Find the derivative of:
(a) $y=( x^{3}-4x+1) ^{100}$ $\quad$ (b) $y=\cos \left( 3x-\dfrac{\pi}{4}\right)$

Solution (a) In the composite function $y=( x^{3}-4x+1)^{100},$ let $u=x^{3}-4x+1.$ Then $y=u^{100}.$ Now $\dfrac{dy}{du}$ and $\dfrac{du}{dx}$ are $$\begin{eqnarray*} \dfrac{dy}{du}=\dfrac{d}{du}u^{100}=100u^{99}\underset{\underset{\color{#0066A7}{\hbox {\(u=x^{3}-4x+1\)}}}{\color{#0066A7}{\uparrow}}}{=}100(x^{3}-4x+1)^{99} \end{eqnarray*}$$

and $$\begin{eqnarray*} \dfrac{du}{dx}=\dfrac{d}{dx}( x^{3}-4x+1) =3x^{2}-4 \end{eqnarray*}$$

We use the Chain Rule to find $\dfrac{dy}{dx}.$ $$\begin{eqnarray*} && \dfrac{dy}{dx}\underset{\underset {\color{#0066A7} {\hbox{ Chain Rule}}} {\color{#0066A7}{{\uparrow}}}}{=}\dfrac{dy}{du}\cdot \dfrac{du}{dx}=100( x^{3}-4x+1) ^{99}(3x^{2}-4)\\ \end{eqnarray*}$$

(b) In the composite function $y=\cos \left( 3x-\dfrac{\pi }{4} \right)$, let $u=3x-\dfrac{\pi }{4}.$ Then $y=\cos u$ and $$\begin{eqnarray*} && \dfrac{dy}{du}=\dfrac{d}{du}\cos u=-\sin u \underset{\underset{\color{#0066A7} {\hbox{\(u=3x-{\dfrac{\pi}{4}}\)}}}{\color{#0066A7}{{\uparrow}}}}{=} -\sin \left( 3x-\dfrac{\pi}{4}\right) \quad\hbox{and}\quad \dfrac{du}{dx}=\dfrac{d}{dx}\left( 3x-\dfrac{\pi }{4}\right) =3\\ \end{eqnarray*}$$

Now we use the Chain Rule. $$\begin{eqnarray*} && \dfrac{dy}{dx}\underset{\underset{\color{#0066A7}{\hbox{Chain Rule}}}{\color{#0066A7}{{\uparrow}}}}{=}\dfrac{dy}{du}\cdot \dfrac{du}{dx}=-\!\sin\! \left(3x-\dfrac{\pi }{4}\right) \cdot 3=-3\sin \left(3x-\dfrac{\pi }{4}\right)\\ \end{eqnarray*}$$