I have a question:

Given: $\displaystyle f(x)=x^7 \cos(x^9)$

Find $\displaystyle f^{79}(0)$, the 79-th derivative of f(x) at x=0

I know the answer is $\displaystyle \frac{79!}{8!}$ but I am not sure of all the steps.

I found the Taylor Series representation to be:

$\displaystyle \sum\frac{(-1)^nx^{18n+7}}{2n!}$

Any enlightenment would be much appreciated.