Find $y^\prime$ if:
(a) $y=\sin ^{-1}(4x^{2})$ $\quad$ (b) $y=e^{\sin ^{-1}x}$

Solution (a) If $y=\sin ^{-1}u$ and $u=4x^{2}$, then $\dfrac{dy}{du}=\dfrac{1}{\sqrt{1-u^{2}}}$ and $\dfrac{du}{dx}=8x.$ By the Chain Rule, $$\begin{eqnarray*} && y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\left( \dfrac{1}{ \sqrt{1-u^{2}}}\right) (8x) = \dfrac{8x}{\sqrt{1-16x^{4}}}\qquad {\color{#0066A7}{\hbox{\(u=4x^{2}\)}}} \end{eqnarray*}$$

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(b) If $y=e^{u}$ and $u=\sin ^{-1}x$, then $\dfrac{dy}{du}=e^{u}$ and $\dfrac{du}{dx}=\dfrac{1}{\sqrt{1-x^{2}}}$. By the Chain Rule, $$\begin{eqnarray*} && y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=e^{u}\cdot \dfrac{ 1}{\sqrt{1-x^{2}}} \underset{\underset{\color{#0066A7}{\hbox{\(u=\sin ^{-1}x\)}}}{\color{#0066A7}{\uparrow}}}{=} \dfrac{ e^{\sin ^{-1}x}}{\sqrt{1-x^{2}}} \\ \end{eqnarray*}$$