,$$\frac{dy}{dx}=\cos u\frac{du}{dx}$$

Since$$\frac{du}{dx}=a$$

we obtain$$\boxed{\frac{dy}{dx}=a\cos(ax+b)}$$

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Question 4

Differentiate:

$$y=\sec(\tan\sqrt{x})$$

This contains three layers:$$x\rightarrow\sqrt{x}\rightarrow\tan(\sqrt{x})\rightarrow\sec(\tan\sqrt{x})$$

Let$$u=\sqrt{x}$$

Then$$v=\tan u$$

and$$y=\sec v$$

Now,$$\frac{dy}{dv}=\sec v\tan v$$$$\frac{dv}{du}=\sec^2u$$

and$$\frac{du}{dx}=\frac{1}{2\sqrt{x}}$$

Therefore,$$\frac{dy}{dx} = \sec v\tan v\cdot\sec^2u\cdot\frac{1}{2\sqrt{x}}$$

Substituting back,$$\boxed{ \frac{dy}{dx} = \frac{\sec(\tan\sqrt{x})\tan(\tan\sqrt{x})\sec^2\sqrt{x}} {2\sqrt{x}} }$$

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Question 5

Differentiate:

$$y=\frac{\sin(ax+b)}{\cos(cx+d)}$$

Use the Quotient Rule:$$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu’-uv’}{v^2}$$

Take$$u=\sin(ax+b)$$

and$$v=\cos(cx+d)$$

Then$$u’=a\cos(ax+b)$$

and$$v’=-c\sin(cx+d)$$

Hence,$$\frac{dy}{dx} = \frac{ \cos(cx+d)\,a\cos(ax+b) – \sin(ax+b)\left[-c\sin(cx+d)\right] } {\cos^2(cx+d)}$$

Therefore,$$\boxed{ \frac{dy}{dx} = \frac{ a\cos(ax+b)\cos(cx+d) +c\sin(ax+b)\sin(cx+d) } {\cos^2(cx+d)} }$$

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Question 6

Differentiate:

$$y=\cos x^3\cdot\sin^2(x^5)$$

Using the Product Rule:$$(uv)’=u’v+uv’$$

Let$$u=\cos(x^3)$$

and$$v=\sin^2(x^5)$$

First,$$u’=-\sin(x^3)\cdot3x^2$$

so$$u’=-3x^2\sin(x^3)$$

For $v$,$$v=[\sin(x^5)]^2$$

Therefore,$$v’=2\sin(x^5)\cos(x^5)\cdot5x^4$$$$v’=10x^4\sin(x^5)\cos(x^5)$$

Using the Product Rule,$$\frac{dy}{dx} = -3x^2\sin(x^3)\sin^2(x^5) + 10x^4\cos(x^3)\sin(x^5)\cos(x^5)$$

Hence,$$\boxed{ \frac{dy}{dx} = -3x^2\sin(x^3)\sin^2(x^5) +10x^4\cos(x^3)\sin(x^5)\cos(x^5) }$$

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Question 7

Differentiate:

$$y=2\sqrt{\cot(x^2)}$$

Rewrite:$$y=2[\cot(x^2)]^{1/2}$$

Using the Chain Rule,$$\frac{dy}{dx} = 2\cdot\frac12[\cot(x^2)]^{-1/2} \frac{d}{dx}[\cot(x^2)]$$

Now,$$\frac{d}{dx}[\cot(x^2)] = -\csc^2(x^2)\cdot2x$$

Therefore,$$\boxed{ \frac{dy}{dx} = -\frac{2x\csc^2(x^2)} {\sqrt{\cot(x^2)}} }$$

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Question 8

Differentiate:

$$y=\cos(\sqrt{x})$$

Let$$u=\sqrt{x}$$

Then$$y=\cos u$$

Therefore,$$\frac{dy}{dx} = -\sin u\frac{du}{dx}$$

Since$$\frac{du}{dx}=\frac{1}{2\sqrt{x}}$$

we get$$\boxed{ \frac{dy}{dx} = -\frac{\sin(\sqrt{x})}{2\sqrt{x}} }$$

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Question 9

Prove that

$$f(x)=|x-1|,\qquad x\in\mathbb R$$

is not differentiable at x=1x=1.

