Differentiate the following w.r.t. x:

Let $y=e^{x^{3}}$

By using the chain rule, we obtain

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

$\Rightarrow \frac{d y}{d x}=e^{\sin ^{-1} x} \cdot \frac{d}{d x}\left(\sin ^{-1} x\right)$

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

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

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