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.

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 $0\leq r<6$.
