# 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.

• You're on the right track; so then, what is the Taylor series for $f$ at $x = 0$? Aug 4, 2015 at 14:06
• Maple says $$928972800\,{\frac {{x}^{8}}{ \left( {x}^{2}+2 \right) ^{9}}}-812851200 \,{\frac {{x}^{6}}{ \left( {x}^{2}+2 \right) ^{8}}}+290304000\,{\frac {{x}^{4}}{ \left( {x}^{2}+2 \right) ^{7}}}-36288000\,{\frac {{x}^{2}}{ \left( {x}^{2}+2 \right) ^{6}}}+725760\, \left( {x}^{2}+2 \right) ^{- 5}-$$ $$371589120\,{\frac {{x}^{10}}{ \left( {x}^{2}+2 \right) ^{10}}}$$ Aug 4, 2015 at 14:06
• i get $$22680$$ Aug 4, 2015 at 14:10
• I know the Taylor series for $ln(1+t)$ but I'm don't know how to convert it to $t=1+x^2$ in a way that can be helpful. Aug 4, 2015 at 14:12
• It's certianly larger and more computationally rigorous then considering $f(x)=\ln(-2+x^{2})$; in that example $f^{(10)}=2^{-4}$. Aug 4, 2015 at 14:22

We have $$\frac{2x}{x^2+2} = \frac{x}{1+\frac{x^2}{2}} = x\bigg(1-\frac{x^2}{2} + \frac{x^4}{4} - \frac{x^6}{8} + \frac{x^8}{16}+\cdots \bigg).$$ Then integrating we get $$\log(x^2+2)-\log2 = \int_0^x \frac{2t}{t^2+2}\; dt = \frac{x^2}{2}-\frac{x^4}{2\cdot 4} + \frac{x^6}{4\cdot 6} - \frac{x^8}{8\cdot 8} + \frac{x^{10}}{10\cdot 16}+\ldots$$
Answer: $22680$ :)
Therefore if we differentiate this series nine times, we see that $f^{(10)}(0)$ is the constant term, which is $\frac{9!}{16}=22680$