Question

If $y=si{n}^{-1}\left[2ta{n}^{-1}\sqrt{\frac{1-x}{1+x}}\right]$ then find $\frac{dy}{dx}$.

Answer 1

$y={\sin }^{-1}\left[2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right]$

$\sin y=2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}$

$\cos y\left(\frac{dy}{dx}\right)=\frac{2}{1+{\left(\sqrt{\frac{1-x}{1+x}}\right)}^2}\times \frac{1}{2}\sqrt{\frac{1+x}{1-x}}\times \frac{\left[\right(-1\left)\right(1+x)-1(1-x\left)\right]}{(1-x{)}^2}$

$\cos y\left(\frac{dy}{dx}\right)=\frac{-2(1+x)}{(1+x+1-x}\times \frac{1}{2}\sqrt{\frac{1+x}{(1-x)}}\times \frac{[1+x+1-x]}{(1+x{)}^2}$

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

$\cos y\left(\frac{dy}{dx}\right)=\frac{-1}{\sqrt{(1+x(1-x)}}$

$\left(\frac{dy}{dx}\right)=\frac{-secy}{\sqrt{(1+x)(1-x)}}$

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