Question

Differentiate the following w.r.t. x:
$e^{{\sin }^{-1}x}$.

Answer 1

let $y=e^{{\sin }^{-1}x}$

now take ${\log }_e$ on both side we get,

${\sin }^{-1}x={\log }_{e}y$

now, differentiate above equation wrt $x$

we get

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

so, $\frac{dy}{dx}=y\times \frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}=\frac{e^{{\sin }^{-1}x}}{\sqrt{1-x^2}}$