Exercise-5.2, Class 12th, Maths, Chapter 5, NCERT

Differentiate the functions with respect to x in Exercises 1 to 8.

1. Differentiate: $\sin (x^2+5)$ with respect to $x$.

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Solution

We need to differentiate the function:

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

This is a composite function, so we will apply the Chain Rule

Let:

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

Differentiate both:

$$\frac{dy}{du}=\cos (u)$$$$\frac{du}{dx}=2x$$

Now apply Chain Rule:

$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$
$$\frac{dy}{dx}=\cos (u)\cdot 2x$$

Substitute $u=x^2+5$:

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

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Question 2

Differentiate with respect to $x$:

$$\cos (\sin x)$$

Solution

Given:

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

This is a composite function, where:

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

Now differentiate step-by-step:

Step 1: Differentiate outer function

$$\frac{dy}{du}=-\sin (u)$$

Step 2: Differentiate inner function

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

Step 3: Apply Chain Rule

$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$
$$\frac{dy}{dx}=(-\sin (u))\cdot \cos x$$

Substitute back $u=\sin x$:

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

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Question 3

Differentiate with respect to $x$:

$$\sin (ax+b)$$

Solution

Let:

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

This is again a composite function, so we apply the Chain Rule.

Step 1:

Let the inner function:

$$u=ax+b$$

Then,

$$y=\sin (u)$$

Step 2: Differentiate

$$\frac{dy}{du}=\cos (u)$$
$$\frac{du}{dx}=a$$

Step 3: Apply Chain Rule

$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$
$$\frac{dy}{dx}=\cos (u)\cdot a$$

Substitute $u=ax+b$:

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

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Question 4

Differentiate with respect to $x$:

$$\boxed{{\displaystyle \frac{d}{dx}[\sec (\tan (\sqrt{x}))}}$$

Solution

Let:

$$u=\tan (\sqrt{x})$$

$$y=\sec (u)$$

Differentiate outer function

$$\frac{dy}{du}=\sec (u)\tan (u)$$

Differentiate inner function

$$u=\tan (v),\text{where }v=\sqrt{x}$$

$$\frac{du}{dv}={\sec }^2(v)$$

Differentiate 

$$v=\sqrt{x}$$
$$\frac{dv}{dx}=\frac{1}{2\sqrt{x}}$$

Apply Chain Rule

$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dv}\cdot \frac{dv}{dx}$$
$$\frac{dy}{dx}=\sec (u)\tan (u)\cdot {\sec }^2(v)\cdot \frac{1}{2\sqrt{x}}$$

Substitute $u=\tan (\sqrt{x})$, $v=\sqrt{x}$:

Final Answer

$$\boxed{{\displaystyle \frac{d}{dx}[\sec (\tan (\sqrt{x}))]=\frac{1}{2\sqrt{x}}\sec (\tan (\sqrt{x}))\tan (\tan (\sqrt{x})){\sec }^2(\sqrt{x})}}$$

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Question 5

Differentiate with respect to $x$:

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

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

This is a quotient, so we use the Quotient Rule:

$$\frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$$

where

$$u=\sin (ax+b),v=\cos (cx+d)$$

Step 1: Differentiate $u$

$$\frac{du}{dx}=\cos (ax+b)\cdot a$$

Step 2: Differentiate $v$

$$\frac{dv}{dx}=-\sin (cx+d)\cdot c$$

Step 3: Apply Quotient Rule

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

Simplify the numerator:

$$=\frac{a\cos (ax+b)\cos (cx+d)+c\sin (ax+b)\sin (cx+d)}{{\cos }^2(cx+d)<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;"></span>}$$

Final Answer

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

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Question 6

Differentiate with respect to $x$:

$$\cos (x^3)\cdot {\sin }^2(x^5)$$

Solution

The given function is a product, so we apply the Product Rule:

$$\frac{d}{dx}(u\cdot v)=u^{\mathrm{\prime }}v+u{v}^{\mathrm{\prime }}$$

Let:

$$u=\cos (x^3)$$
$$v={\sin }^2(x^5)$$

Step 1: Differentiate $u=\cos (x^3)$

Using chain rule:

$$u^{\mathrm{\prime }}=-\sin (x^3)\cdot 3x^2$$

$$u^{\mathrm{\prime }}=-3x^2\sin (x^3)$$

Step 2: Differentiate $v={\sin }^2(x^5)$

Rewrite:

$$v=(\sin (x^5){)}^2$$

Let $t=\sin (x^5)$, then $v=t^2$

$$\frac{dv}{dt}=2t=2\sin (x^5)$$

$$\frac{dt}{dx}=\cos (x^5)\cdot 5x^4$$

So,

$$v^{\mathrm{\prime }}=2\sin (x^5)\cdot 5x^4\cos (x^5)$$
$$v^{\mathrm{\prime }}=10x^4\sin (x^5)\cos (x^5)$$

Step 3: Apply Product Rule

$$\frac{dy}{dx}=u^{\mathrm{\prime }}v+u{v}^{\mathrm{\prime }}$$

Substitute:

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

Final Answer

$$\boxed{{\displaystyle \frac{d}{dx}[\cos (x^3){\sin }^2(x^5)]=-3x^2\sin (x^3){\sin }^2(x^5)+10x^4\cos (x^3)\sin (x^5)\cos (x^5)}}$$

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y=2cot(x2)​

Let’s differentiate step by step.

