Exercise-5.4, Class 12th, Maths, Chapter 5, NCERT

Differentiate the following w.r.t. x:

Question 1

Differentiate:

$$y=\frac{e^x}{\sin x}$$Solution

This is a quotient, so use the Quotient Rule:

$$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{u}^{\mathrm{\prime }}-u{v}^{\mathrm{\prime }}}{v^2}$$

Let:

$$u=e^x,v=\sin x$$

Then:

$$u^{\mathrm{\prime }}=e^x,v^{\mathrm{\prime }}=\cos x$$

Apply the rule:

$$\frac{dy}{dx}=\frac{\sin x\cdot e^x-e^x\cdot \cos x}{(\sin x{)}^2}$$

Factor out $e^x$:

$$\frac{dy}{dx}=\frac{e^x(\sin x-\cos x)}{{\sin }^{2}x}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{e^x(\sin x-\cos x)}{{\sin }^{2}x}}}$$

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Question 2

Differentiate w.r.t. $x$:

$$y=e^{{\sin }^{-1}x}$$Solution

Let:

$$y=e^{{\sin }^{-1}x}$$

Use the chain rule:

$$\frac{dy}{dx}=e^{{\sin }^{-1}x}\cdot \frac{d}{dx}({\sin }^{-1}x)$$

We know:

$$\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$$

Therefore:

$$\frac{dy}{dx}=e^{{\sin }^{-1}x}\cdot \frac{1}{\sqrt{1-x^2}}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{e^{{\sin }^{-1}x}}{\sqrt{1-x^2}}}}$$

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Question 3

Differentiate:

$$y=e^{x^3}$$Solution

Use the chain rule:

Let:

$$y=e^u,u=x^3$$

Then:

$$\frac{dy}{du}=e^u,\frac{du}{dx}=3x^2$$

Apply chain rule:

$$\frac{dy}{dx}=e^u\cdot 3x^2$$

Substitute $u=x^3$:

$$\frac{dy}{dx}=3x^2{e}^{x^3}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=3x^2{e}^{x^3}}}$$

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Question 4

Differentiate with respect to $x$:

$$y=\sin ({\tan }^{-1}(e^{-x}))$$Solution

Let:

$$u={\tan }^{-1}(e^{-x})$$

So:

$$y=\sin (u)$$

Step 1: Differentiate outer function

$$\frac{dy}{du}=\cos u$$

Step 2: Differentiate inner function

$$u={\tan }^{-1}(e^{-x})$$

Derivative of ${\tan }^{-1}t$ is $\frac{1}{1+t^2}\cdot t^{\mathrm{\prime }}$

So,

$$\frac{du}{dx}=\frac{1}{1+(e^{-x}{)}^2}\cdot \frac{d}{dx}(e^{-x})$$

$$\frac{du}{dx}=\frac{1}{1+e^{-2x}}\cdot (-e^{-x})$$

Step 3: Apply chain rule

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

$$\frac{dy}{dx}=\cos ({\tan }^{-1}(e^{-x}))\cdot (-\frac{e^{-x}}{1+e^{-2x}})$$

We know the identity:

$$\cos ({\tan }^{-1}t)=\frac{1}{\sqrt{1+t^2}}$$

So:

$$\cos ({\tan }^{-1}(e^{-x}))=\frac{1}{\sqrt{1+e^{-2x}}}$$

Final Simplified Answer

$$\frac{dy}{dx}=-\frac{e^{-x}}{(1+e^{-2x}{)}^{3\mathrm{/}2}}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=-\frac{e^{-x}}{(1+e^{-2x}{)}^{3\mathrm{/}2}}}}$$

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Question 5

Differentiate w.r.t. $x$:

$$y=\log (\cos (e^x))$$Solution

Use the chain rule multiple times.

Let:

$$y=\log (\cos (e^x))$$

Step 1: Differentiate outer logarithm

$$\frac{dy}{dx}=\frac{1}{\cos (e^x)}\cdot \frac{d}{dx}(\cos (e^x))$$

Step 2: Differentiate $\cos (e^x)$

$$\frac{d}{dx}(\cos (e^x))=-\sin (e^x)\cdot \frac{d}{dx}(e^x)$$
$$=-\sin (e^x)\cdot e^x$$

Combine the results

$$\frac{dy}{dx}=\frac{1}{\cos (e^x)}\cdot (-e^x\sin (e^x))$$
$$\frac{dy}{dx}=-e^x\frac{\sin (e^x)}{\cos (e^x)}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=-e^x\tan (e^x)}}$$

