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, \[ \begin{eqnarray*} y^\prime =\frac{d}{dx}[(1+x^{2}) e^{x}] \underset{\underset{\color{#0066A7}{\text{Product Rule}}}{\color{#0066A7}{\uparrow}}}{=} (1+x^{2}) \left[ \frac{d}{dx}e^{x}\right] +\left[\frac{d}{dx}(1+x^{2})\right] e^{x}=( 1+x^{2}) e^{x}+2xe^{x}\\ \end{eqnarray*} \]
At this point, we have found the derivative, but it is customary to simplify the answer. Then \[ \begin{eqnarray*} y\prime \underset{\underset{\color{#0066A7}{\text{Factor out}~e^{x}.}}{\color{#0066A7}{\uparrow }}}{=} ( 1+x^{2}+2x) e^{x} \underset{\underset{\color{#0066A7}{\text{Factor}.}}{\color{#0066A7}{\uparrow }}}{=} ( x+1) ^{2}e^{x} \end{eqnarray*} \]