How find the 2001th and 2003th derivatives of $f(x)= \frac{x^5}{1+x^6}$ Lef $ f: \mathbb{R} \rightarrow \mathbb{R} $ be defined by $f(x)= \dfrac{x^5}{1+x^6}$. I want to find the 2001th and 2003th derivatives of $f$ at the point $x=0$. I tried to use the chain rule but I do not see the pattern and I do not know what theorem to use.
 A: Hint: Using the formula for the sum of a geometric series, we can write $$f(x) = \dfrac{x^5}{1+x^6} = \sum_{n = 0}^{\infty}x^5(-x^6)^n = \sum_{n = 0}^{\infty}(-1)^nx^{6n+5}.$$
This converges absolutely for $|x| < 1$, so you can differentiate term by term and plug in $x = 0$. After you differentiate $2001$ or $2003$ times and plug in $x = 0$, at most one of those terms will be non-zero.
A: The Taylor series expansion for $$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n.$$ On the other hand (as others have also expressed)
$$f(x)=\underbrace{x^5(1+x^{6})^{-1}=\sum_{k=0}^{\infty}(-1)^kx^{6k+5}}_{\text{geometric series}}.$$
Now we can compare the two series and obtain:
$$f^{(n)}(0)=\begin{cases}0 & \text{ if } n \neq 6k+5\\ n!(-1)^{k}& \text{ if } n = 6k+5\end{cases}.$$
Since $2001=6(333)+3$ and $2003=6(333)+5$, so
$$f^{(2001)}(0)=0 \quad \text{ and } \quad f^{(2003)}(0)=(-1)^{333}(2003!)=-2003! $$
A: Note that $\frac{x^5}{1+x^6}=x^5 \sum_{k=0}^\infty (-1)^k x^{6k}=\sum_{k=0}^\infty x^{6k+5}$ by using the geometric series expansion of $\frac1{1+y}$ for $y=x^6$. Then try to write 2001 and 2003 in the form $6k+r$ with $0\leq r<6$.
