Using Michael Hardy's answer we have: $$\begin{align} (f'*g)(x) & = \left.\phantom{\frac{}{}}f(u)g(x-u)\right|_{u\to-\infty}^{u\to\infty} + (f*g')(x)\end{align}$$ And if we differentiate $(f*g)(x)$, then: $$(f * g)' (x) = \frac{d}{dx}\int_{-\infty}^{\infty} f(u)g(x-u)du = \frac{d}{dx}\int_{-\infty}^{\infty} f(x-u)g(u)du= (f*g')(x) = (f'*g)(x)$$
So my question is about the constant. With the first method $(f'*g)(x) = (f*g')(x) + C$
($C$ could be infinite) while the second method gives $(f'*g)(x) = (f*g')(x)$. I think it has to do with the conditions on the functions in the second method and because of that $C$ is always zero for those functions. I don't know whether it's true or not.
In order for this integration to work in the classical sense, you must require that $f(u)g(x-u)\to 0$ as $u\to \pm\infty.$ Otherwise, the cases are:
1) $f(u)g(x-u)\to c(x)\neq 0$ as $u\to\infty$ or $u\to -\infty$ for some $x.$ In this case, $f(u)g(x-u)$ cannot be an integrable function, so you cannot define convolution of $f$ and $g$ in this case, if you want $h=f*g$ to be defined and therefore differentiable everywhere. If you don't care about this, then the property just might not hold, as $f*g$ might be infinity (... is this a constant function whose derivative is zero...? Maybe in some universe.), and $f'*g$ might be zero. For example if $f=1$ and the integral of $g$ is infinity.
2) The limit of $f(u)g(x-u)$ does not exist as $u\to\infty$ or $u\to-\infty$. In this case again the expression does not make sense.
In an analysis course, convolution is usually first defined for functions that have compact support, so that the limit as $u\to\pm\infty$ is $0$, or it is defined for Schwartz functions -- infinitely differentiable functions that go to zero very fast at infinity. Then the definition is extended by a limiting process to more general functions for which that constant you are asking about is still zero.
For the "second method" you mentioned to work, you need to be able to move derivatives inside the integral. But you are only allowed to do that if certain decay conditions are met (such as both functions go to zero rapidly at infinity).