Tag: Miscellaneous Exercise on Chapter 5 class 12th maths NCERT

Question 1

Differentiate w.r.t. x:

$$y=(3x^2-9x+5{)}^9$$

Solution

Using the chain rule, differentiate the outer function first and then multiply by the derivative of the inner function:

$$\frac{dy}{dx}=9(3x^2-9x+5{)}^8\times \frac{d}{dx}(3x^2-9x+5)$$

Now differentiate the bracket:

$$\frac{d}{dx}(3x^2-9x+5)=6x-9$$

So,$$\frac{dy}{dx}=9(3x^2-9x+5{)}^8(6x-9)$$

$$\boxed{{\displaystyle \frac{dy}{dx}=27\mathrm{}(2x-3)(3x^2-9x+5{)}^8}}$$

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Question 2

Differentiate w.r.t. $x$:

$$y={\sin }^{3}x+{\cos }^{6}x$$

Solution

Differentiate each term separately. Using the chain rule:

$$\frac{d}{dx}({\sin }^{3}x)=3{\sin }^{2}x\cdot \frac{d}{dx}(\sin x)=3{\sin }^{2}x\cos x$$
$$\frac{d}{dx}({\cos }^{6}x)=6{\cos }^{5}x\cdot \frac{d}{dx}(\cos x)=6{\cos }^{5}x(-\sin x)$$

So,$$\frac{dy}{dx}=3{\sin }^{2}x\cos x-6{\cos }^{5}x\sin x$$

Now factor common terms: $3\sin x\cos x$

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

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Question 3

Differentiate w.r.t. $x$:

$$y=(5x{)}^{3\cos 2x}$$

Solution

Take log on both sides:

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

Differentiate w.r.t. $x$ (using product rule):

$$\frac{1}{y}\frac{dy}{dx}=3\ln (5x)\frac{d}{dx}(\cos 2x)+3\cos 2x\frac{d}{dx}(\ln (5x))$$

$$=3\ln (5x)(-2\sin 2x)+3\cos 2x\cdot \frac{1}{x}$$

$$=-6\ln (5x)\sin 2x+\frac{3\cos 2x}{x}$$

Multiply both sides by $y$:

$$\frac{dy}{dx}=(5x{)}^{3\cos 2x}(\frac{3\cos 2x}{x}-6\ln (5x)\sin 2x)$$

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Question 4

Differentiate w.r.t. $x$:

$$y={\sin }^{-1}(x\sqrt{x}),0\le x\le 1$$

Solution

First rewrite the function inside:

$$x\sqrt{x}=x^{3\mathrm{/}2}$$

$$y={\sin }^{-1}(x^{3\mathrm{/}2})$$

Now differentiate:

$$\frac{dy}{dx}=\frac{1}{\sqrt{1-(x^{3\mathrm{/}2}{)}^2}}\cdot \frac{d}{dx}(x^{3\mathrm{/}2})$$

$$\frac{d}{dx}(x^{3\mathrm{/}2})=\frac{3}{2}{x}^{1\mathrm{/}2}$$

So,$$\frac{dy}{dx}=\frac{\frac{3}{2}{x}^{1\mathrm{/}2}}{\sqrt{1-x^3}}$$

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{3\sqrt{x}}{2\sqrt{1-x^3}},0\le x\le 1}}$$

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Question 5$$y=\frac{{\cos }^{-1}\left(\frac{x}{2}\right)}{\sqrt{2x+7}},-2<x<2$$Answer :

To Find: $\frac{dy}{dx}$

Let:$$y={\cos }^{-1}\left(\frac{x}{2}\right)(2x+7{)}^{-1\mathrm{/}2}$$

Use Product Rule:

$$\frac{dy}{dx}=\frac{du}{dx}\cdot v+u\cdot \frac{dv}{dx}$$

Where:$$u={\cos }^{-1}\left(\frac{x}{2}\right),v=(2x+7{)}^{-1\mathrm{/}2}$$

Step 1: Differentiate $u={\cos }^{-1}(x\mathrm{/}2)$

$$\frac{du}{dx}=-\frac{1}{\sqrt{1-{\left(\frac{x}{2}\right)}^2}}\cdot \frac{d}{dx}\left(\frac{x}{2}\right)$$
$$\frac{du}{dx}=-\frac{1}{\sqrt{1-\frac{x^2}{4}}}\cdot \frac{1}{2}$$
$$\frac{du}{dx}=-\frac{1}{2\sqrt{1-\frac{x^2}{4}}}$$

