# How do you find the derivative of arcsin(2x^2)?

Mar 9, 2018

(4x)/sqrt(1-4x^4
$y = {\sin}^{- 1} \left(2 {x}^{2}\right)$, take , $u = 2 {x}^{2}$
$y = {\sin}^{- 1} u$
$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{\sqrt{1 - {u}^{2}}} \mathmr{and} \frac{\mathrm{du}}{\mathrm{dx}} = 4 x$
$\textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}} = \frac{1}{\sqrt{1 - {u}^{2}}} \cdot 4 x$
=>(dy)/(dx)=1/(sqrt(1-(2x^2)^2))*4x=(4x)/sqrt(1-4x^4