KC Sinha Mathematics Solution Class 12 Chapter 11 अवकलन (Differentiation) Exercise 11.1 (Q41-Q44)

Exercise 11.1

Question 41

$\sin \sqrt{1-x^{2}}+x^{2} \cos 4 x$
Sol :
Let y=

$\sin \sqrt{1-x^{2}}+x^{2} \cos 4 x$

Differentiating with respect to x

$\frac{d y}{d n}=\frac{d(\sin \sqrt{1-x^{2}})}{d(\sqrt{1-x^{2}})} \times \frac{d(\sqrt{1-x^{2}})}{d\left(1-x^{2}\right)} \times d \frac{\left(1-x^{2}\right)}{d x}$

$+\frac{d\left(x^{2}\right)}{dx} \cdot \cos 4 x+x^{2} \cdot \frac{d(\cos 4 x)}{d(4 x)} \times \frac{d(4 x)}{d x}$

$=\cos \sqrt{1-x^{2}} \cdot \frac{1}{2 \sqrt{1-x^{2}}} \cdot(-2x)+2 x \cos 4 x+x^2.(-sin^4x)\times 4$

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

Question 42

$\frac{1}{4 \sqrt{4 x^{3}-1}}+\cos ^{2}(5 x+8)$
Sol :
Let y=

$\frac{1}{4 \sqrt{4 x^{3}-1}}+\cos ^{2}(5 x+8)$

$y=\frac{1}{4} \cdot\left(4 x^{3}-1\right)^{-\frac{1}{2}}+\cos ^{2}(5 x+8)$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{1}{4} \cdot \frac{d\left(4 x^{3}-1\right)^{-\frac{1}{2}}}{d\left(4 x^{3}-1\right)}\times \frac{d\left(4 x^{3}-1\right)}{d x}+\frac{d\left(\cos ^{2}(5 x+8)\right.}{\left.d(cos(5x+8)\right)}\times\frac{d(\cos(5x+8))}{d(5x+8)}\times \frac{d(5x+8)}{dx}$

$=\frac{1}{4} \cdot\left(-\frac{1}{2}\right) \cdot\left(4 x^{3}-1\right)^{-\frac{3}{2}} \times 12 x^{2}+2\cos(5x+8).[-\sin(5x+8)].5$

$=-\frac{3}{2} \cdot \frac{x^{2}}{\left(4 x^{3}-1\right)^{3/2}}-5 \sin 2(5 x+8)$

Question 43

$f(x)=\sqrt{\frac{x-1}{x+1}}$ show that

$f^{\prime}(x)=\frac{1}{(x+1) \sqrt{x^{2}-1}}$
Sol :

$f(x)=\sqrt{\frac{x-1}{x+1}}$

Differentiating with respect to x

$f^{\prime}(x)=\frac{d(\sqrt{\frac{x-1}{x+1}})}{d\left(\frac{x-1}{x+1}\right)} \times \frac{d\left(\frac{x-1}{x+1}\right)}{dx}$

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

$=\frac{1}{2} \sqrt{\frac{x+1}{x-1}} \times \frac{x+1-x+1}{(x+1)^{2}}$

$=\frac{1}{2} \sqrt{\frac{x+1}{x-1}} \times \frac{2}{(x+1)^{2}}$

$=\frac{\sqrt{x+1}}{\sqrt{x-1}} \times \frac{1}{(x+1) \cdot(\sqrt{x+1})^{2}}$

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

Question 44

$y=\frac{\cos x+\sin x}{\cos x-\sin x}$ show that

$\frac{d y}{d x}=\sec ^{2}\left(\frac{\pi}{4}+x\right)$
Sol :

$y=\frac{\cos x+\sin x}{\cos x-\sin x}$

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$y=\frac{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos }}$

$y=\frac{1+\tan x}{1-\tan x}$

$y=\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \tan x}$

$y=\tan \left(\frac{\pi}{4}+x\right)$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left(\tan \left(\frac{\pi}{4}+x\right)\right)}{d\left(\frac{\pi}{4}+x\right)} d \frac{\left(\frac{\pi}{4}+x\right)}{d x}$

$=\sec ^{2}\left(\frac{\pi}{4}+x\right)(0+1)$

$\frac{d y}{d x}=\sec^{2}\left(\frac{\pi}{4}+x\right)$