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Q)

# Differentiate $$\tan^{-1} \{ \large\frac{\sqrt{1+x^2}-1}{x} \}$$ w.r.t. $$x$$.

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A)
Toolbox:
• $1+\tan^2\theta=\sec^2\theta$
• $\large\frac{d}{dx}$$(\tan^{-1}x)=\large\frac{1}{1+x^2} • \sin\theta=2\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2}$
Step 1:
$y=\tan^{-1}\big[\large\frac{\sqrt{1+x^2}-1}{x}\big]$
Put $x=\tan\theta$
$\Rightarrow 1+x^2=1+\tan^2\theta=\sec^2\theta$
$\therefore y=\tan^{-1}\bigg[\large\frac{\sec\theta-1}{\tan\theta}\bigg]$
$\Rightarrow y=\tan^{-1}\bigg[\large\frac{(1/\cos\theta)-1}{\sin\theta/\cos\theta}\bigg]$
$\Rightarrow y=\tan^{-1}\bigg[\large\frac{1-\cos\theta}{\sin\theta}\bigg]$
Step 2:
But $1-\cos \theta=2\sin^2\large\frac{\theta}{2}$
$\sin\theta=2\sin\large\frac{\theta}{2}$$\cos\large\frac{\theta}{2} y=\tan^{-1}\bigg[\large\frac{2\sin^2\large\frac{\theta}{2}}{2\sin\large\frac{\theta}{2}\cos\large\frac{\theta}{2}}\bigg] \therefore y=\tan^{-1}[\tan\large\frac{\theta}{2}] y=\large\frac{\theta}{2} Step 3: But \theta=\tan^{-1}x Hence y=\large\frac{1}{2}$$\tan^{-1}x$
$\large\frac{dy}{dx}=\frac{1}{2}\frac{1}{1+x^2}$
$\quad\;\;=\large\frac{1}{2(1+x^2)}$