Now,

$$\sqrt{1-\frac{x^2}{4}}=\frac{\sqrt{4-x^2}}{2}$$

So,

$$\frac{du}{dx}=-\frac{1}{2\cdot \frac{\sqrt{4-x^2}}{2}}=-\frac{1}{\sqrt{4-x^2}}$$

Step 2: Differentiate $v=(2x+7{)}^{-1\mathrm{/}2}$

$$\frac{dv}{dx}=-\frac{1}{2}(2x+7{)}^{-3\mathrm{/}2}\cdot 2$$
$$\frac{dv}{dx}=-(2x+7{)}^{-3\mathrm{/}2}$$

Apply Product Rule

$$\frac{dy}{dx}=(-\frac{1}{\sqrt{4-x^2}})(2x+7{)}^{-1\mathrm{/}2}+\left({\cos }^{-1}\left(\frac{x}{2}\right)\right)(-(2x+7{)}^{-3\mathrm{/}2})$$
$$\frac{dy}{dx}=-\frac{1}{\sqrt{4-x^2}\sqrt{2x+7}}-\frac{{\cos }^{-1}(x\mathrm{/}2)}{(2x+7{)}^{3\mathrm{/}2}}$$

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Question 6

$$y={\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right),0<x<\frac{\pi }{2}$$

Solution

Let

$$A=\sqrt{1+\sin x},B=\sqrt{1-\sin x}$$

So the expression inside ${\cot }^{-1}$ becomes:

$$\frac{A+B}{A-B}$$

Simplify the expression

Multiply numerator and denominator by $(A+B)$:

$$\frac{A+B}{A-B}\cdot \frac{A+B}{A+B}=\frac{(A+B{)}^2}{A^2-B^2}$$Expand numerator:

$$(A+B{)}^2=A^2+2AB+B^2$$And denominator:

$$A^2-B^2=(1+\sin x)-(1-\sin x)=2\sin x$$

Now calculate numerator:

$$A^2+B^2=(1+\sin x)+(1-\sin x)=2$$
$$AB=\sqrt{(1+\sin x)(1-\sin x)}=\sqrt{1-{\sin }^{2}x}=\cos x$$

So:$$(A+B{)}^2=2+2\cos x$$

Thus:$$\frac{A+B}{A-B}=\frac{2+2\cos x}{2\sin x}=\frac{1+\cos x}{\sin x}$$

Use Trigonometric Identity

$$\frac{1+\cos x}{\sin x}=\cot \frac{x}{2}$$

So:$$y={\cot }^{-1}(\cot \frac{x}{2})$$

Given domain $0<x<\frac{\pi }{2}$, we have:

$$0<\frac{x}{2}<\frac{\pi }{4}$$

In this domain, ${\cot }^{-1}(\cot \theta )=\theta$
Thus:$$y=\frac{x}{2}$$

Differentiate

$$\frac{dy}{dx}=\frac{1}{2}$$

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Question 7

$$y=(\log x{)}^{\log x},x>1$$

We need to find:

$$\frac{dy}{dx}$$Solution

Take logarithm on both sides

$$y=(\log x{)}^{\log x}$$
$$\ln y=\log x\cdot \ln (\log x)$$

Differentiate both sides w.r.t. $x$

Use implicit differentiation:

Left side:

$$\frac{1}{y}\frac{dy}{dx}$$

Right side — Product rule:

$$\frac{d}{dx}[\log x\cdot \ln (\log x)]$$

Let $u=\log x$ and $v=\ln (\log x)$

$$u^{\mathrm{\prime }}=\frac{1}{x},v^{\mathrm{\prime }}=\frac{1}{\log x}\cdot \frac{1}{x}$$

Apply product rule:

$$\frac{d}{dx}[uv]=u^{\mathrm{\prime }}v+u{v}^{\mathrm{\prime }}$$
$$=\frac{1}{x}\ln (\log x)+\log x\cdot \frac{1}{x\log x}$$

$$=\frac{\ln (\log x)}{x}+\frac{1}{x}$$

Equate both sides

$$\frac{1}{y}\frac{dy}{dx}=\frac{\ln (\log x)}{x}+\frac{1}{x}$$

Multiply both sides by $y$:

$$\frac{dy}{dx}=y(\frac{\ln (\log x)}{x}+\frac{1}{x})$$

Substitute $y=(\log x{)}^{\log x}$

$$\frac{dy}{dx}=(\log x{)}^{\log x}(\frac{\ln (\log x)}{x}+\frac{1}{x})$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=(\log x{)}^{\log x}(\frac{\ln (\log x)}{x}+\frac{1}{x})}}$$

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Question 8

$$y=\cos (a\cos x+b\sin x)$$

where $a$ and $b$ are constants.

