In the course of using the Cauchy Residue Theorem for a particular problem, I have a pole of order $n$ that yields the residue
$$ R(z)= \lim_{s\rightarrow z} \frac{(s-z)^n}{s[A+Bf(s)]^n} $$
where $A$, $B$, and $z$ which is a zero of $[A+Bf(s)]$ are real constants, $s$ is the Laplace variable, and $f(s)$ is a known infinitely differentiable function of $s$. The indeterminate limit has to be resolved with $n$ applications of L'Hôpital's Rule which requires the $n$th derivative of the term in the denominator of the expression above.
In short therefore, I would like a compact (or at least fairly compact) expression for
$$ \frac{d^n}{dx^n}\Bigl[x\Bigl(A+Bf(x)\Bigr)^n\Bigr] = A^n\frac{d^n}{dx^n}\Bigl[x\Bigl(1+rf(x)\Bigr)^n\Bigr] $$
where $r=B/A$. All of my attempts have resulted in one horrendous book-keeping mess after another. Any help will as always be deeply appreciated.
With a binomial expansion, I've managed to get it down to
$$ \frac{d^n}{dx^n}\Bigl[x\Bigl(A+Bf(x)\Bigr)^n\Bigr] = A^n\sum_{k=0}^{n} {n\choose k} r^{n-k} \frac{d^n}{dx^n}\Bigl[xf(x)\Bigr]^{n-k} $$
but then each term by term derivative inside appears to open up another monstrosity!
UPDATE: I was WAY overthinking this and thanks to user Raoul pointing it out, the Limit of a power is the power of the limit rule CONSIDERABLY simplifies my life!
