If my math is correct:
$$\Gamma(f(x))=\displaystyle \int_0^\infty e^{-t}t^{f(x)-1}dt$$
$$\frac{\partial}{\partial x}\Gamma(f(x))=\displaystyle \int_0^\infty e^{-t}f^\prime (x)\textrm{ln}(t)t^{f(x)-1}dt$$
$$=f^\prime (x)\displaystyle \int_0^\infty e^{-t}\textrm{ln}(t)t^{f(x)-1}dt$$
But here I get stuck. The integral is tantalizingly close to the definition of $\Gamma(f(x))$, except for the $\textrm{ln}(t)$ term. I've tried integration by parts to resolve the integral but no luck so far.
I'm hoping to find a formula for $\frac{\partial}{\partial x}\Gamma(f(x))$ that can be expressed from the terms: $\Gamma(f(x))$, $\Gamma^\prime(x)$, or $\Gamma(x)$ and therefore can be calculated in the R program I'm running.
By the chain rule for derivatives. $$\partial_x\Gamma(f(x))=\Gamma^\prime(f(x))f^\prime(x). $$ Now using the digamma function defined by $\psi(x)=\partial_x\log\Gamma(x)$ we easily derive $\Gamma^\prime(x)=\Gamma(x)\psi(x)$; hence, $$\partial_x\Gamma(f(x))=\Gamma(f(x))\psi(f(x))f^\prime(x). $$
Edit:
Based on OPs comments we may also use this approach:
By the chain rule and definition of digamma $$\partial_x\log\Gamma(f(x))=\psi(f(x))f^\prime(x). $$ But it is also true that $$\partial_x\log\Gamma(f(x))=\frac{\partial_x\Gamma(f(x))}{\Gamma(f(x))}. $$ Combining these equalities and rearranging to solve for $\partial_x\Gamma(f(x))$ gives the desired result.