Find the 8th derivative of the function $h(x) = xe^x $using sequences
Solution 1:
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?
Solution 2:
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
$$ (xe^x)^{(8)}(x)=(x+8)e^x. $$