What is the derivative of $x!^{x!^{x!^{x!^{x!^{x!^{x!^{.{^{.^{.}}}}}}}}}}$

What is the derivative of $$x!^{x!^{x!^{x!^{x!^{x!^{x!^{.{^{.^{.}}}}}}}}}}$$

My effort:

Let $$g(x)=x!^{x!^{x!^{x!^{x!^{x!^{x!^{.{^{.^{.}}}}}}}}}}\implies g(x)=x!^{g(x)}$$ Taking natrual log on both sides, $$\ln(g(x))=g(x)\cdot\ln(x!)$$ Differentiating, $$\frac{1}{g(x)}\cdot g'(x)=g'(x)\cdot\ln(x!)+g(x)\cdot\frac{1}{x!}\cdot x!\cdot\psi^{(0)}(x+1)$$ $$\implies g'(x)\left[\frac{1}{g(x)}-ln(x!)\right]=g(x)\cdot\psi^{(0)}(x+1)$$ So does isolating $g'(x)$ give me the correct solution? If not, how can I solve for the differential?

Edit: The gamma function is indeed implicitly assumed when the factorial function is used.

Your definition of $g(x)$ does not make sense for $x! \gt e^{(1/e)} \approx 1.44467$ because it does not converge as seen in the answer to this question and this question. It is less than this in about the ranges $-4.970 \lt x \lt -4.103$ and $-0.380 \lt x \lt 1.614$. Within those ranges, you are doing fine.

I would say that everything looks fine, as long as you define that $x! = \Gamma(x+1)$.