The derivative of $x!$ and its continuity

Solution 1:

The factorial function is only defined for nonnegative integers, there however a function that is continouos and has the same properties as the factorial function for positive integers (or zero), that is the Gamma function:

$$\Gamma(n+1)=\int_0^\infty x^{n-1}e^{-x}dx=n!$$

More information about this function, and its derivatives here: http://mathworld.wolfram.com/GammaFunction.html

Solution 2:

The pi function is an extension of the factorial, defined by $$\Pi (z)=\int_0^\infty t^ze^{-t} \,dt$$ By repeated Integration by Parts, we can see that it does indeed satisfy $\Pi(z+1)=z\Pi(z)$. By plugging in $z=0$, we see that $0!=1$., so its initial value is correct. By taking the derivative of this function wrt z, we get that $\Pi'(z)=\displaystyle\int_0^\infty t^z e^{-t}\log z \, dt$. This derivative is not used often.

A more commonly used extension of the factorial is the Gamma function, defined as$$\Gamma (z)=\int_0^\infty t^{z-1}e^{-t} \,dt$$ Note that $\Pi(z)=\Gamma(z+1)$. Its derivative is not used often, but its logarithmic derivative, $\frac{d}{dz}\log\Gamma(z)=\frac{\Gamma'(z)}{\Gamma(z)}$, is commonly used and is called the digamma function $\psi_0 (z).$

Solution 3:

As said in previous answers, you can relate the factorial to the gamma function in the sense that $$x!=\Gamma(x+1)$$ The gamma function is undefined at negative integer values of $x$ (vertical asymptotes) but it is defined everywhere else.

Its derivative is given by $$\frac{d(x!)}{dx}=\frac{d\Gamma (x+1)}{dx}= \Gamma (x+1)\, \,\,\psi ^{(0)}(x+1)=x! \,\,\, \psi ^{(0)}(x+1)$$ where appears the digamma function.

http://en.wikipedia.org/wiki/Gamma_function is also a good page to look at.