Using the Power Rule for Functions to Find a Derivative

1. If \(f( x) =(3-x^{3}) ^{-5}\), then \[ \begin{eqnarray*} f^\prime ( x) &=&\dfrac{d}{dx}(3-x^{3}) ^{-5} \underset{\underset{\color{#0066A7}{\hbox{for Functions}}}{\underset{\color{#0066A7}{\hbox{Power Rule}}}{{\color{#0066A7}{\uparrow}}}}}{=} -5(3-x^{3}) ^{-5-1}\cdot \dfrac{d}{dx}(3-x^{3}) \\ &=& -5(3-x^{3}) ^{-6}\cdot ( -3x^{2}) = 15x^{2}(3-x^{3}) ^{-6}=\dfrac{15x^{2}}{( 3-x^{3}) ^{6}} \end{eqnarray*} \]
2. If \(f( \theta ) =\cos ^{3}\theta \), then \(f( \theta ) =( \cos \theta ) ^{3},\) and \[ \begin{eqnarray*} f^\prime ( \theta ) &=&\dfrac{d}{d\theta }( \cos \theta ) ^{3} \underset{\underset{\color{#0066A7}{\hbox{for Functions}}}{\underset{\color{#0066A7}{\hbox{Power Rule}}}{{\color{#0066A7}{\uparrow}}}}}{=} 3( \cos \theta ) ^{3-1}\cdot \dfrac{d}{ d\theta }\cos \theta =3\cos ^{2}\theta \cdot ( -\sin \theta)\\ &=&-3\cos ^{2}\theta \sin \theta \end{eqnarray*} \]