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Question 7

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

Rewrite using exponent form:

$$y=2(\cot (x^2){)}^{1\mathrm{/}2}$$Solution

Let:

$$u=\cot (x^2)$$
$$y=2u^{1\mathrm{/}2}$$

Step 1: Differentiate outer function

$$\frac{dy}{du}=2\cdot \frac{1}{2}{u}^{-1\mathrm{/}2}=u^{-1\mathrm{/}2}$$

Step 2: Differentiate inner function

$$u=\cot (x^2)$$

Derivative of $\cot t$ is $-{\csc }^{2}t$

So, by chain rule:

$$\frac{du}{dx}=-{\csc }^2(x^2)\cdot 2x$$

Apply Chain Rule

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

$$\frac{dy}{dx}=u^{-1\mathrm{/}2}\cdot (-2x{\csc }^2(x^2))$$

Substitute back $u=\cot (x^2)$:

$$\frac{dy}{dx}=-2x{\csc }^2(x^2)\cdot (\cot (x^2){)}^{-1\mathrm{/}2}$$

Rewrite using square root:

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

Final Answer

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

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Question 8

Differentiate with respect to $x$:

$$\cos (\sqrt{x})$$Solution

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

This is a composite function, so we will apply the Chain Rule.

Step 1: Identify inner and outer functions

$$u=\sqrt{x}=x^{1\mathrm{/}2}$$
$$y=\cos (u)$$

Step 2: Differentiate each part

$$\frac{dy}{du}=-\sin (u)$$

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

Step 3: Apply Chain Rule

$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$
$$\frac{dy}{dx}=(-\sin (u))\cdot \frac{1}{2\sqrt{x}}$$

Substitute $u=\sqrt{x}$:

Final Answer

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

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Question 9

Prove that the function

$$f(x)=\mathrm{\mid }x-1\mathrm{\mid },x\in \mathbb{R}$$

is not differentiable at $x=1$.

Solution

First, write the function in piecewise form:

$$f(x)=\{\begin{array}{ll}1-x,& \text{if }x<1\\ x-1,& \text{if }x>1\end{array}$$

Also,

$$f(1)=\mathrm{\mid }1-1\mathrm{\mid }=0$$

To check differentiability at $x=1$, evaluate the left-hand derivative and the right-hand derivative.

Left-hand derivative (LHD) at $x=1$

For $x<1$, $f(x)=1-x$

$$\frac{d}{dx}(1-x)=-1$$So,

$$\text{LHD at }x=1=\underset{h\to 0^{-}}{\lim }\frac{f(1+h)-f(1)}{h}=-1$$

Right-hand derivative (RHD) at $x=1$

For $x>1$, $f(x)=x-1$

$$\frac{d}{dx}(x-1)=1$$

So,

$$\text{RHD at }x=1=\underset{h\to 0^{+}}{\lim }\frac{f(1+h)-f(1)}{h}=1$$

Conclusion

$$\text{LHD}=-1,\text{RHD}=1$$

Since:$$\text{LHD}\mathrm{\ne }\text{RHD}$$

$$\boxed{{\displaystyle \therefore f(x)=\mathrm{\mid }x-1\mathrm{\mid }\text{ is not differentiable at }x=1.}}$$

Reason

The graph of $\mathrm{\mid }x-1\mathrm{\mid }$ has a sharp corner (cusp) at $x=1$, which makes the slope undefined there.

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Question 10

Prove that the greatest integer function defined by

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

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

Solution:

Understanding the function

The function $[x]$ defines the greatest integer less than or equal to $x$.

For $0<x<3$, the function behaves as follows:

$$f(x)=\{\begin{array}{ll}0,& 0<x<1\\ 1,& 1<x<2\\ 2,& 2<x<3\end{array}$$

Continuity & Differentiability at $x=1$

Left-hand limit approaching 1:

For $x<1$, $f(x)=0$

$$\underset{x\to 1^{-}}{\lim }f(x)=0$$

Right-hand limit approaching 1:

For $x>1$, $f(x)=1$

$$\underset{x\to 1^{+}}{\lim }f(x)=1$$

Since

$$\underset{x\to 1^{-}}{\lim }f(x)\mathrm{\ne }\underset{x\to 1^{+}}{\lim }f(x),$$
$$f(x)\text{ is discontinuous at }x=1.$$Differentiability result

A function must be continuous at a point to be differentiable there.

Since $f(x)$ is not continuous at $x=1$, it cannot be differentiable at $x=1$.

$$\boxed{{\displaystyle f(x)\text{ is not differentiable at }x=1.}}$$

Similarly at $x=2$

Left-hand limit approaching 2

For $x<2$, $f(x)=1$

$$\underset{x\to 2^{-}}{\lim }f(x)=1$$

Right-hand limit approaching 2

For $x>2$, $f(x)=2$

$$\underset{x\to 2^{+}}{\lim }f(x)=2$$

Since

$$\underset{x\to 2^{-}}{\lim }f(x)\mathrm{\ne }\underset{x\to 2^{+}}{\lim }f(x),$$
$$f(x)\text{ is discontinuous at }x=2.$$Thus,

$$\boxed{{\displaystyle f(x)\text{ is not differentiable at }x=2.}}$$

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