KC Sinha: Exercise 11.1- Mathematics Solution Class 12 Chapter 11 अवकलन

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Question 1

x के सापेक्ष के फलन अर्थात् संयुक्त फलनों के अवकलन पर आधारित प्रश्नः

[Differentiate the following functions with respect to x]

(i) sin(ax+b)
Sol :
Let y= sin(ax+b)

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d[\sin (a x+b)]}{d\left(a{x}+b\right)} \times \frac{d\left(a{x}+b\right)}{d{x}}$

= cos(ax+b)a

=acos(ax+b)

(ii) sin x²
Sol :
Let y=sin x²

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left(\sin x^{2}\right)}{d\left(x^{2}\right)} \times \frac{d\left(x^{2}\right)}{dx}$

=cos x²2x

=2x.cos x²

(iii) tan(5x+9)
Sol :
Let y=tan(5x+9)

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d[\tan (5 x+1)]}{d(5 x+9)} \times \frac{d(5 x+9)}{d{x}}$

=sec²(5x+9)5

=5sec²(5x+9)

(iv) cos(sin x²)
Sol :
Let y=cos(sin x²)

Differentiating with respect to x

$\frac{d y}{dx}=\frac{\left.d [ \cos \left(\tan x^{2}\right)\right]}{d\left(\sin x^{2}\right)} \times \frac{d\left(\sin x^{2}\right)}{d\left(x^{2}\right)} \times \frac{d\left(x^{2}\right)}{dx}$

=-sin(sin x²) .cos x².2x

=-2x.sin(sin x²) .cos x²

(v) sin³x
Sol :
Let y=sin³x

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left(\sin ^{3} x\right)}{d(\sin x)}=\frac{d(\sin x)}{d x}$

=3.sin²x.cosx

(vi) $\sqrt{x^{2}+x+1}$
Sol :
Let y=$\sqrt{x^{2}+x+1}$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d(\sqrt{x^{2}+x+1})}{d\left(x^{2}+x+1\right)} \times \frac{d\left(x^{2}+x+1\right)}{dx}$

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

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

Differentiate the following functions with respect to x

Question 2

tan(xⁿ)
Sol :
Let y=tan(xⁿ)

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left[\tan \left(x^{n}\right)\right]}{d\left(x^{n}\right)} \times \frac{d\left(x^{n}\right)}{d x}$

=sec²(xⁿ).nx^n-1

=nx^n-1.sec²(xⁿ)

Question 3

cosec(cosec x)
Sol :
Let y=cosec(cosec x)

Differentiating with respect to x

$\frac{d{y}}{dx}=\frac{d[\operatorname{cosec}(\operatorname{cosec} x)]}{d[\operatorname{cosec} x)} \times \frac{d(\cos x)}{dx}$

=-cosec(cosec x).cot(cosec x).(-coec x . cot x)

=cosec(cosec x).cot(cosec x).cosec x. cot x

Question 4

tan(x²+3)
Sol :
Let y=tan(x²+3)

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left[\tan \left(x^{2}+3\right)\right]}{d\left(x^{2}+3\right)} \times \frac{d( x^{2}+3)}{d x}$

=sec²(x²+3).2x

=2x.sec²(x²+3)

Question 5

tan x⁰
Sol :
Let y=tan x⁰

$=\tan x \times \frac{\pi}{180}$

$y=\tan \frac{\pi x}{180}$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left(\tan \frac{\pi x}{180}\right)}{d\left(\frac{\pi x}{180}\right)} \times \frac{d\left(\frac{\pi x}{180}\right)}{d x}$

$=\sec ^{2} \frac{\pi x}{120} \times \frac{\pi}{180}$

$=\frac{\pi}{180} \cdot \sec ^{2} x^{0}$

Question 6

$\left(3 x^{2}+6 x+5\right)^{\frac{7}{2}}$
Sol :
Let y=$\left(3 x^{2}+6 x+5\right)^{\frac{7}{2}}$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left(3 x^{2}+6 x+5\right)^{\frac{7}{2}}}{d\left(3 x^{2}+6 x+5\right)} \times \frac{d\left(3 x^{2}+6 x+5\right)}{dx}$

