Directional derivative of of $H^1(\Omega) \ni u\mapsto \int_{\Omega} \sqrt{1+|\nabla u|^2}$ 
Show that the directional derivative of $F:H^1(\mathbb{R})\to \mathbb{R}, u\mapsto \int_{\Omega} \sqrt{1+|\nabla u|^2}$ is $$\partial_e F(u)=\int_{\Omega} \frac{\nabla{u} \cdot \nabla{e}}{\sqrt{1+|\nabla u|^2}}.$$

By definition of directional derivative I think we have $$\partial_e F(u)=\lim_{t\to0}\frac{F(u+te)-F(u)}{t}=\lim_{t\to0}\frac{F(u+te)-F(u)}{te}e=F'(u)e$$ or rather $F'(u)\cdot e$? 
Differentiating naively I get $F'(u)=\int_\Omega\frac{\nabla{u}}{\sqrt{1+|\nabla u|^2}}$ and so I would have $\partial_e F(u)=\int_{\Omega} \frac{\nabla{u} \cdot e}{\sqrt{1+|\nabla u|^2}}$, missing a gradient.
I also tried to expand $F(u+te)$ but I am not sure how to simply $F(u+te)-F(u)= \sqrt{1+|\nabla{u}+t\nabla{e}|^2}-\sqrt{1+|\nabla{u}|^2}$.
 A: As noted by @mattos, the first variation of a functional gives the Gateaux derivative (directional derivative): $$\partial F_e(u)=\lim_{t\to 0} \frac{F(u+te)-F(u)}{t}= \frac{d}{dt} F(u+te)\bigg\rvert_{t=0}=\frac{d}{dt} 
\int_{\Omega}\sqrt{1+|\nabla{u}+t\nabla{e}|^2}dx\bigg\rvert_{t=0}$$
Taking the derivative inside the integral and using the chain rule we get $$ \partial F_e(u)=\int_{\Omega} \frac{(\nabla{u}+t\nabla{e})\cdot \nabla{e}}{\sqrt{1+|\nabla{u}+t\nabla{e}|^2}}dx\bigg\rvert_{t=0}=\int_{\Omega} \frac{(\nabla{u}+t\nabla{e})\cdot \nabla{e}}{\sqrt{1+|\nabla{u}+t\nabla{e}|^2}}dx\bigg\rvert_{t=0}=\int_{\Omega} \frac{\nabla{u}\cdot \nabla{e}}{\sqrt{1+|\nabla{u}|^2}}dx.$$
A: In your original attempt, your manipulation of
$$
\lim_{t \to 0} \frac{F(u+te) - F(u)}{t}
$$
does not make sense. Remember that $e$ is a function in $H^1$ and $F$ is a functional, it doesn't make sense to (for example) divide by $e$ here. You need to think of 
$$
f(t) := F(u + te)
$$
as your scalar function. Find the derivative of this at $t=0$, i.e. we have
$$
f'(0) = \lim_{t \to 0} \frac{F(u+te) - F(u)}{t}
$$
So thinking more along the lines of your second attempt will get to the right answer.
