Find $y^\prime$ if:
(a) $y=5\cos ^{2}(3x+2)$ $\quad$ (b) $y=\sin ^{3}\left( \dfrac{\pi }{2}x\right)$

Solution (a) For $y=5\cos ^{2}(3x+2) ,$ we use the Chain Rule with $y=5u^{2}$, $u=\cos v$, and $v=3x+2.$ Then $y=5u^{2}=5\cos ^{2}v=5\cos ^{2}(3x+2)$ and $$\begin{eqnarray*} \dfrac{dy}{du} &=&\dfrac{d}{du}(5u^{2}) =10u \underset{\underset{\color{#0066A7}{{\hbox{\(v=3x+2\)}}}}{\underset{\color{#0066A7}{{\hbox{\(u=cos v\)}}}}{{\color{#0066A7}{\uparrow}}}}} {=} 10\cos (3x+2) \\ \dfrac{du}{dv} &=&\dfrac{d}{dv}\cos v=-\sin v \underset{\underset{{\color{#0066A7}{{\hbox{\(v=3x+2\)}}}}}{{\color{#0066A7}{\uparrow}}}}{=} -\sin (3x+2) \\ \dfrac{dv}{dx} &=&\dfrac{d}{dx}(3x+2) =3 \end{eqnarray*}$$

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Then $$\begin{eqnarray*} y^\prime &=& \dfrac{dy}{dx} \underset{\underset{\color{#0066A7}{\scriptsize\hbox{Chain Rule}}}{{\color{#0066A7}{\uparrow}}}}{=} \dfrac{dy}{du}\cdot \dfrac{du}{dv}\cdot \dfrac{dv}{dx}=[10\cos (3x+2)] [ -\!\sin (3x+2)] [ 3]\\ &=& -30\cos (3x+2) \sin (3x+2) \end{eqnarray*}$$

(b) For $y=\sin ^{3}\left( \dfrac{\pi }{2}x\right) ,$ we use the Chain Rule with $y=u^{3}$, $u=\sin v\,$, and $v=\dfrac{\pi }{2}x$. Then $y=u^{3}=\left( \sin v\right) ^{3}=\left[ \sin \left( \dfrac{\pi }{2}x\right) \right] ^{3}=\sin ^{3}\left( \dfrac{\pi }{2}x\right)$, and $$\begin{eqnarray*} \dfrac{dy}{du} &=&\dfrac{d}{du}u^{3}=3u^{2} \underset{\underset{\underset{\color{#0066A7}{{\hbox{\(v=\dfrac{\pi}{2}x\)}}}}{\color{#0066A7}{{\hbox{\(u=\sin v\)}}}}}{{\color{#0066A7}{\uparrow}}}}{=} 3\left[ \sin \left( \dfrac{\pi }{2}x\right) \right] ^{2} = 3 \sin^{2}\left(\frac{\pi}{2}x\right)\\ \dfrac{du}{dv} &=&\dfrac{d}{dv}\sin v=\cos v \underset{\underset{\color{#0066A7}{{\hbox{\(v=\dfrac{\pi}{2}x\)}}}}{{\color{#0066A7}{\uparrow}}}}{=}\cos \left( \dfrac{\pi}{2}x\right) \\ \dfrac{dv}{dx} &=&\dfrac{d}{dx}\left( \dfrac{\pi }{2}x\right) = \dfrac{\pi }{2} \end{eqnarray*}$$

Then $$\begin{eqnarray*} y^\prime &=& \dfrac{dy}{dx} \underset{\underset{\color{#0066A7}{\scriptsize\hbox{Chain Rule}}}{{\color{#0066A7}{\uparrow}}}}{=} \dfrac{dy}{du}\cdot \dfrac{du}{dv}\cdot \dfrac{dv}{dx}=3\sin ^{2}\left( \dfrac{\pi }{2}x\right) \cdot \cos \left( \dfrac{\pi }{2}x\right) \cdot \left( \dfrac{\pi }{2}\right) \\ &=&\dfrac{3\pi }{2}\sin ^{2}\!\left( \dfrac{\pi }{2}x\right) \cos \left( \dfrac{\pi }{2}x\right) \\ \end{eqnarray*}$$