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Question 6

Differentiate w.r.t. $x$:

$$y=e^x+e^{x^2}+e^{x^3}+e^{x^4}+e^{x^5}$$

Solution

Differentiate term-by-term:

    1. $\frac{d}{dx}(e^x)=e^x$

   2. $\frac{d}{dx}(e^{x^2})=e^{x^2}\cdot \frac{d}{dx}(x^2)=e^{x^2}\cdot 2x$

   3. $\frac{d}{dx}(e^{x^3})=e^{x^3}\cdot \frac{d}{dx}(x^3)=e^{x^3}\cdot 3x^2$

   4. $\frac{d}{dx}(e^{x^4})=e^{x^4}\cdot \frac{d}{dx}(x^4)=e^{x^4}\cdot 4x^3$

   5. $\frac{d}{dx}(e^{x^5})=e^{x^5}\cdot \frac{d}{dx}(x^5)=e^{x^5}\cdot 5x^4$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=e^x+2x{e}^{x^2}+3x^2{e}^{x^3}+4x^3{e}^{x^4}+5x^4{e}^{x^5}}}$$

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Question 7

Differentiate w.r.t. $x$:

$$y=\sqrt{e^{\sqrt{x}}},x>0$$Solution

Rewrite the function:

$$y={\left(e^{\sqrt{x}}\right)}^{1\mathrm{/}2}=e^{\frac{1}{2}\sqrt{x}}$$

Let:$$u=\frac{1}{2}\sqrt{x}=\frac{1}{2}{x}^{1\mathrm{/}2}$$

So:

$$y=e^u$$

Differentiate

$$\frac{dy}{du}=e^u$$
$$\frac{du}{dx}=\frac{1}{2}\cdot \frac{1}{2}{x}^{-1\mathrm{/}2}=\frac{1}{4\sqrt{x}}$$

Apply chain rule:

$$\frac{dy}{dx}=e^u\cdot \frac{du}{dx}$$

$$\frac{dy}{dx}=e^{\frac{1}{2}\sqrt{x}}\cdot \frac{1}{4\sqrt{x}}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{e^{\frac{1}{2}\sqrt{x}}}{4\sqrt{x}},x>0}}$$

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Question 8

Differentiate w.r.t. $x$:

$$y=\log (\log x),x>1$$Solution

This is a composition of two logarithmic functions, so apply the chain rule.

Let:

$$u=\log x$$
$$y=\log u$$

Differentiate step-by-step

$$\frac{dy}{du}=\frac{1}{u}$$
$$\frac{du}{dx}=\frac{1}{x}$$

Apply chain rule

$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$$
$$\frac{dy}{dx}=\frac{1}{\log x}\cdot \frac{1}{x}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{1}{x\log x},x>1}}$$

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Question 9

Differentiate w.r.t. $x$:

$$y=\frac{\cos x}{\log x},x>0$$Solution

This is a quotient, so apply the Quotient Rule:

$$\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v{u}^{\mathrm{\prime }}-u{v}^{\mathrm{\prime }}}{v^2}$$Let:

$$u=\cos x,v=\log x$$Then:

$$u^{\mathrm{\prime }}=-\sin x$$

$$v^{\mathrm{\prime }}=\frac{1}{x}$$

Apply quotient rule

$$\frac{dy}{dx}=\frac{(\log x)(-\sin x)-(\cos x)\left(\frac{1}{x}\right)}{(\log x{)}^2}$$
$$\frac{dy}{dx}=\frac{-\sin x\log x-\frac{\cos x}{x}}{(\log x{)}^2}$$Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{-\sin x\log x-\frac{\cos x}{x}}{(\log x{)}^2},x>0}}$$

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Question 10

Differentiate w.r.t. $x$:

$$y=\cos (\log x+e^x),x>0$$Solution

Use the chain rule.

Let:

$$u=\log x+e^x$$So:

$$y=\cos (u)$$

Differentiate outer function

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

Differentiate inner function

$$\frac{du}{dx}=\frac{d}{dx}(\log x)+\frac{d}{dx}(e^x)=\frac{1}{x}+e^x$$

Apply chain rule

$$\frac{dy}{dx}=-\sin (u)(\frac{1}{x}+e^x)$$

Substitute back $u=\log x+e^x$:

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=-\sin (\log x+e^x)(\frac{1}{x}+e^x),x>0}}$$