Question

If $y=\sin \left\{{\tan }^{-1}\left[\frac{1-x}{1+x}\right]\right\}$find
$\frac{dy}{dx}$

Answer 1

$\begin{array}{c}y=\sin \left\{{\tan }^{-1}\left\{\frac{1-x}{1+x}\right\}\right\}\\ Let\theta ={\tan }^{-1}x\to \left(i\right)\\ y=\sin \left[{\tan }^{-1}\left\{\frac{\tan \frac{\pi }{4}-\tan \theta }{1+\tan \frac{\pi }{4}\times \tan \theta }\right\}\right]\\ y=\sin \left[{\tan }^{-1}\left\{\tan \left(\frac{\pi }{4}-\theta \right)\right\}\right]\\ y=\sin \left(\frac{\pi }{4}-\theta \right)\\ diff.w.r.t\theta \\ \frac{dy}{d\theta }=-\cos \left(\frac{\pi }{4}-\theta \right).\left(1\right)\to \left(ii\right)\\ diff.equation\left(i\right)w.r.t\theta \\ \frac{dx}{d\theta }={\sec }^2\theta \to \left(iii\right)\\ From\left(ii\right),\left(iii\right)\\ \frac{dy}{dx}=\frac{-\cos \left(\frac{\pi }{4}-\theta \right)}{{\sec }^2\theta }\\ =-\left\{\frac{\cos \frac{\pi }{4}.\cos \theta +\sin \frac{\pi }{4}.\sin \theta }{1+{\tan }^2\theta }\right\}\\ \frac{dy}{dx}=-\frac{1}{\sqrt{2}}\left[\frac{\cos \theta +\sin \theta }{1+x^2}\right]\\ =-\frac{1}{\sqrt{2}}\left[\frac{\frac{1}{\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1}}}{1+x^2}\right]\\ =-\frac{1}{\sqrt{2}}\left[\frac{x+1}{\left(\sqrt{x^2+1}\left(1+x^2\right)\right)}\right]\end{array}$