Find the 8th derivative of the function $h(x) = xe^x $using sequences How do you find the 8th derivative of $h(x) = x e^x $ without doing it "manually".
I know that $\displaystyle e^x = \sum_{i=0}^n \frac{x^n}{n!} $
so that $\displaystyle h(x) = x \sum_{i=0}^n \frac{x^n}{n!}  = \sum_{i=0}^n \frac{x^{n+1}}{n!}  $ 
I can't figure out what to do from here. 
Edit: I forgot to mention that we want to find the derivative evaluated at $x=0$, so $h^{(8)}(0)$.
 A: Hint:
Look at the expanded form of $xe^x$:
$$x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + \frac{x^5}{4!} + \ldots$$
Since you're only evaluating at $x = 0$, the only terms in the power series that matter after you take derivatives are the ones that don't equal zero when you put in $x = 0$. Which terms are that?
A: One may use the Cauchy formula

$$
(fg)^{(n)}(x)=\sum_{k=0}^n\binom{n}{k}f^{(n-k)}(x)g^{(k)}(x)
$$ 

with
$$
f(x)=e^x,\quad f^{(n-k)}(x)=e^x,\quad g(x)=x,\quad g'(x)=1,\, g^{(k)}(x)=0, \, k>1
$$ giving easily

$$
(xe^x)^{(8)}(x)=(x+8)e^x.
$$

A: If you differentiate $xe^x$ you get $(x+1)e^x$, so differentiating again will always produce an expression of the form $(x+c_n)e^x$.
Let's find a recursion formula:
$$
D\bigl((x+c_n)e^x\bigr)=e^x+(xe^x+c_n)e^x=(x+c_n+1)e^x
$$
so $c_{n+1}=c_n+1$ and, since $c_0=0$, we have
$$
c_n=n
$$
Thus the $n$-th derivative of $f(x)=xe^x$ at $0$ is
$$
f^{(n)}(0)=n
$$
This is confirmed by the Taylor expansion:
$$
xe^x=\sum_{n\ge1}\frac{x^{n}}{(n-1)!}=
\sum_{n\ge0}\frac{f^{(n)}(0)}{n!}x^n
$$
so
$$
f^{(n)}(0)=\frac{n!}{(n-1)!}=n
$$
for $k\ge1$.
