Taking the derivative of definite integral?

I'm having trouble understanding the derivative of definite integral. For example, why is the following true?

$\frac{d}{dx}\displaystyle\int_{0}^{x}F_{1}(x-v)f_{1}(v)\, \mathrm{d}v = \displaystyle\int_{0}^{x}f_{1}(x-v)f_{1}(v)\, \mathrm{d}v$

I mean I know that $\frac{d}{dx}F_{1}(x) = f_{1}(x)$ by the fundamental theorem of calculus, but the example I gave confuses me...what if the integration limit $x$ is a constant? Doesn't the derivative then equal $0$?

Could someone give me intuitive explanation? I understand it somehow, but I don't really get the intuition, perhaps the limits of integration or the notation confuses me...

here is a link to my original question, where I was asking about the $\chi^2$-distribution:

Help with understanding the $\chi^2$-distribution

I'd be thankful if someone could maybe write it out step by step so I could see what's going on :)

To illustrate (slightly heurustucally), you may contemplate what happens as $x$ is replaced with $x+h$ for small $h$.
We want to compute $$\frac{\displaystyle\int_0^{x+h}F(x+h-v)f(v)\, \mathrm dv-\int_0^{x}F(x-v)f(v)\, \mathrm dv}h\\ =\frac1h\left(\int_0^{x}(F(x+h-v)f(v)-F(x-v)f(v))\, \mathrm dv +\int_x^{x+h}F(x+h-v)f(v)\, \mathrm dv\right)$$ Using $F(x+h-v)-F(x-v)=h\cdot f(x-v+\theta(x)h)$ for some $\theta(x)\in(0,1)$, and also that $F(x+h-v)f(v)$ is nearly constant on the tiny interval $[x,x+h]$, this is $$\approx\frac1h\left(\int_0^{x}h\cdot f(x-v+\theta(x)h)f(v)\, \mathrm dv +h\cdot F(0)f(x)\right)\\ =\int_0^{x} f(x-v+\theta(x)h)f(v)\, \mathrm dv \approx\int_0^{x} f(x-v)f(v)\, \mathrm dv$$ So we see (but should justify a bit more rigorously) that part of the result comes from differentiating the integrand, and part comes from the integrand at the upper limit (because the upper limit varies with $x$). In general, another contribution by the lower limit would be needed, but here that limit is constantly $0$, hence no change occurs: $$\frac{\mathrm d}{\mathrm dx}\int_{f(x)}^{g(x)}h(x,t)\,\mathrm dt=g'(x)h(x,g(x))-f'(x)h(x,f(x))+\int_{f(x)}^{g(x)}\frac{\partial}{\partial x}h(x,t)\,\mathrm dt.$$ The special thing that happens in your original problem is that the contribution by the upper limit is always $0$ because the factor $F(0)=\int_0^0f(x)\mathrm dx$ occurs.