The n-th derivative of x² e^x

Idea. The Leibniz formula is the "binomial theorem for derivatives". With $u=x^2$ we have $u'=2x$, $u''=2$ and $u^{(k)}=0$ for $k\ge3$. Also $v=e^x$ satisfies $v^{(j)}=e^x$ for all $j$.

Computation. Only $k=0,1,2$ survive:
$$f^{(n)}(x)=\binom n0x^2e^x+\binom n1\,2x\,e^x+\binom n2\,2\,e^x=e^x\left(x^2+2nx+n(n-1)\right).$$

Check. For $n=1$: $e^x(x^2+2x)$, which is the product rule. For $n=2$: $e^x(x^2+4x+2)$.

Remark. The same method shows that the $n$-th derivative of $P(x)e^x$, with $P$ a polynomial, is $e^x$ times a polynomial of the same degree.