Chain rule with exponential and arcsine

Identify the outer function as $f(u) = e^u$ and the inner function as $u = \arcsin(x)$. Since $\frac{d}{du}[e^u] = e^u$, applying the chain rule gives $g'(x) = e^{\arcsin(x)} \cdot \frac{d}{dx}[\arcsin(x)]$.

Recall that $\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}$. Substituting this in gives $g'(x) = \dfrac{e^{\arcsin(x)}}{\sqrt{1-x^2}}$.

The outer function was f of u equals e to the u. Since the derivative of e to the u with respect to u is again e to the u, this exponential form carried straight through the chain rule unchanged. That is why the final derivative still contains e raised to arcsine of x. The inner function, arcsine of x, contributed the other factor.