Find the derivative of:
(a) $f( x) =e^{x}(x^{2}+1) ^{3}$ $\quad$ (b) $g( x) =\left( \dfrac{3x+2}{4x^{2}-5}\right)^{5}$

Solution (a) The function $f$ is the product of $e^{x}$ and $(x^{2}+1) ^{3}$, so we first use the Product Rule. $$\begin{eqnarray*} && f^\prime (x) \underset{\underset{\color{#0066A7}{\hbox{Product Rule}}}{{\color{#0066A7}{\uparrow}}}}{=} e^{x}\left[ \dfrac{d}{dx}(x^{2}+1) ^{3}\right] +\left[ \dfrac{d}{dx}e^{x}\right] (x^{2}+1) ^{3}\\ \end{eqnarray*}$$

204

To complete the solution, we use the Power Rule for Functions to find $\dfrac{d}{dx}(x^{2}+1) ^{3}$. $$\begin{eqnarray*} f^\prime ( x) &=& e^{x}\left[ 3(x^{2}+1) ^{2}\cdot \dfrac{d}{dx}(x^{2}+1) \right] +e^{x}(x^{2}+1) ^{3}\qquad {{\color{#0066A7}{\hbox{Power Rule for Functions}}}} \\ &=& e^{x}[3(x^{2}+1) ^{2}\cdot 2x] + e^{x}(x^{2}+1) ^{3} = e^{x}[6x(x^{2}+1) ^{2}+(x^{2}+1) ^{3}]\\ &=& e^{x}(x^{2}+1) ^{2}[6x+x^{2}+1] =e^{x}(x^{2}+1) ^{2}( x^{2}+6x+1) \end{eqnarray*}$$

(b) $g$ is a function raised to a power, so we begin with the Power Rule for Functions. $$\begin{eqnarray*} g^\prime ( x) &=&\dfrac{d}{dx}\left( \dfrac{3x+2}{4x^{2}-5}\right) ^{5} = 5\left( \dfrac{3x+2}{4x^{2}-5}\right) ^{4}\left[ \dfrac{d}{dx}\left( \dfrac{3x+2}{4x^{2}-5}\right) \right] \,\, {{\color{#0066A7}{\hbox{Power Rule for Functions}}}}\\ &=&5\left( \dfrac{3x+2}{4x^{2}-5}\right) ^{4}\left[ \dfrac{( 3) ( 4x^{2}-5) -(3x+2) (8x) }{(4x^{2}-5) ^{2}}\right]\quad \qquad {{\color{#0066A7}{\hbox{Quotient Rule}}}}\\ &=&\dfrac{5(3x+2) ^{4}[ (12x^{2}-15) -( 24x^{2}+16x) ] }{(4x^{2}-5) ^{6}}\\ &=&\dfrac{5(3x+2) ^{4}[ -12x^{2}-16x-15] }{( 4x^{2}-5) ^{6}} \end{eqnarray*}$$