I was investigating the factorial function due to their importance in the binomial expansion. They play an important role in calculating the coefficients of the terms of the binomial expansion.


What is the derivative of the function $y = x!$ ?

My (failed) attempt:
I first plotted some points on a graph paper and tried to join them. I tried smoothing out the graph and then I tried to connect them with straight lines.
Both techniques seemed to not work and produced something I understood was wrong as the derivative had gone beserk.
Then I tried it on Desmos: https://www.desmos.com/calculator/6opi7buw86.

The graph made me want to publish an official complaint against Desmos for intentionally scamming me but thought better of it.

I googled up $x!$ and learnt about the $\Gamma$ function which produced this unexplainable graph. I understand the basics of it (or so I believe) but I am not confident that I can continue this quest.

Reason for posting this question:
What is this $\Gamma$ function?

What is the meaning of $\frac 12!$ ? How does it have a value, as well as all the other non-integer values? Especially the negative values which makes no sense to my incompetent brain.

Can I use the $\Gamma$ function to find an equation for the derivative of $x!$ (or $\Gamma (x)$ for this matter)?

I would prefer explanations as simple and elaborate as possible. Please excuse me for my apparent lack of knowledge. Thanks in advance!

Was this question already posted? I could not find this question. Could anyone direct me to such a question (if it exists)? And please try to avoid Youtube links as I learned that they are not always true (specially Numberphile).

  • $\begingroup$ I would appreciate any help, though I would prefer it to be simple. I have to understand what you want to say! $\endgroup$ Dec 31, 2017 at 16:57
  • $\begingroup$ So $x!$ is the usual factorial and is only for nonnegative integer values, right? How one defines a derivative for a function on integer values? $\endgroup$
    – Shashi
    Dec 31, 2017 at 16:58
  • $\begingroup$ In order for a derivative to make sense, the factorial would make to be a proper "function" (as opposed isolated points). To do this, we need to extend the definition to include non-integer values, a process called analytic continuation. The Gamma function is the result of this. $\endgroup$
    – Dylan
    Dec 31, 2017 at 16:59
  • 2
    $\begingroup$ That is precisely why. I suggest you start here en.wikipedia.org/wiki/Gamma_function. Look at its integral definition. Evaluating it over integer values is not too complicated. $\endgroup$
    – Dylan
    Dec 31, 2017 at 17:05
  • 1
    $\begingroup$ You can start to find that out by reading the link provided in the comment of @Dylan. $\endgroup$
    – Lee Mosher
    Dec 31, 2017 at 17:07

4 Answers 4


The derivative of a function is given as a limit: $$f'(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}.$$ In order for this limit to have meaning, the point $x$ must be a limit point of the domain of $f$. That is, if we look at any small interval containing $x$, the function $f$ must be defined for infinitely many points in that interval.

The factorial function is defined for only the nonnegative integers. The nonnegative integers have no limit points, so we cannot immediately talk about the derivative of the factorial function at any point. This one of the reasons why the gamma function is interesting and important.

The gamma function is defined for positive numbers $x$ as $$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}\,dt.$$ If you have studied integration by parts, it is easy to see that $$\Gamma(x+1)=x\,\Gamma(x)$$ for positive numbers $x$. This is the same properties satisfied by the factorial. Furthermore, for any nonnegative integer $n$, $$n!=\Gamma(n+1).$$ So clearly the gamma function is clearly a close relative of the factorial. We may therefore think of the gamma function (or more properly, $\Gamma(x+1)$) as the ``smoothed out'' or interpolated version of the factorial function you drew on graph paper.

I hope that answers your first question. For your second question, we may think of factorials of noninteger values such as $1/2$ using the gamma function: $$\Gamma(1/2+1)=\frac{1}{2}\Gamma(1/2)=\frac{1}{2}\int_0^\infty t^{-1/2} e^{-t}\,dt.$$ Make the substitution $u=\sqrt{t}$ in the integral to get $$\Gamma(1/2+1)=\frac{1}{2}\int_0^\infty 2e^{-t^2}\,dt.$$ This last integral can be solved using a couple of tricky methods, but we will simply give the result: $$\Gamma(1/2+1)=\frac{1}{2}\sqrt{\pi}.$$

Now to your main question: the derivative of the factorial is better stated as the derivative of the gamma function: $$\frac{d}{dx}\Gamma(x+1)=\frac{d}{dx}\int_0^\infty t^x e^{-t}\,dt=\int_0^\infty \frac{d}{dx}(t^x)e^{-t}\,dt=\int_0^\infty \ln(t)t^x e^{-t}\,dt.$$

Unfortunately, there is not a good way to summarize this derivative in terms of functions that we already understand without an integral.

Note that in the above calculation of the derivative that the derivative is taken with respect to $x$, while the integral is taken with respect to $t$, so the derivative and the integral do not simply cancel each other out.

  • 1
    $\begingroup$ Should the last differential operator in the derivative of the gamma function become a partial when it enters the integral? $\endgroup$
    – Biggs
    Jan 7, 2018 at 6:08

To address this question, we need to talk about the continuous version of the factorial function, which is the Gamma function. In particular, $x!=\Gamma(x+1)$.

Now, the definition of this function is in terms of an integral:

$$\Gamma(x):=\int_0^\infty t^{x-1}e^{-t}\,dt$$

This function is differentiable, and its derivative (for positive real $x$) is given by:

$$\Gamma'(x)=\int_0^\infty t^{x-1}e^{-t}\ln(t)\,dt$$

For more details, check out the Wiki article: Gamma function.

If you want to work with $\Gamma(x+1)$, that's no problem; the chain rule lets you simply replace $x$ with $x+1$.

Does this answer your question?

  • 1
    $\begingroup$ Thanks a lot Tony! ($ \not= \Gamma (Tony)$). $\endgroup$ Dec 31, 2017 at 17:15

The Gamma function is defined on the reals as $$\Gamma(z)=\int_0^{\infty}x^{z-1}e^{-x}dx$$You can use integration by parts to show that for the naturals $\Gamma(n+1)=n!$. The Gamma function is differentiable, so its derivative is a natural way to look at the derivative of the factorial function.


$x!$ is defined for integer $x$. On reals $x$, this is a discontinuous function. Thus the derivative doesn't exists.

Nevertheless, one can consider the extension: $$x!=\Gamma(x+1)$$ $$\frac{d}{dX}\Gamma(X)=\Gamma(X)\psi(X)$$ $\psi(X)$ is the function digamma.

So, by extension of the definition of $x!$ : $$\frac{d(x!)}{dx}=\Gamma(x+1)\psi(x+1)$$


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