Exercise 11.1
Question 33
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
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
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
Sol :
Let y=$\frac{\sin ^{2} x}{1+\cos ^{2} x}$
Differentiating with respect to x
$\frac{dy}{d x}=\frac{\frac{d\left(\sin ^{2} x\right)}{d(\sin x)} \times \frac{d(\sin x)}{dx} \cdot\left(1+\cos ^{2} x\right)-\sin ^{2} x \cdot\left[\frac{d ({1})}{dx}+\frac{d (\cos^2x )}{d(\cos x)}\times \frac{d(\cos x)}{dx}\right]}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{2 \sin x \cdot \cos x\left(1+\cos ^{2} x\right)-\sin ^{2} x[2 \cos x(-\sin x)]}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{\sin 2 x\left(1+\cos ^{2} x\right)+\sin ^{2} x \sin^2 x}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{\sin 2 x+\sin 2 x \cos ^{2} x+\sin ^{2} 2 \sin 2 x}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{\sin 2 x+\sin 2 x\left(\cos ^{2} x+\sin ^{2} x\right)}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{\sin 2 x+\sin 2 x}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{2 \sin 2x}{(1+\cos ^{2} x)^2}$
$\frac{dy}{d x}=\frac{\frac{d\left(\sin ^{2} x\right)}{d(\sin x)} \times \frac{d(\sin x)}{dx} \cdot\left(1+\cos ^{2} x\right)-\sin ^{2} x \cdot\left[\frac{d ({1})}{dx}+\frac{d (\cos^2x )}{d(\cos x)}\times \frac{d(\cos x)}{dx}\right]}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{2 \sin x \cdot \cos x\left(1+\cos ^{2} x\right)-\sin ^{2} x[2 \cos x(-\sin x)]}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{\sin 2 x\left(1+\cos ^{2} x\right)+\sin ^{2} x \sin^2 x}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{\sin 2 x+\sin 2 x \cos ^{2} x+\sin ^{2} 2 \sin 2 x}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{\sin 2 x+\sin 2 x\left(\cos ^{2} x+\sin ^{2} x\right)}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{\sin 2 x+\sin 2 x}{\left(1+\cos ^{2} x\right)^{2}}$
$=\frac{2 \sin 2x}{(1+\cos ^{2} x)^2}$
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