# 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.

• @Brenton no, he isnt. He multiplies by the derivative to the right, which is correct. He says the derivative is $\frac{1}{x!}\cdot x!\cdot\psi^{(0)}(x+1)$ Dec 27 '15 at 5:37
• @BrevanEllefsen Sorry, I missed the $\psi$ term. Dec 27 '15 at 5:38
• What is $x!$ for $x$, say, $\sqrt2$? Dec 27 '15 at 5:40
• @PrzemysławScherwentke $x!=\Gamma (x+1)$ (Gamma function). Dec 27 '15 at 5:40
• @PrzemysławScherwentke he clearly defines it as $\Gamma(\sqrt{2}+1)$... he uses the derivative of the gamma function in his proof Dec 27 '15 at 5:41

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.

• Clear. Thanks for the heads up! Dec 27 '15 at 11:21
• What if he wanted to allow complex numbers for $x$? Dec 27 '15 at 15:04
• @GEdgar: I don't know where the tower converges in the complex plane, but presumably somebody does. Once it converges you have to worry about branches of the log function as well. Dec 27 '15 at 15:07
• @GEdgar if$$|W(-\ln(x!))|<1$$it converges, where $W$ is the Lambert $W$ function. math.stackexchange.com/questions/1674027/… Jun 30 '16 at 16:29

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

• That indeed is my implicit assumptions. Thanks! Dec 27 '15 at 11:24

If you define $x!$ to be equal to $\int_{0}^{\infty}e^{-t}t^x \mathrm{d}t$, then you can say the following:

$y = (x!)^{y} = (\int_{0}^{\infty}e^{-t}t^x \mathrm{d}t)^y$

Raising both sides to the power of $\frac{1}{y}$:

$y^\frac{1}{y} = \int_{0}^{\infty}e^{-t}t^x \mathrm{d}t$.

Since the derivative of $y^\frac{1}{y}$ with respect to $y$ is $-y^\frac{1}{y}(\ln(y)-1)$, and by the Leibniz rule, the derivative of $\int_{0}^{\infty}e^{-t}t^x \mathrm{d}t$ with respect to $x$ is $\int_{0}^{\infty}\ln(t)e^{-t}t^x \mathrm{d}t$, implicit differentiation tells us that:

$-y^\frac{1}{y}(\ln(y)-1)\mathrm{d}y = (\int_{0}^{\infty}\ln(t)e^{-t}t^x \mathrm{d}t)\mathrm{d}x$

We can solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\int_{0}^{\infty}\ln(t)e^{-t}t^x \mathrm{d}t}{-y^\frac{1}{y}(\ln(y)-1)}$

We can simplify the denominator. Since $y=(x!)^y$, we can conclude that $y^\frac{1}{y}=x!$ and also that $\ln(y)=y\ln(x!)=x!^{x!^{x!^{.^{.^.}}}}\ln(x!)$. Substituting this into the denominator gives:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\int_{0}^{\infty}\ln(t)e^{-t}t^x \mathrm{d}t}{-x!(x!^{x!^{x!^{.^{.^.}}}}\ln(x!)-1)}$

And there's the derivative.