TITLE: 10th derivative of a function
QUESTION [10 upvotes]: I want to find $f^{(10)}(0)$ where $f(x)=\ln(2+x^2)$.  
I know that it can be done "by hand", but I believe there is a smarter way.  
I think I should use Taylor series and the fact that $f^{(n)}(0)=a_n*n!$ , but I'm not sure how.

REPLY [6 votes]: \begin{align}
f'(x)=\frac{x}{1+\frac{x^{2}}{2}}=\sum_{k=0}^{\infty }(-1)^{k}\left ( \frac{x^{2k+1}}{2^{k}} \right )=x-\frac{x^{3}}{2}+\cdots+\frac{x^{9}}{2^{4}}-\frac{x^{11}}{2^{5}}+\cdots 
                \end{align}
Therefore if we differentiate this series nine times, we see that $f^{(10)}(0)$ is the constant term, which is $\frac{9!}{16}=22680$