I was practising to differentiate various standard functions of $x$, such as $\sin x, \ln x, e^x$, etc. , when I got stuck on $\dfrac {\mathrm d}{\mathrm dx} x!$ .
I tried several approachs such as differentiating $x(x-1)(x-2)...1$ . I tried using $x!=x(x-1)!$ , but for that obviously I had to know the derivative of factorial function, which is essentially I am trying to find.
Can anyone help me out?
It does not a priori make sense to differentiate $x!$ because the domain of $x\mapsto x!$ is $\mathbf N$, not $\mathbf R$ (or anything else supporting a good notion of differentiation, like $\mathbf C$). As it turns out, you can extend $x\mapsto x!$ to a smooth function on the real line in an essentially unique manner, and the result is called the (Euler) gamma function, $x\mapsto\Gamma(x+1)$. (The shift in argument between $-!$ and $\Gamma(-)$ is notoriously confusing, but conventional.)
I am not sure it is much use to you, though, because the function is not easily expressed algebraically in terms of anything else, and neither is its derivative. For the latter, one usually defines the digamma function $$\psi(x) = \Gamma'(x) / \Gamma(x)\equiv[\log\Gamma(x)]',$$ so one could say the answer to your question is that $$\Gamma'(x+1) = \psi(x+1)\Gamma(x+1),$$ but most likely the expected answer is “$x!$ is not differentiable”. (Although would be possible to use $x!$ as a synonym for $\Gamma(x+1)$, usually the notation $x!$ presupposes that $x\in\mathbf N$.)
There's an issue with defining $(x!)'$, in that the factorial function $f(x) = x!$ isn't continuous, so can't be differentiable.
To fix this, we can define the Gamma function. For integers, we have that $\Gamma(x) = (x-1)!$, but by defining this as: $$\Gamma(z) = \int_0^\infty x^{z-1}e^{-x}dx$$ You can verify that this gives the correct result for integers by repeatedly integrating by parts.
Now, we can ask, what's the derivative of $\Gamma(z)$? This is defined in terms of something called the polygamma function, which (unfortunately) is defined as: $$\psi^{(n)}(z) = \frac{d^{m+1}}{dz^{m+1}}\ln(\Gamma(z))$$ Especially, we have that: $$\psi^{(0)}(z) = \frac{\Gamma'(z)}{\Gamma(z)}$$ So, we have that: $$\Gamma'(z) = \psi^{(0)}(z) \Gamma(z)$$ In terms of the definition I gave you, this is unenlightening. But, on integer arguments we can recover the following (by using an alternative expression for $\psi^{(0)}$, similar to the expression $x! = \Gamma(x+1)$ for $\Gamma$): $$(n!)' = \Gamma'(n+1) = \underbrace{\left(-\gamma+\sum_{k = 1}^{n}\frac{1}{k}\right)}_{\psi^{(0)}(n+1)}(n+1)!$$ where $\gamma\approx 0.57721$ is the Euler-Mascheroni constant.
For further reading, the Gamma function and Polygamma function articles are fairly useful.
For $\displaystyle \psi_0 \left ( x \right )=\frac{\Gamma '\left ( x \right )}{\Gamma \left ( x \right )}$, hence we have $\psi_0 \left ( x+1 \right )\Gamma \left ( x+1 \right )=\Gamma '\left ( x+1 \right )=\left ( x! \right )'.$
where $\psi \left ( \cdot \right )$ is digamma function and $\Gamma \left ( \cdot \right )$ is gamma function.
\Gammavs.\gamma). And in that case, have you tried differentiating that function instead of the factorial? – Arthur Jan 14 '17 at 07:44