Question

Differentiate ${\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$ with refract to ${\cos }^{-1}\left(2x\sqrt{1-n^2}\right)$

Answer 1

$x=\cos \theta$

$y={\tan }^{-1}\left(\frac{\sqrt{1-\cos \theta }}{\cos \theta }\right)={\tan }^{-1}\left(\tan \theta \right)$

$y={\cos }^{-1}\left(x\right)$

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

$z={\cos }^{-1}\left(2x\sqrt{1-x^2}\right)$

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

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

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

$={\cos }^{-1}\left(\cos (\frac{\pi }{2}-2\theta )\right)=\frac{\pi }{2}-2\theta =\frac{\pi }{2}-2{\cos }^{-2}\left(x\right)$

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

$\frac{dy}{dz}=\frac{-1}{2}$