Differentiate ^ -1 ( 1+x^2 -1 x ) w.r.t. ^ -1 x when x 0. - Vidyadip

Question

Differentiate $\tan^{-1}\Big(\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big)$ w.r.t. $\tan^{-1}\text{x}$ when $\text{x}\neq0.$

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Answer

Let $\text{u}=\tan^{-1}\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}$ and $\text{y}=\tan^{-1}\text{x}$
Put $\text{x}=\tan\theta$
$\Rightarrow\ \text{u}=\tan^{-1}\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}$
$=\tan^{-1}\frac{\sec\theta-1}{\tan\theta}=\tan^{-1}\Big(\frac{1-\cos\theta}{\sin\theta}\Big)$
$=\tan^{-1}\bigg[\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\bigg]$ $=\tan^{-1}\Big[\tan\frac{\theta}{2}\Big]=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}\text{x}$
$\therefore\ \frac{\text{du}}{\text{dx}}=\frac{1}{2}\cdot\frac{1}{1+\text{x}^2}\ \ \dots(\text{i})$
and $\frac{\text{dv}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\tan^{-1}\text{x}=\frac{1}{1+\text{x}^2}\ \ \dots(\text{ii})$
$\therefore\ \frac{\text{du}}{\text{dv}}=\frac{\frac{\text{du}}{\text{dx}}}{\frac{\text{dv}}{\text{dx}}}=\frac{1}{2}$

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