Question

Differentiate: $e^{{\sin }^{-1}x}$

Answer 1

${\sin }^{-1}x=t$

$\frac{d}{dx}{e}^t=e^t\frac{dt}{dx}=e^t\frac{d}{dx}{\sin }^{-1}x=e^t\frac{1}{\sqrt{1-x^2}}=e^{{\sin }^{-1}x}\frac{1}{\sqrt{1-x^2}}$

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