Question

Differentiate $\tan$ ${}^{-1}\left(\frac{2x}{1-x^2}\right)$ w.r.to $\cos$ ${}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.

Answer 1

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

or, $y=2{\tan }^{-1}x$ [ using formula]

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

or, $z=2{\tan }^{-1}z$ [ using direct formula]

Then we've $y=z$.

Then $\frac{dy}{dz}=1$.