Question

Differentiate ${\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$ with respect to ${\cos }^{-1}\left(2x\sqrt{1-x^2}\right)$, where $x\ne 0.$

Answer 1

Let $u={\tan }^{-1}\left[\frac{\sqrt{1-x^2}}{x}\right]$

Put $x=\cos \theta$

$u={\tan }^{-1}\left[\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }\right]$

$\Rightarrow u={\tan }^{-1}\left[\frac{\sin \theta }{\cos \theta }\right]$

$\Rightarrow u={\tan }^{-1}\left[\tan \theta \right]$

$\Rightarrow u=\theta$

$\Rightarrow u={\cos }^{-1}x$

$\Rightarrow \frac{du}{dx}=-\frac{1}{\sqrt{1-x^2}}$

Let $v={\cos }^{-1}\left[2x\sqrt{1-x^2}\right]$

Put $x=\sin \theta$

So,

$v={\cos }^{-1}\left[2\sin \theta \sqrt{1-{\sin }^2\theta }\right]$

$\Rightarrow v={\cos }^{-1}\left[2\sin \theta \cos \theta \right]$

$\Rightarrow {\cos }^{-1}\left[\sin 2\theta \right]$

$\Rightarrow {\cos }^{-1}\left[\cos (\frac{\pi }{2}-2\theta )\right]$

$\Rightarrow v=\frac{\pi }{2}-2\theta$

$\Rightarrow v=\frac{\pi }{2}-2{\sin }^{-1}x$

$\Rightarrow v=-\frac{2}{\sqrt{1-x^2}}$

Now,

$\frac{du}{dv}=\frac{du}{dx}\times \frac{dx}{dv}=-\frac{1}{\sqrt{1-x^2}}\times -\frac{\sqrt{1-x^2}}{2}=\frac{1}{2}$