$=\frac{7}{2}\left(3 x^{2}+6 x+5\right)^{5 / 2} \cdot(6 x+6)$

$=21(x+1) \cdot\left(3 x^{2}+6 x+5\right)^{5 / 2}$

Question 7

$\sqrt{5+2 x-4 x^{5}}$
Sol :
Let y=$\sqrt{5+2 x-4 x^{5}}$

Differentiating with respect to x

$\frac{dy}{dx}=\frac{d(\sqrt{5+2 x-4x^5} )}{d(5+2x-4x^5)} \times d\left(\frac{5+2x-4 x^{5}}{dx}\right)$

$=\frac{1}{2 \sqrt{5+2 x-4 x^{5}}} \times\left(2-20 x^{4}\right)$

$=\frac{2\left(1-10 x^{4}\right)}{2 \sqrt{5+2 x-4 x^{5}}}$

$=\frac{1-10 x^{4}}{\sqrt{5+2 x-4 x^5}}$

Question 8

sin(cos x³)

Differentiating with respect to x

$\frac{d{y}}{dx}=\frac{d\left[\sin \left(\cos x^{3}\right)\right]}{d\left(\cos x^{3}\right)} \times \frac{d\left(\cos x^{3}\right)}{d\left(x^{3}\right)} \times \frac{d\left(x^{2}\right)}{d x}$

=cos(cos x³).(-sin x³).3x²

Question 9

cos(sin x³)

Question 10

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

Differentiating with respect to x

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

$=\cos \sqrt{1+x^{2}} \cdot \frac{1}{2 \sqrt{1+x^{2}}} \times 2 x$

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

Question 11

$\sqrt{\tan 2 x}$
Sol :
Let y=$\sqrt{\tan 2 x}$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d(\sqrt{\tan 2 x})}{d(\tan 2 x)} \times \frac{d(\tan 2 x)}{d(2 x)} \times \frac{d(2 x)}{d x}$

$=\frac{1}{2 \sqrt{\tan 2 x}} \times \sec ^{2} 2 x \cdot 2$

$=\frac{\sec ^{2} 2 x}{\sqrt{\tan 2 x}}$

Question 12

$\sqrt{\sin x^{2}}$
Sol :
Let y=$\sqrt{\sin x^{2}}$

Differentiating with respect to x

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

$=\frac{1}{2 \sqrt{\sin x^{2}}} \times \cos x^{2} \times 2 x$

$=\frac{x \cos x^{2}}{\sqrt{\sin x^{2}}}$

Question 13

sin²(3x+4)
Sol :
Let y=sin²(3x+4)

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left[\sin^{2}(3 x+4)\right]}{d[\sin (3 x+4)]} \times \frac{d[\sin(3 x+4)]}{d(3 x+4)}\times\frac{d(3x+4)}{dx}$

=2.sin(3x+4).cos(3x+4).3

=3sin2(3x+4)

Question 14

$\sec ^{3}\left(\frac{x}{2}\right)$
Sol :
Let y=$\sec ^{3}\left(\frac{x}{2}\right)$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left[\sec ^{3}\left(\frac{x}{2}\right)\right]}{d\left[\sec \left(\frac{3}{2}\right)\right]} \times \frac{d\left(\operatorname{sec} \frac{x}{2}\right)}{d\left(\frac{x}{2}\right)} \times \frac{d\left(\frac{x}{2}\right)}{dx}$

$=3\sin^{2}\left(\frac{x}{2}\right) \cdot \sec \frac{x}{2} \cdot \tan \frac{x}{2} \times \frac{1}{2}$

