TITLE: Find the tenth derivative of $x^2e^x$
QUESTION [5 upvotes]: If $$f(x) = x^2e^x$$
find $$f^{({10})}x$$
I'm not familiar with doing the series expansion for this and the solution that was provided to me did not help me at all, as it wasn't explained. I think it's "Higher Leibnitz Rule", but I'm unsure.
I could differentiate the function 10 times, but the solutions uses summation. How would I go about doing this?

REPLY [5 votes]: For fun, since good ways have been described, we do not so good. First a small change of notation. Let $f(x)=x^2e^x$. We find $f^{(10)}(a)$, or more generally
$f^{(n)}(a)$.
Expand $e^x$ in a power series about $x=a$. We get 
$$e^x=\sum_0^\infty \frac{e^a}{n!}(x-a)^n.\tag{1}$$ 
Expand $x^2$ about $x=a$. We get 
$$x^2=a^2+2a(x-a)+(x-a)^2.\tag{2}$$
For $n\ge 2$, the coefficient of $(x-a)^n$ in the product of (1) and (2) is
$$e^a\left(\frac{1}{n!}a^2+\frac{2}{(n-1)!}a+\frac{1}{(n-2)!}\right).$$
The above coefficient is $\frac{1}{n!}$ times the derivative of $x^2f(x)$ at $x=a$. It follows that the required derivative is
$$e^a\left(a^2 +2na+n(n-1)\right).$$
Remark: By expanding $x^k$ about $x=a$, we can in the same way find the $n$-th derivative of $x^ke^x$ at $x=a$.