Compute the derivatives of $\frac{d^{2\ell}}{dx^{2\ell}}\tanh(x)^{2k}$ in $x=0$ I would like to compute the derivatives $\frac{d^{2\ell}}{dx^{2\ell}}_{\vert x=0}\tanh(x)^{2k}$ at $x=0$ where $k,\ell\in \mathbb{N}$ positive integers with $\ell\geq k$.
I am not sure how to attack this problem seriously. Can you give me an instruction what to do in this situation?
Best wishes
Edit:
In case of $k>\ell$ we have $\frac{d^{2\ell}}{dx^{2\ell}}_{\vert x=0}\tanh(x)^{2k}=0$. (see the commentary and answer below). 
 A: We have:
$$\left.\frac{d^{2l}}{dx^{2l}}\tanh^{2k}(x)\right|_{x=0} =\frac{(2l)!}{2\pi}\int_{0}^{2\pi}\tanh^{2k}(e^{i\theta})e^{-2li\theta}\,d\theta=\frac{(2l)!}{2\pi i}\oint \frac{\tanh^{2k}(z)}{z^{2l+1}}\,dx$$
but if $k>l$, the last integrand function has no singularity inside the unit disk, so the RHS is zero.

To compute the whole Taylor series of $\tanh(z)$ in a neighbourhood of the origin, one usually exploits the Weierstrass product for the (hyperbolic) cosine function:
$$ \cosh(z) = \prod_{n\geq 0}\left(1+\frac{4z^2}{(2n+1)^2 \pi^2}\right)$$
hence by taking the logarithmic derivative of both sides:
$$\begin{eqnarray*} \frac{\tanh(z/2)}{4z} &=& \sum_{n\geq 0}\frac{1}{z^2+(2n+1)^2\pi^2}=\sum_{n\geq 0}\sum_{m\geq 0}\frac{(-1)^m z^{2m}}{\pi^{2m+2}(2n+1)^{2m+2}}\\&=&\sum_{m\geq 0}(-1)^m \frac{z^{2m}}{\pi^{2m+2}}\sum_{n\geq 0}\frac{1}{(2n+1)^{2m+2}}\\&=&\sum_{m\geq 0}(-1)^m \frac{z^{2m}}{\pi^{2m+2}}\left(1-\frac{1}{4^{m+1}}\right)\zeta(2m+2)\end{eqnarray*}$$
then:

$$ \tanh(z) = \sum_{m\geq 0}(-1)^m z^{2m+1}\left(4^{m+1}-1\right)\frac{2\,\zeta(2m+2)}{\pi^{2m+2}}.\tag{1}$$


Since $\frac{d}{dz}\tanh(z)=1-\tanh(z)^2$, if we compute the Taylor series of $\tanh^2(z)$ we also have the Taylor series of $\tanh^3(z)$, since $\frac{d}{dz}\tanh^2(z) = 2(1-\tanh^2 z)\tanh(z) = 2\left(\tanh(z)-\tanh^3(z)\right).$
Once we have the Taylor series of $\tanh(z),\tanh^2(z),\tanh^3(z)$ we also have the Taylor series of $\tanh^4(z)$ and so on. So the problem in the case $l\geq k$ boils down to computing the Taylor series of $\tanh^2(z)$ and going through a recurrence relation. Now $(1)$ gives:

$$\begin{eqnarray*} \tanh^2(z)=1-\frac{d}{dz}\tanh(z) &=& 1-\sum_{m\geq 0}(-1)^m z^{2m}\left(4^{m+1}-1\right)\frac{(4m+2)\,\zeta(2m+2)}{\pi^{2m+2}}\\&=&\sum_{m\geq 1}(-1)^{m+1} z^{2m}\left(4^{m+1}-1\right)\frac{(4m+2)\,\zeta(2m+2)}{\pi^{2m+2}}.\end{eqnarray*} $$