$=\frac{3}{2} \sec ^{3}\left(\frac{x}{2}\right) \cdot \tan \frac{x}{2}$

Question 15

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

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d[\sin \{\cos (\tan \sqrt{x})\}]}{d[\cos (\tan \sqrt{2})]} \times d \frac{[\cos (\tan \sqrt{2})]}{d(\tan \sqrt{x})} \frac{d(\tan \sqrt{2})}{d(\sqrt{x})}\times \frac{d(\sqrt x)}{dx}$

=cos{cos(tan√x)}{-sin(tan√x)}.sec²((√x)).$\frac{1}{2 \sqrt{x}}$

$=-\dfrac{1}{2 \sqrt{x}} \sec ^{2} \sqrt{x} \sin (\tan \sqrt{x}) \cdot \cos \left[\cos(\tan\sqrt{x})\right]$

Question 16

sin[cos{tan(cot x)}]
Sol :
Let y=sin[cos{tan(cot x)}]

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d[\sin (\cos)(\tan (\cot x)]\}]}{d[\cos \{\tan (\cot x)\}]}+\frac{d[\cos \{\tan (\cot x)\}]}{d\left[\tan \left(\cot x\right)\right]}\times \frac{d[\tan(\cot x)]}{d(\cot x)}\times \frac{d(\cot x)}{dx}$

=cos[cos{tan(cot x)}].[-sin{tan(cot x)}].sec²(cot x).(-cosec²x)

=cosec²x.sec²(cot x).sin{tan(cot x)}.cos[cos{tan(cot x)}]

Question 17

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

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d(\sqrt{\tan (\tan x)})}{d(\tan (\tan x))}=\frac{d(\tan (\tan x))}{d(\tan x)} \frac{d(\tan x)}{dx}$

$=\frac{1}{2 \sqrt{\tan (\tan x)}} \times \sec^{2}(\tan x) \cdot \sec ^{2} x$

$=\frac{\sec ^{2} x \cdot \sec ^{2}(\tan x)}{2 \sqrt{\tan (\tan x)}}$

Question 18

$\sqrt{1+\sin x}$
Sol :
Let y=$\sqrt{1+\sin x}$

Differentiating with respect to x

$\frac{d{y}}{dx}=\frac{d(\sqrt{1+\sin x})}{d(1+\sin x)} \times \frac{d\left(1+\sin x\right)}{d x}$

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

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

Question 19

$\sqrt{\tan \left(1+x^{2}\right)}$
Sol :
Let y=$\sqrt{\tan \left(1+x^{2}\right)}$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d(\sqrt{\tan \left(1+x^{2}\right)})}{d\left(\tan \left(1+x^{2}\right)\right)} \times \left.\frac{\left.d(\tan \left(1+x^{2}\right)\right)}{d\left(1+x^{2}\right)} \times \frac{d\left(1+x^{2}\right)}{dx}\right.$

$=\frac{1}{2 \sqrt{\tan \left(1+x^{2}\right)}} \times \sec ^{2}\left(1+x^{2}\right) \cdot 2 x$

$=\frac{x \sec ^{2}\left(1+x^{2}\right)}{\sqrt{\tan \left(1+x^{2}\right)}}$

Question 20

$\cot \sqrt{\cos \sqrt{x}}$
Sol :
Let y=$\cot \sqrt{\cos \sqrt{x}}$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d[\cot\sqrt{\cos \sqrt{x}})}{d[\sqrt{\cos \sqrt{x}}]} \times \frac{d \sqrt{\cos \sqrt{x}}}{d(\cos \sqrt{x})} \times \frac{d( \cos \sqrt{x})}{d(\sqrt{x})}\times\frac{d(\sqrt{x})}{dx}$

$=-\operatorname{cosec}^{2} \sqrt{\cos \sqrt{x}}\times \frac{1}{2 \sqrt{\cos \sqrt{x}}} \times (-\sin \sqrt{x}) \times \frac{1}{2 \sqrt{x}}$

