Question

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

Answer 1

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

By using the chain rule, we obtain

$\frac{dy}{dx}=\frac{d}{dx}(e^{{\sin }^{-1}x})$

$\Rightarrow \frac{dy}{dx}=e^{si{n}^{-1}x}.\frac{d}{dx}\left({\sin }^{-1}x\right)$

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

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

$\therefore \frac{dy}{dx}=\frac{e^{{\sin }^{-1}x}}{\sqrt{1-x^2}},x\in (-1,1)$