Question

Find the derivative of $y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.

Answer 1

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

Putting $x=\tan \theta$

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

$y={\cos }^{-1}\left(\cos 2\theta \right)$ $[\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }]$

$y=2\theta$

Putting value of $\theta ={\tan }^{-1}x$.

$y=2\left({\tan }^{-1}x\right)$ Since $x=\tan \theta$

$\therefore \theta ={\tan }^{-1}x$

Differentiating both sides w.r.t. x

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

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

$\frac{dy}{dx}=2\cdot \left(\frac{1}{1+x^2}\right)$ $[{\tan }^{-1}x=\frac{1}{1+x^2}]$

$\frac{dy}{dx}=\frac{2}{1+x^2}$.