$=\frac{\sin \sqrt{x} \operatorname{cosec}^{2} \sqrt{\cos \sqrt{x}}}{4 \sqrt{x} \sqrt{\cos \sqrt{x}}}$

Question 21

$\sin \sqrt{\sin \sqrt{x}}$
Sol :
Let y=$\sin \sqrt{\sin \sqrt{x}}$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d[\sin \sqrt{\sin \sqrt{x}}]}{d(\sqrt{\sin \sqrt{x}})} \times \frac{d(\sqrt{\sin \sqrt{x}})}{d(\sin \sqrt{x})} \times \frac{d(\sin \sqrt{x})}{d(\sqrt{x})}\times \frac{d(\sqrt x)}{dx}$

$=\cos \sqrt{\sin \sqrt{x}} \cdot \frac{1}{2 \sqrt{\sin \sqrt{x}}} \times \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}$

$=\frac{\cos \sqrt{\sin \sqrt{x}} \cdot \cos \sqrt{x}}{4 \sqrt{x} \sqrt{\sin \sqrt{x}}}$

Question 22

$\sin \sqrt{\cos \sqrt{a x}}$
Sol :
Let y=$\sin \sqrt{\cos \sqrt{a x}}$

Differentiating with respect to x

$\frac{d y}{d}=\frac{d(\sin \sqrt{\cos \sqrt{a x}})}{d(\sqrt{\cos \sqrt{a x})}} \times \frac{d(\sqrt{\cos \sqrt{a{x}}})}{d(\cos \sqrt{a{x}})}\times \frac{d(\cos \sqrt{a{x}}}{d(\sqrt{a{x}})} \times \frac{d (\sqrt{a{x})}}{dx}\times \frac{d(ax)}{dx}$

$=\cos \sqrt{\cos \sqrt{a x}}\times \frac{1}{2 \sqrt{\cos \sqrt{4 x}}} \times(-\sin \sqrt{a x}) \times \frac{1}{2 \sqrt{ax}}$

$=-\frac{1}{4} \times\frac{a}{\sqrt{a} \sqrt{x}} \cdot \frac{\sin \sqrt{a x} \cos \sqrt{\cos \sqrt{a x}}}{\sqrt{\cos \sqrt{ax}}}$

$=\frac{-1}{4} \sqrt{\frac{a}{x}} \cdot \frac{\sin \sqrt{a} x}{\sqrt{\cos \sqrt{a x}}} \times \cos \sqrt{\cos \sqrt{ax}}$

Question 23

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

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d(\sqrt{\sin (\sqrt x)})}{d(\sin (\sin \sqrt{x}))} \cdot \frac{d(\sin (\sin \sqrt{x}))}{d(\sin \sqrt{x})} \times \frac{d(\sin \sqrt{x})}{d(\sqrt{2})}\times\frac{d(\sqrt{x})}{d x}$

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

$=\frac{\cos \sqrt{x} \cdot \cos (\sin\sqrt{x})}{4 \sqrt{x} \sqrt{\sin (\sin \sqrt{x})}}$

$\frac{d{y}}{d x}=\frac{d(\cos (\tan \sqrt{x+1}))}{d(\tan \sqrt{x+1})} \times d\left(\frac{\tan \sqrt{x+1}}{d(\sqrt{x+1})}\right) \times \frac{d(\sqrt{x+1})}{d(x+1)}\times \frac{d(x+1)}{dx}$

$\frac{d y}{d t}=-\sin (\tan \sqrt{x+1}) \cdot \sec ^{2} \sqrt{x+1} \times \frac{1}{2 \sqrt{x+1}}\times 1$

$=\frac{-\sec ^{2} \sqrt{x+1}. \sin (\tan \sqrt{x+1})}{2 \sqrt{x+1}}$

Question 24

$\cos (\tan \sqrt{x+1})$
Sol:
Let y=$\cos (\tan \sqrt{x+1})$

Differentiating with respect to x

Question 25

$\sin \sqrt{\cos \sqrt{\tan m x}}$
Sol :
Let y=$\sin \sqrt{\cos \sqrt{\tan m x}}$

