How can I show this? This is part of an assigned task for which I am to find the Taylor Series expansion of $f(x) = 1 / (5x^5 + 1)$ around $0$. I have noticed by studying the $n$th derivatives of this function that for $n$ is not a multiple of $5$, the derivative evaluates to $0$.
I believe this is useful but how can I prove this? I believe it may be related to the fact that odd derivatives of even functions at $x=0$ are $0$. Is there a theorem for the function that is $1/(5x^5 + 1)$ that says its derivatives evaluate to $0$ around $0$?