Derivative for arbitrary number of times

Try

 f[t_] := Exp[k t] Sinh[t/n]/(Cosh[t/n] - a) 

and for example the third derivative f'''[0] evaluates like this

Derivative[3][f][0] (*-(3/((1 - a)^2 n^3)) + 1/((1 - a) n^3) + (3 k^2)/((1 - a) n)*) 

You can proceed to some extent, although not fully, by using the Leibniz formula $$ (f g)^{(n)}|_{t=0}=\sum_{p=0}^{n}\left(\begin{array}{l}{n} \\ {p}\end{array}\right) f^{(n-p)}|_{t=0} ~g^{(p)}|_{t=0} $$ Your $f(t)=e^{kt}$ and therefore $f^{(n-p)}|_{t=0}=k^{n-p}$. I couldn't find a simple form for the $p$th derivative of your $g(t)= \left(\frac{\sinh (t / n)}{\cosh (t / n)-a }\right) $, other than the fact that an even number of derivatives vanish at $t=0$. Therefore,

$$ \frac{d^n}{dt^n} \left(e^{k t} \times\frac{\sinh (t / n)}{\cosh (t / n)-a )}\right)\bigg|_{t=0}=\sum_{p=1~, ~p\in 2\mathbb{Z}^++1}^n \binom{n}{p}k^{n-p} \frac{d^p}{dt^p} \left(\frac{\sinh (t / n)}{\cosh (t / n)-a }\right)\bigg|_{t=0}~. $$ [Not sure whether this is a question for Mathematica though. At times you can get lucky – given a sequence of functions as a list, you can get the $n$th one in the sequence using FindSequenceFunction but this specific case is too complicated.]

Edit

We can go a step further with Leibniz rule : $$\frac{d^p}{dt^p} \left(\frac{\sinh (t / n)}{\cosh (t / n)-a }\right)\bigg|_{t=0}=\sum_{q=0~, ~q\in 2\mathbb{Z}^++1}^{q=p}\binom{p}{q} n^{q-p} \frac{d^q}{dt^q} \left(\frac{1}{\cosh (t / n)-a }\right)\bigg|_{t=0}$$

Therefore, $$ \frac{d^n}{dt^n} \left(e^{k t} \times\frac{\sinh (t / n)}{\cosh (t / n)-a )}\right)\bigg|_{t=0}=\sum_{p=1~, ~p\in 2\mathbb{Z}^++1}^n \binom{n}{p}k^{n-p} \sum_{q=0~, ~q\in 2\mathbb{Z}^++1}^{q=p}\binom{p}{q} n^{q-p} \frac{d^q}{dt^q} \left(\frac{1}{\cosh (t / n)-a }\right)\bigg|_{t=0}. $$ This is probably as general as it gets for arbitrary $n$.