Differentiating with respect to x

$\frac{d y}{d}=\frac{d(\sin \sqrt{\cos \sqrt{\tan m x}})}{d(\sqrt{\cos \sqrt{\tan m x}})} \times \frac{d(\sqrt{\cos \sqrt{\tan m x}})}{d(\cos \sqrt{\tan m x})}\times\frac{{d}(\cos \sqrt{m x})}{d(\sqrt{\tan m x})}\times \frac{d(\sqrt{\tan mx})}{d(mx)}\times \frac{d(\tan mx)}{d(mx)}\times\frac{d(mx)}{dx}$

$=\cos \sqrt{\cos \sqrt{\tan mx}} \cdot \frac{1}{2 \sqrt{\cos \sqrt{\tan m x}}} \times - \sin \sqrt{\tan m x} \times \frac{1}{2 \sqrt{\tan mx}\times \sec^2 mx \times m}$

$=\frac{-m \sec ^{2} m x \sin \sqrt{\tan m x} \cdot \cos \sqrt{\cos \sqrt{\tan mx}}}{4 \sqrt{\tan mx} \cdot \sqrt{\cos \sqrt{\tan m x}}}$

Question 26

$\frac{1}{\left(1+\tan ^{3} x\right)^{2}}$
Sol :
Let $y=\frac{1}{\left(1+\tan ^{3} x\right)^{2}}$

$y=\left(1+\tan ^{3} x\right)^{-2}$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left[\left(1+\tan^{3} x\right)^{-2}\right]}{d[1+\tan ^3 x]} \times \frac{d\left(1+\tan ^{3} x\right)}{d x}$

$=-2\left(1+\tan ^{3} x\right)^{-3} \cdot\left[\frac{d(1)}{dx}+\frac{d\left(\tan ^{3} x\right)}{d(\tan x)} \times \frac{d(\tan x)}{dx}\right]$

$=\frac{-2}{\left(1+\tan ^{3} x\right)^{3}} \cdot 3 \tan ^{2} x \cdot \sec ^{2} x$

$=\frac{-6 \tan ^{2} x \cdot \sec ^{2} x}{\left(1+\tan ^{3} x\right)^{3}}$

Question 27

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

Differentiating with respect to x

$\frac{dy}{dx}=\frac{d\left[\cos \left(\frac{1-x^{2}}{1-x^{2}}\right)\right]}{d\left(\frac{1-x^{2}}{1+x^{2}}\right)} \times \frac{d\left(\frac{\left(-x^{2}\right)}{1+x^{2}}\right)}{dx}$

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

$=-\sin \left(\frac{1-x^{2}}{1+x^{2}}\right) \cdot \frac{-2 x\left(1+x^{2}\right)-\left(1-x^{2}\right){2 x}}{\left(1+x^{2}\right)^{2}}$

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

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

Question 28

$\cos \left(\frac{x}{1+\sqrt{x}}\right)$
Sol :
Let y=$\cos \left(\frac{x}{1+\sqrt{x}}\right)$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left[\cos \left(\frac{x}{1+\sqrt{x}}\right)\right]}{d\left(\frac{x}{1+\sqrt{x}}\right)} \times \frac{d\left(\frac{x}{1+\sqrt{x}}\right)}{dx}$

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

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

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

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

Question 29

$\frac{1+\sqrt{x}}{1-\sqrt{x}}$
Sol :
Let y=$\frac{1+\sqrt{x}}{1-\sqrt{x}}$

Differentiating with respect to x

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

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

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

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

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

Question 30

$\sqrt{\frac{1-x}{1+x}}$
Sol :
Let y=$\sqrt{\frac{1-x}{1+x}}$

Differentiating with respect to x

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

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

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

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

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

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

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

Question 31

$\tan \left(\frac{x-x^{-1}}{x+x^{-1}}\right)$
Sol :
Let y=$\tan \left(\frac{x-x^{-1}}{x+x^{-1}}\right)$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left[\tan \left(\frac{x-x^{-1}}{x+x^{-1}}\right)\right]}{d\left(\frac{x-x^{-1}}{x+x^{-1}}\right)} \times \frac{d\left(\frac{x-x^{-1}}{x+x^{-1}}\right)}{dx}$

