Good day, I need derivate $f(x)=g(x)+\int_{a}^{x}K(x,y,f(y))dy$ and show that $f{}'(x)=g{}'(x)+K(x,x,f(x))+\int_{a}^{x}\frac{\partial }{\partial x}K(x,y,f(y))dy$.
I think the next but I do not sure: Take x=y=t and $u(x,y)=\int_{a}^{x}K(x,y,f(y))dy$ then $\frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}$. And $\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}\int_{a}^{x}K(x,x,f(x))dx$, $\frac{\partial u}{\partial y}=\frac{\partial }{\partial y}\int_{a}^{x}K(y,y,f(y))dy=K(x,x,f(x))$.
Is correct the processe? Also, $\frac{\partial }{\partial x}\int_{a}^{x}K(x,x,f(x))dx,\int_{a}^{x}\frac{\partial }{\partial x}K(x,x,f(x))dx$ can differ by constant because a limit of integration is a, or not?
It's not correct. You have to apply directly the leibniz integral rule. Consider that in your integral $f(x)=g(x)+\int_{a}^{x}K(x,y,f(y))dy$ the argument and the limits of integration depend on $x$ and the simple partial derivative into the integral doesn't work (further, not important, but $u$ is not function of $y$, $y$ is only a dummy variable to integrate).
$$\require{cancel}f'(x)=g'(x)+K(x,x,f(x))(x)'-K(x,a,f(a))\cancelto{0}{(a)'}+\int_{a}^{x}\frac{\partial }{\partial x}K(x,y,f(y))dy$$
$$f'(x)=g'(x)+K(x,x,f(x))+\int_{a}^{x}\frac{\partial }{\partial x}K(x,y,f(y))dy$$
And it's done.