Using the Chain Rule with the Inverse Sine Function

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*} \]