Question

If $y={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ then find $\frac{dy}{dx}$

Answer 1

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

Differentiating above equation w.r.t. $x$, we have

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

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

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

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

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