Directional derivative of norm of gradient. I want to show that
$$ \partial_v\left( \frac{1}{2} | \nabla  u| ^2\right)=\nabla u^T \nabla v  .$$
I was using the standard formula to calculate directional derivatives, i.e. $\nabla f(x) \cdot v$, but it doesn't seem to give the desired result.
Any help in helping me understand this is appreciated!
 A: Okay, let me guess: $u$ is a function (probably from the Sobolev space $H^1$) and you want to differentiate the functional $E \colon H^1 \to \mathbb{R}$ defined by $E(u)=\frac{1}{2} \int |\nabla u|^2 \, dx$. Am I (essentially) right?
If so, you want to compute
$$
\lim_{\varepsilon \to 0} \frac{E(u+\varepsilon v) - E(u)}{\varepsilon} =
\lim_{\varepsilon \to 0} \frac{1}{2\varepsilon} \left( \int 2\varepsilon (\nabla u)^T \nabla v\, dx + \varepsilon^2 \int |\nabla v|^2 \, dx \right),
$$
which is similar to your expression.
A: You should use the following facts for $F:\mathbb R^n \to \mathbb R^m,G: \mathbb R^m \to \mathbb R$, and $H = G \circ F$. 


*

*If $F$ is differentiable then its partial derivative at $x$ in the direction $y$, $\delta F(x,y): \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^m$. And if $F$ is differentiable then $(\delta F(x,y))_i = \langle \nabla F_i(x), y\rangle$ with $F(x)=(F_1(x), \ldots, F_m(x))$.

*If the directional derivatives $\delta F(x,y)$ and $\delta G(F(x),y)$ are well defined, it follows that the directional derivative of $H$ is given by
$\delta H(x,y)= \delta G(F(x),\delta G(x,y))$.
Now it should be easy to compute your directional derivative.
