Question

If $y={\sin }^{-1}\sqrt{\frac{1-x}{1+x}},$ then $-\frac{3}{\sqrt{2}}\frac{dy}{dx}{|}_{x=\frac{1}{2}}$ is equal to

Answer 1

Given $y={\sin }^{-1}\sqrt{\frac{1-x}{1+x}}$
$\Rightarrow {\sin }^{2}y=\frac{1-x}{1+x}$
$\Rightarrow 2\sin y.\cos y\frac{dy}{dx}=\frac{-(1+x)-(1-x)}{(1+x{)}^2}$
$\Rightarrow \sin y.\cos y\frac{dy}{dx}=-\frac{1}{(1+x{)}^2}$
Now at $x=\frac{1}{2}\Rightarrow y=\frac{1}{\sqrt{3}}$;
Thus at $x=\frac{1}{2},\frac{dy}{dx}=-\frac{1}{(1+\frac{1}{2}{)}^2(\frac{1}{\sqrt{3}}\left)\right(\frac{2}{\sqrt{3}})}$
$\Rightarrow -\frac{3}{\sqrt{2}}\frac{dy}{dx}{\mid }_{x=\frac{1}{2}}=2$