Given a bounded open domain $\Omega \subset \mathbb{R}^2=\{x={(x_1,x_2):x_i\in \mathbb{R}}\}$ and a function $F: V \to \mathbb{R}$ such that:
$$F(u) = \int_{\Omega}{K(u)\nabla u\cdot\nabla v \,dx},\qquad\forall v\in V$$
where $V=H^1_0(\Omega)$
Here, we can consider that $v$ is given and K(u) is a nonlinear function, for example, $K(u)=u^2$.
What is the differential $DF(u)\delta u$ of $F$ at $u$?
Let's clear this up, I would like to mention that the reason this all got confused is because F is a FUNCTIONAL rather than a function.
$$ F'(u)=\lim_{h\to 0}\frac {F(u+hf)-F(u)}{h}$$
Now: $$\frac {F(u+hf)-F(u)}{h}=\int\frac {K(u+hf)\nabla (u+fh)\cdot\nabla v-K(u)\nabla (u)\cdot\nabla v}{h}dx=\int\frac {(K(u+fh)\nabla (u)-K(u)\nabla (u))\cdot\nabla v+hK(u+hf)\nabla (f)\cdot\nabla v}{h}dx$$
Taking the limit we get:
$=\int K'(u)\nabla u\cdot\nabla v+K(u)\nabla f\cdot\nabla v dx$
Thus $F'(u)=\int K'(u)\nabla u\cdot\nabla v+K(u)\nabla f\cdot\nabla v dx$ where $f\in V $
Now the DIFFERENTIAL is actually the first variation of F where $hf=\delta u $
So the differential is:
$\int K'(u)\nabla u\cdot\nabla v+K(u)\nabla \delta u\cdot\nabla(v)dx $