Find $y^\prime$ if:

(a) $y=e^{x^{2}-4}$ $\quad$ (b) $y=\sin (4e^{x})$

Solution (a) For $y=e^{x^{2}-4},$ we let $u=x^{2}-4.$ Then $y=e^{u}$ and $$\begin{eqnarray*} && \dfrac{dy}{du}=\dfrac{d}{du}e^{u}=e^{u}\underset{\underset{\color{#0066A7}{{{\hbox{\(u=x^{2}-4\)}}}}} {\color{#0066A7}{\uparrow}}}{=}e^{x^{2}-4} \quad\hbox{and}\quad \dfrac{du}{dx}=\dfrac{d}{dx}( x^{2}-4) =2x\\ \end{eqnarray*}$$

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Using the Chain Rule, we get $$y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=e^{x^{2}-4}\cdot 2x=2xe^{x^{2}-4}$$

(b) For $y=\sin (4e^{x})$, we let $u=4e^{x}$. Then $y=\sin u$ and $$\begin{eqnarray*} && \dfrac{dy}{du}=\dfrac{d}{du}\sin u=\cos u \underset{\underset{\color{#0066A7}{{{\hbox{\(u=4e^{x}\)}}}}}{\color{#0066A7}{\uparrow}}}{=}\cos (4e^{x})\quad\hbox{and}\quad\dfrac{du}{dx}=\dfrac{d}{dx}(4e^{x})=4e^{x}\\ \end{eqnarray*}$$

Using the Chain Rule, we get $$y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\cos (4e^{x}) \cdot 4e^{x}=4e^{x}\cos (4e^{x})$$