I would like to calculate the $n$-th derivative of $\dfrac{e^{ax}}{\ln(cx)}$
I tried to calculate it in this way:
$$(fg)^{(n)}(x)=\sum_{k=0}^{n}\binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$$
$$\frac{\mathrm{d}^n}{\mathrm{d}x^n}\frac{1}{\ln(cx)}=\frac{1}{x^n\ln(cx)^{n+1}}\sum_{k=0}^{n}(-1)^{k}k!\left[{n\atop k}\right]\ln(cx)^{n-k}$$
And $$\frac{\mathrm{d}^n}{\mathrm{d}x^n}e^{ax}=a^n e^{ax}$$
So
$$\frac{\mathrm{d}^n}{\mathrm{d}x^n}\frac{e^{ax}}{\ln(cx)}=e^{ax}\sum_{k=0}^{n}\binom{n}{k}\ \frac{a^{n-k}}{x^{k}\ln\left(cx\right)^{k+1}}\sum_{j=0}^{k}\left(-1\right)^{j}j! \left[{k\atop j}\right]\ln\left(cx\right)^{k-j}$$
Is there a way to eliminate one of the summations?
