Question

Find $\frac{dy}{dx}$ for the following
i) $y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$, $-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}$.
ii) $y={\sin }^{-1}\left(\frac{1-x}{1+x}\right)$, $0<x<1$.

Answer 1

i) Given $y={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),\frac{-1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}};\frac{-\pi }{6}<\theta <\frac{\pi }{6};\Rightarrow \frac{-\pi }{2}<3\theta <\frac{\pi }{2}$

Let $x=\tan \theta$

$\therefore y={\tan }^{-1}\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)$

$={\tan }^{-1}\left(\tan 3\theta \right)$

$y=-3\theta \frac{-1}{\sqrt{3}}<x<0$

$3\theta 0\le x<\frac{1}{\sqrt{3}}$

$x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta \Rightarrow \frac{d\theta }{dx}=\frac{1}{{\sec }^2\theta }$

$\frac{dy}{dx}=-3\frac{d\theta }{dx}\frac{-1}{\sqrt{3}}<x<0$

$3\frac{d\theta }{dx},0\le x<\frac{1}{\sqrt{2}}$

$\therefore \frac{dy}{dx}=\{\begin{array}{c}\frac{-3}{1+x^2},\frac{-1}{\sqrt{3}}<x<0\\ \frac{3}{1+x^2},0\le x<\frac{1}{\sqrt{3}}\end{array}$

ii) Given, $y={\sin }^{-1}\left(\frac{1-x}{1+x}\right)0<x<1$

Differentiate on both sides w.r.t. x

$\frac{dy}{dx}=\frac{|1+x|}{\sqrt{1-{\left(\frac{1-x}{1+x}\right)}^2}}\frac{-2}{{(1+x)}^2}$

$=\frac{-2(1+x)}{{(1+x)}^2.2\sqrt{x}}(\because 1+x>0)$

$\frac{dy}{dx}=\frac{-1}{\sqrt{x}(1+x)}$