I am trying to find a formula for the $n$'th derivative of $f(x)^{g(x)}$. I have tried using the formula bellow along with Leibniz rule without success.
$$D^n(f(g(x)))=\sum_{k=0}^{n-1} \binom {n-1}{k}D^{\left(n-k\right)}g\left(x\right)D^{\left(k\right)}\left(f'\left(g\left(x\right)\right)\right)$$
This is how I went about it:
Let $y=f(x)^{g(x)}$
This implies $\frac{y^\prime}{y}=g^\prime(x)\ln(f(x))+g(x)f^\prime(x)f(x)^{-1}$
Now, using Leibniz rule on the LHS as well as on the RHS (multiple times on the RHS since there is more than two functions multiplied) followed by the equation above for the $\ln(f(x))$ I got a long formula wich can be rearranged to obtain the value of $D^ny$, however it could be easily seen it didn't work for a lot of functions in the form $f(x)^{g(x)}$.
Is there a well known formula for what I am trying to find? Or does the method I am trying to use fail at some stage?
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