Find $y^\prime$ if $y=( 1+x^{2}) e^{x}\!$.

Solution The function $y$ is the product of two functions: a polynomial, $f( x) =1+x^{2}$, and the exponential function $g( x) =e^{x}$. By the Product Rule, $$\begin{eqnarray*} &&y^\prime =\frac{d}{dx}[(1+x^{2}) e^{x}] {=} (1+x^{2}) \left[ \frac{d}{dx}e^{x}\right] +\left[\frac{d}{dx}(1+x^{2})\right] e^{x}=( 1+x^{2}) e^{x}+2xe^{x}\\[-6.4pt] &&\hspace{4.9pc}\underset{\color{#0066A7}{\scriptsize \hbox{Product Rule}}}{\color{#0066A7}{\uparrow}} \end{eqnarray*}$$

At this point, we have found the derivative, but it is customary to simplify the answer. Then $$\begin{eqnarray*} &&y\prime {=} ( 1+x^{2}+2x) e^{x} {=} ( x+1) ^{2}e^{x}\\[-6.4pt] &&\hspace{-0.7pc}\underset{\color{#0066A7}{\scriptsize \hbox{Factor out \(e^{x}\).}}}{\color{#0066A7}{\uparrow }}\hspace{2.5pc}\underset{\color{#0066A7}{\scriptsize \hbox{Factor}.}}{\color{#0066A7}{\uparrow }} \end{eqnarray*}$$