$=\sec ^{2}\left(\frac{x-x^{-1}}{x+x^{-1}}\right) \times \frac{d \left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)}{d}$

$=\sec ^{2}\left(\frac{x-x^{-1}}{x+x^{-1}}\right) \cdot \frac{d\left(\frac{\frac{x^2-1}{x}}{\frac{x^{2}+1}{x}}\right)}{dx}$

$=\sec ^{2}\left(\frac{x-x^{-1}}{x+x^{-1}}\right) \cdot \frac{d\left(\frac{x^{2}-1}{x^{2}+1}\right)}{d x}$

$=\sec ^{2}\left(\frac{x-x^{-1}}{x+x^{-1}}\right) \cdot \frac{2 x \cdot\left(x^{2}+1\right)-\left(x^{2}-1\right) \cdot 2 x}{\left(x^{2}+1\right)^{2}}$

$=\sec^{2}\left(\frac{x-{x}^{-1}}{x+x^{-1}}\right) \cdot \frac{2 x^{3}+2 x-2 x^{3}+2 x}{\left(x^{2}+1\right)^{2}}$

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

Question 32

$\sin \sqrt{\sin x+\cos x}$
Sol :
Let y=$\sin \sqrt{\sin x+\cos x}$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left(\sin \sqrt{\sin x+\cos x}\right)}{d(\sqrt{\sin x+\cos{x}})} \times \frac{d(\sqrt{\sin x+\cos x}) d(\sin x)}{d(\sin x+\cos x)}\times \frac{d(\sin x+\cos x)}{dx}$

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

$-\frac{(\cos x-\sin x) \cdot \cos \sqrt{\sin x+\cos x}}{2 \sqrt{\sin x+\cos x}}$

Question 33

$\sqrt{\left(\frac{1-\tan x}{1+\tan x}\right)}$
Sol :
Let y=$\sqrt{\left(\frac{1-\tan x}{1+\tan x}\right)}$

Differentiating with respect to x

$\frac{dy}{dx}=\frac{d(\sqrt{\frac{-\tan x}{1+\tan x}})}{d\left(\frac{1-\tan x}{1+\tan }\right)} \cdot \frac{d\left(\frac{1-\tan x}{1+\tan x}\right)}{d x}$

$=\frac{1}{2 \sqrt{\frac{1-\tan x}{1+\tan x}}} \times \frac{d\left(\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} (\tan x)}\right)}{dx}$

$=\frac{1}{2} \cdot \sqrt{\frac{1+\tan x}{1-\tan x}} \cdot \frac{d\left(\tan \left(\frac{\pi}{4}-x\right)\right)}{dx}$

$=\frac{1}{2} \sqrt{ \frac{1+\tan x}{1-\tan x}}=\frac{d\left(\tan \left(\frac{\pi}{4}-x\right)\right)}{d\left(\frac{\pi}{4}-x\right)}\times \frac{d\left(\frac{\pi}{4}-x\right)}{d x}$

$=\frac{1}{2} \sqrt{\frac{1+\tan }{1-\tan x}} \times \sec ^{2}\left(\frac{\pi}{4}-2\right) \times(-1)$

$=-\frac{1}{2} \sqrt{\frac{1+\tan x}{1-\tan x}} \cdot \sec ^{2}\left(\frac{\pi}{4}-x\right)$

Question 34

$\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)$
Sol :
Let y=$\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left[\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)\right]}{d\left(\frac{1+x^{2}}{1-x^{2}}\right)} \times \frac{\left(\frac{1+x^{2}}{1-x^{2}}\right)}{d x}$