We have to find:

$$\frac{dy}{dx}$$Solution

Let:

$$u=a\cos x+b\sin x$$

So the function becomes:

$$y=\cos u$$

Differentiate using Chain Rule

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

Now differentiate $u$:

$$\frac{du}{dx}=a\frac{d}{dx}(\cos x)+b\frac{d}{dx}(\sin x)$$

$$\frac{du}{dx}=a(-\sin x)+b(\cos x)$$

$$\frac{du}{dx}=-a\sin x+b\cos x$$

Substitute back into derivative

$$\frac{dy}{dx}=-\sin (a\cos x+b\sin x)\cdot (-a\sin x+b\cos x)$$

Final Answer

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

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Question 9

$$y=(\sin x-\cos x{)}^{(\sin x-\cos x)},\frac{\pi }{4}<x<\frac{3\pi }{4}$$

We must find:

$$\frac{dy}{dx}$$

Solution

Let:

$$y=(\sin x-\cos x{)}^{(\sin x-\cos x)}$$

Take natural log on both sides (logarithmic differentiation):

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

Differentiate both sides w.r.t $x$:

Left side:

$$\frac{1}{y}\frac{dy}{dx}$$

Right side (product rule):

Let $u=\sin x-\cos x$ and $v=\ln (\sin x-\cos x)$

$$u^{\mathrm{\prime }}=\cos x+\sin x$$
$$v^{\mathrm{\prime }}=\frac{1}{\sin x-\cos x}(\cos x+\sin x)$$

Apply product rule:

$$\frac{d}{dx}[uv]=u^{\mathrm{\prime }}v+u{v}^{\mathrm{\prime }}$$

$$=(\cos x+\sin x)\ln (\sin x-\cos x)+(\sin x-\cos x)\frac{\cos x+\sin x}{\sin x-\cos x}$$

Simplify the second term:

$$=(\cos x+\sin x)\ln (\sin x-\cos x)+(\cos x+\sin x)$$

Factor out $(\cos x+\sin x)$

$$=(\cos x+\sin x)[\ln (\sin x-\cos x)+1]$$

Now equate both sides

$$\frac{1}{y}\frac{dy}{dx}=(\cos x+\sin x)[\ln (\sin x-\cos x)+1]$$

Multiply both sides by $y$:

$$\frac{dy}{dx}=y(\cos x+\sin x)[\ln (\sin x-\cos x)+1]$$

Substitute $y=(\sin x-\cos x{)}^{(\sin x-\cos x)}$

$$\boxed{{\displaystyle \frac{dy}{dx}=(\sin x-\cos x{)}^{(\sin x-\cos x)}(\cos x+\sin x)[\ln (\sin x-\cos x)+1]}}$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=(\sin x-\cos x{)}^{(\sin x-\cos x)}(\cos x+\sin x)[\ln (\sin x-\cos x)+1]}}$$

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Question 10

$$y=x^x+x^a+a^x+a^a,a>0,x>0$$

We need to find:

$$\frac{dy}{dx}$$

Differentiate term by term

1. $x^x$

Use logarithmic differentiation:

$$\frac{d}{dx}(x^x)=x^x(\ln x+1)$$

       2. $x^a$

Here $a$ is constant:

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

        3. $a^x$

Exponential with constant base:

$$\frac{d}{dx}(a^x)=a^x\ln a$$

        4. $a^a$

Constant term:

$$\frac{d}{dx}(a^a)=0$$

Combine all results

$$\frac{dy}{dx}=x^x(\ln x+1)+a{x}^{a-1}+a^x\ln a+0$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=x^x(\ln x+1)+a{x}^{a-1}+a^x\ln a}}$$