<s>[INST] 10th derivative of a function 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.  
 [/INST] \begin{align}
f(x) &= \ln(2 + x^{2}) = \ln 2 + \ln\left( 1 + \frac{x^{2}}{2}\right) \\
&= \ln 2 + \sum_{k=1}^{\infty} \frac{(-1)^{k-1} \, x^{2k}}{2^{k} \, k} \\
&= \ln 2 + \frac{x^{2}}{2} - \frac{x^{4}}{8} + \frac{x^{6}}{24} - \frac{x^{8}}{64} + \frac{x^{10}}{160} - \cdots  \\
f(x) &= \ln 2 + \frac{x^{2}}{2!} - 3 \, \frac{x^{4}}{4!} + 30 \, \frac{x^{6}}{6!} - 630 \, \frac{x^{8}}{8!} + 22680 \, \frac{x^{10}}{(10)!} - 1247400 \, \frac{x^{12}}{(12)!} + \cdots
\end{align}
From this series expansion it can be determined that
\begin{align}
f^{(2n+1)}(0) &= 0 \hspace{10mm} \text{for} \quad  n \geq 0 \\
f^{(2n)}(0) &= \frac{(-1)^{n-1} \, (2n)!}{2^{n} \, n} = \frac{(-1)^{n-1} \, (2n-1)!}{2^{n-1}} \hspace{10mm} \text{for} \quad n \geq 1 \\
f(0) &= \ln 2.
\end{align}
</s>