$=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right) \cdot \frac{\frac{d\left(1+x^{2}\right)}{d x} \cdot\left(1-x^{2}\right)-\left(1+x^{2}\right) \frac{d\left(1-x^{2}\right)}{d x}}{\left(1-x^{2}\right)^{2}}$

$=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right) \cdot \frac{2x\left(1-x^{2}\right)-\left(1+x^{2}\right)(-2 x)}{\left(1-x^{2}\right)^{2}}$

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

$=\frac{4 x}{\left(1-x^{2}\right)^{2}} \cos \left(\frac{1+x^{2}}{1-x^{2}}\right)$

Question 35

$\sqrt{\left(\frac{\sec x-1}{\sec x+1}\right)}$
Sol :
Let y=$\sqrt{\frac{\sec x-1}{\sec x+1}}$

$y=\sqrt{\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1}}$

$=\sqrt{\frac{\frac{1-\cos x}{\cos x}}{\frac{1+\cos x}{\cos x}}}$

$y=\sqrt{\frac{1-\cos x}{1+\cos x }}$

$y=\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}}$

$y=\sqrt{\tan ^{2} \frac{x}{2}}$

$y=\tan \frac{x}{2}$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left(\tan \frac{x}{2}\right)}{d\left(\frac{x}{2}\right)} \times \frac{d\left(\frac{x}{2}\right)}{dx}$

$=\sec ^{2} \frac{x}{2} \times \frac{1}{2}$

$=\frac{1}{2} \sec^{2} \frac{x}{2}$

Question 36

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

Differentiating with respect to x

$\frac{dy}{d x}=\frac{\frac{d\left(\sin ^{2} x\right)}{d(\sin x)} \times \frac{d(\sin x)}{dx} \cdot\left(1+\cos ^{2} x\right)-\sin ^{2} x \cdot\left[\frac{d ({1})}{dx}+\frac{d (\cos^2x )}{d(\cos x)}\times \frac{d(\cos x)}{dx}\right]}{\left(1+\cos ^{2} x\right)^{2}}$

$=\frac{2 \sin x \cdot \cos x\left(1+\cos ^{2} x\right)-\sin ^{2} x[2 \cos x(-\sin x)]}{\left(1+\cos ^{2} x\right)^{2}}$

$=\frac{\sin 2 x\left(1+\cos ^{2} x\right)+\sin ^{2} x \sin^2 x}{\left(1+\cos ^{2} x\right)^{2}}$

$=\frac{\sin 2 x+\sin 2 x \cos ^{2} x+\sin ^{2} 2 \sin 2 x}{\left(1+\cos ^{2} x\right)^{2}}$

$=\frac{\sin 2 x+\sin 2 x\left(\cos ^{2} x+\sin ^{2} x\right)}{\left(1+\cos ^{2} x\right)^{2}}$

$=\frac{\sin 2 x+\sin 2 x}{\left(1+\cos ^{2} x\right)^{2}}$

$=\frac{2 \sin 2x}{(1+\cos ^{2} x)^2}$

Question 37

$\sqrt{\frac{1+\sin x}{1-\sin x}}$
Sol :
Let y=$\sqrt{\frac{1+\sin x}{1-\sin x}}$

$y=\sqrt{\frac{1+\sin x}{1-\sin x} \times \frac{1+\sin x}{1+\sin x}}$

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

$y=\sqrt{\frac{(1+\sin x)^{2}}{\cos ^{2} x}}$

$y=\frac{1+\sin x}{\cos x}$

$y=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$

y=sec x+tan x

Differentiating with respect to x

$\frac{d y}{d x}=\sec x \tan x+\sec ^{2} x$

=sec x(tan x+sec x)

Question 38

$\left(\frac{2 \tan x}{\tan x+\cos x}\right)^{2}$
Sol :
Let y=$\left(\frac{2 \tan x}{\tan x+\cos x}\right)^{2}$

