$$f(x) = \exp(Ax^{-2/\alpha}\rho(x,\alpha,r)) $$
where $$\rho(x,\alpha,r) = \int^\infty_{r^2x^{-2/\alpha}} \frac{1}{1+u} du$$
I have to find the $n_{th}$ derivate of $f(x)$, i.e. $$(-1)^n \frac{d^n f(x)}{fx^n}$$
Notice, the tricky part is the $x$ in the limit of the $rho$ function.
All other elements in the equations are assumed constant except $x$
The function $\rho(x,\alpha,r)$ is defined only if $x\ge 0$ and is convergent if so. Then it would be$$\rho(x,\alpha,r)=\ln(1+r^2x^{-\dfrac{2}{\alpha}})$$by substituting in $f(x)$ we have$$f(x)=(1+r^2x^{-\dfrac{2}{\alpha}})e^{Ax^{-\dfrac{2}{\alpha}}}$$the nth differentiation can be calculated through Taylor's series but not explicitly. Therefore:$$f(x)=1+\sum_{n=1}^{\infty}A^{n-1}\dfrac{A+nr^2}{n!}x^{-\dfrac{2n}{\alpha}}$$The case where $-\dfrac{2}{\alpha}\in\Bbb N$ is simple and for the other cases we have$$(-1)^n\frac{d^n f(x)}{dx^n}=(-1)^n\sum_{k=1}^{\infty}(-\dfrac{2k}{\alpha})(-\dfrac{2k}{\alpha}-1)...(-\dfrac{2k}{\alpha}-n+1)A^{k-1}\dfrac{A+kr^2}{k!}x^{-\dfrac{2k}{\alpha}-n}$$and using the Gamma function we finally obtain$$(-1)^n\frac{d^n f(x)}{dx^n}=\sum_{k=1}^{\infty}\dfrac{\Gamma(-\dfrac{2k}{\alpha}+1)}{\Gamma(-\dfrac{2k}{\alpha}-n+2)}A^{k-1}\dfrac{A+kr^2}{k!}x^{-\dfrac{2k}{\alpha}-n}$$