Solution

For $x<1$,$$|x-1|=1-x$$

For $x>1$,$$|x-1|=x-1$$

At $x=1$,$$f(1)=0$$

Left-hand derivative

$$f’_-(1) = \lim_{h\to0^-} \frac{f(1+h)-f(1)}{h}$$

For $h<0$,$$f(1+h)=|h|=-h$$

Therefore,$$f’_-(1) = \lim_{h\to0^-}\frac{-h}{h} =-1$$

Right-hand derivative

For $h>0$,$$f(1+h)=h$$

Hence,$$f’_+(1) = \lim_{h\to0^+}\frac{h}{h} =1$$

Thus,$$f’_-(1)\ne f’_+(1)$$

Therefore the derivative does not exist at $x=1$.$$\boxed{\text{Hence, }f(x)=|x-1|\text{ is not differentiable at }x=1.}$$

This follows the textbook’s treatment of the modulus function: continuity does not necessarily imply differentiability.

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Question 10

Prove that the greatest integer function

$$f(x)=[x],\qquad 0<x<3$$

is not differentiable at $x=1$ and $x=2$.

Solution

The greatest integer function gives the greatest integer less than or equal to $x$.

Around $x=1$:

For $x<1$ and sufficiently close to 1,$$[x]=0$$

For $x>1$ and sufficiently close to 1,$$[x]=1$$

Therefore,$$\lim_{h\to0^-}\frac{f(1+h)-f(1)}{h}$$

and$$\lim_{h\to0^+}\frac{f(1+h)-f(1)}{h}$$

cannot give the same finite value. Hence $f$ is not differentiable at $x=1$.

Similarly, around $x=2$:

For $x<2$,$$[x]=1$$

while for $x>2$,$$[x]=2$$

Thus the left-hand and right-hand derivatives at $x=2$ are not equal.

Therefore,$$\boxed{\text{The greatest integer function is not differentiable at }x=1\text{ and }x=2.}$$

The textbook also explains that the greatest integer function has jumps at integral points, which is why it is discontinuous there.

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Memory Tricks

Trick 1: “Outside × Inside”

For$$y=f(g(x))$$

remember:

Derivative of Outside × Derivative of Inside

Example:$$\frac{d}{dx}\sin(x^2) = \cos(x^2)\times2x$$

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Trick 2: Three-layer functions

For:$$y=\sec(\tan\sqrt{x})$$

read from outside to inside:

SEC → TAN → √

Then differentiate from inside to outside:

√ → TAN → SEC

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Trick 3: Modulus = Corner

A modulus graph often creates a sharp corner.

At a corner:$$\text{LHD}\ne\text{RHD}$$

Therefore, the derivative does not exist.

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Trick 4: Greatest Integer = Jump

Remember:

GI function jumps at integers.

At a jump, differentiability fails.

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Case Study

Case Study: Modelling the Position of a Moving Object

Suppose the position of an object is represented by$$s(t)=\sin(t^2+5)$$

where $t$ represents time.

To determine its instantaneous velocity, differentiate $s(t)$:$$v(t)=\frac{ds}{dt}$$

Using the Chain Rule,$$v(t)=\cos(t^2+5)\cdot2t$$

Therefore,$$\boxed{v(t)=2t\cos(t^2+5)}$$

Questions

1. Which differentiation rule is essential here?

Answer: Chain Rule.

2. What is the inner function?$$t^2+5$$

3. What is the derivative of the inner function?$$2t$$

4. What is the final velocity function?$$\boxed{2t\cos(t^2+5)}$$

This case study demonstrates how the Chain Rule converts a nested mathematical expression into a manageable differentiation process. The textbook explicitly introduces this rule for composite functions.

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Quiz

Test your understanding of Exercise 5.2 with this interactive quiz:

Differentiate y = sin(x² + 5) with respect to x.

1 of 5

A

2x sin(x² + 5)

B

x cos(x² + 5)

C

2x cos(x² + 5)

D

cos(x² + 5)Next

Give feedback

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CTA

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