What's the derivative of order 2013 of $f(x)=\frac{x^5}{1+x^6}$? What's the derivative of order 2013 of $f:\mathbb R\to \mathbb R$, $f(x)=\frac{x^5}{1+x^6}$ at the point 0? This question seems really difficult to solve directly, I need some others ideas how to solve it.
I need help!
thanks a lot.
 A: Hint: The Maclaurin series expansion of $\dfrac{1}{1+x^6}$ is
$$1-x^6 +x^{12}-x^{18}+x^{24}-x^{30}+ \cdots.$$
Multiply by $x^5$. Now you have the Maclaurin series expansion of your function.
If you know the derivatives at $x=0$, then you know the Maclaurin series expansion. So if you know the Maclaurin series expansion $\dots$.
Note that $2013$ is not of the form $6k+5$, making the calculation particularly easy. 
A: This function $f$ is the product of two functions $x \mapsto x^5$ and $x \mapsto (1 + x^6)^{-1}$. Then you can use the formula $$(fg)^{(n)} = \sum_{i=0}^n \mathrm C_n^i f^{(i)}g^{(n-i)}$$
A: $$\frac{x^{5}}{1+x^{6}}=x^{5}(1+x^{6})^{-1}=x^{5}\sum_{n=0}^{\infty}(-x^{6})^{n}=\sum_{n=0}^{\infty}(-1)^{n}x^{6n+5}=x^{5}-x^{11}+x^{17}-\ldots$$
Differentiating this $2013$ times will decrement each power by $2013$. Only a term of degree $2013$ will contribute to this derivative at zero, because any lower powers will have been differentiated to zero as constants, and any higher powers will evaluate to zero at zero.  
Does such a term exist?  
NB: This expansion is only valid for $|x|<1$, but since the derivative is being evaluated at zero we're okay.
A: The same formula defines a function on the complexes. The denominator factors into
$$x^6 + 1 = \prod_{n=0}^5 (x - \zeta^{2n+1}) $$
where $\zeta$ is a primitive 12-th root of unity, so $f(x)$ has simple poles. The residue at each pole is given by the formula
$$ g(x) = \frac{x^5}{(x^6 + 1)'} = \frac{1}{6} $$
and so we have the partial fractions decomposition
$$f(x) = \sum_{n=0}^5 \frac{1}{6} \frac{1}{x - \zeta^{2n+1}} $$
Using the formula
$$ \frac{d^k}{dx^k} \frac{1}{x-a} = (-1)^k \frac{k!}{(x-a)^{k+1}} $$
we therefore get
$$f^{(2013)}(x) = -\frac{2013!}{6} \sum_{n=0}^5 \frac{1}{(x - \zeta^{2n+1})^{2014}} $$
