Question

Find $\frac{dy}{dx}$ where $y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right);0<x<1$

Answer 1

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

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

$=\frac{-1}{\sqrt{\frac{1+x^4-2x^2-1-x^4+2x^2}{(1+x^2{)}^2}}}\times \frac{(1+x^2)(-2x)-(1-x^2)\left(2x\right)}{(1+x^2{)}^2}$

$=\frac{-(1+x^2)}{2x}\times \frac{-2x(1+x^2+1-x^2)}{(1+x^2{)}^2}$

$=\frac{2}{(1+x^2)}$