Differentiating with respect to x

$\frac{d y}{dx}=\frac{d\left(\frac{2 \tan x}{\tan x+\cos x}\right)^{2}}{d\left(\frac{2 \tan x}{\tan x+\cos x}\right)} \times\frac{d \left(\frac{2 \tan x}{\tan x+\cos x}\right)}{d x}$

$=2\left(\frac{2 \tan x}{\tan x+\cos x}\right) \cdot \frac{2\left[\sec ^{2} x \cdot(\tan x+\cos x)-\tan \left(\sec ^{2} x-\sin x)\right]\right.}{(\tan x+\cos x)^{2}}$

$=\frac{8 \tan x\left[\sec ^{2} x\tan x+\sec^{2} x \cos x-\sec^{2} x \tan x+ \tan x \sin x\right.]}{(\tan x+\cos x)^3}$

$=\frac{8 \tan x \cdot\left(\sec ^{2} x \cos x+\tan x \sin x\right)}{(\tan x+\cos x)^{3}}$

$=\frac{8 \tan x(\sec x+\tan x \sin x)}{(\tan x+\cos x)^{3}}$

Question 39

$\sqrt{x} \sin x+\sin \sqrt{x}$
Sol :
Let y=$\sqrt{x} \sin x+\sin \sqrt{x}$

Differentiating with respect to x

$\frac{d y}{dx}=d\left(\frac{\sqrt{x} \sin x}{d x}\right)+\frac{d(\sin \sqrt{x})}{d x}$

$=\frac{d(\sqrt{x})}{d} \cdot \sin x+\sqrt{x} \cdot \frac{d(\sin x)}{dx}+\frac{d(\sin \sqrt{x})}{d(\sqrt{x})} \times \frac{d(\sqrt{x})}{dx}$

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

$=\frac{1}{2 \sqrt{2}}\left[\sin x+2 x \cos x+\cos \sqrt{x}\right]$

Question 40

$\cos \left(a x^{2}+b x+c\right)+\sin ^{3} \sqrt{a x^{2}+b x+c}$
Sol :
Let y=$\cos \left(a x^{2}+b x+c\right)+\sin ^{3} \sqrt{a x^{2}+b x+c}$

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d\left[\cos\left(a x^{2}+b x+c\right)\right]}{d\left(a x^{2}+b x+c\right)} \cdot \frac{d\left(a x^{2}+b x+c\right)}{d x}+\frac{d(\sin^ 3 \sqrt{ax^ 2+b x+c})}{d(\sin \sqrt{a x^{2}+b x+c})} \times \frac{d(\sin \sqrt{a x^{2}+b x+c})}{d(\sqrt{ax^{2}+b x+c})}\times \frac{d(\sqrt{a x^{2}+b x+c})}{d\left(a x^{2}+b x+c\right)} \times \frac{d\left(a x^{2}+b x+c\right)}{d x}$

$+3 \sin ^{2} \sqrt{a x^{2}+b x+c} \times \cos \sqrt{a x^{2}+b x+c}\times \frac{1}{2 \sqrt{ax^2+bc+c}}\times(2ax+b)$

$=-(2 a x+b) \sin \left(a x^{2}+b x+c\right)+\frac{3}{2} \frac{(2 a x+b) \cdot \cos \sqrt{a x^{2}+b x+c} \cdot \sin ^{2} \sqrt{a x^{2}+b x+c}}{\sqrt{a x^{2}+b x+c}}$

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

If $f(x)=\sqrt{\frac{x-1}{x+1}}$ 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

If $y=\frac{\cos x+\sin x}{\cos x-\sin x}$ show that $\frac{d y}{d x}=\sec ^{2}\left(\frac{\pi}{4}+x\right)$
Sol :
$y=\frac{\cos x+\sin x}{\cos x-\sin x}$

[]

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

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

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

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

Differentiating with respect to x

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

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

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