Find the second and third derivatives of $y=( 1+x^{2})e^{x}$.

Solution In Example 1, we found that $y^\prime =(1+x^{2}) e^{x}+2xe^{x}=(x^{2}+2x+1) e^{x}$. To find $y''$, we use the Product Rule with $y^\prime$. $$\begin{eqnarray*} y'' &=&\dfrac{d}{dx}[ (x^{2}+2x+1) e^{x}] \underset{\underset{\color{#0066A7}{\text{Product Rule}}}{\color{#0066A7}{\uparrow }}}{=} (x^{2}+2x+1) \left( \dfrac{d}{dx}e^{x}\right) +\left[ \dfrac{d}{dx} ( x^{2}+2x+1) \right] e^{x} \\ &=&(x^{2}+2x+1) e^{x}+( 2x+2) e^{x}=(x^{2}+4x+3) e^{x}\\[5pt] y''' &=&\dfrac{d}{dx}[ ( x^{2}+4x+3) e^{x}] \underset{\underset{\color{#0066A7}{\text{Product Rule}}}{\color{#0066A7}{\uparrow }}}{=}(x^{2}+4x+3) \dfrac{d}{dx}e^{x}+\left[ \dfrac{d}{dx}(x^{2}+4x+3) \right] e^{x}\\ &=&( x^{2}+4x+3) e^{x}+\left( 2x+4\right) e^{x}=( x^{2}+6x+7) e^{x} \end{eqnarray*}$$