Question

Differentiate the following functions w.r.t $x$
$\left(i\right){\cos }^{-1}\left(\frac{2x}{1+x^2}\right)$
$\left(ii\right){\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

Answer 1

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

Let $x=\tan \theta$

$dx=se{c}^2\theta d\theta \Rightarrow \frac{d\theta }{dx}=\frac{1}{{\sec }^2\theta }=\frac{1}{1+x^2}$

Differentiate (1) w.r.t x

$\frac{dy}{dx}=\frac{d}{dx}\left({\cos }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)\right)$

$=\frac{d}{dx}\left({\cos }^{-1}\right(\sin 2\theta \left)\right)$

$=\frac{d(\frac{\pi }{2}\pm 2\theta )}{dx}$

$=\pm 2\frac{d\theta }{dx}$

$\therefore \frac{dy}{dx}=\frac{2}{1+x^2}$ $x\in (-\infty ,-1)\cup (1,\infty )$

$=-\frac{2}{1+x^2}$ $x\in [-1,1]$

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

Let $x=\sin \theta$

$dx=\cos \theta d\theta \Rightarrow \frac{d\theta }{dx}=\frac{1}{\sqrt{1-x^2}}$

$y={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)$

$={\sin }^{-1}\left(\sin 2\theta \right)$

$\therefore \frac{dy}{dx}=2\frac{d\theta }{dx}(-1{)}^n=\frac{2(-1{)}^n}{\sqrt{1-x^2}}$

$\therefore \frac{dy}{dx}=\frac{2}{\sqrt{1-x^2}}$ $\forall x\in [\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$

$=\frac{-2}{\sqrt{1-x^2}}$ $\forall x\in R-[\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}]$