Second derivative via repeated product rule

Using the product rule on $f(x) = x^2 e^x$ with $u = x^2$, $u' = 2x$, $v = e^x$, $v' = e^x$, we get $f'(x) = 2x e^x + x^2 e^x$. This can be written as $f'(x) = (x^2 + 2x)e^x$.

Now differentiate $f'(x) = (x^2+2x)e^x$ using the product rule again: with $u = x^2+2x$, $u' = 2x+2$, $v = e^x$, $v' = e^x$, we get $f''(x) = (2x+2)e^x + (x^2+2x)e^x$. Combining like terms gives $f''(x) = (x^2 + 4x + 2)e^x$.