Let we have $F(u)=\int_A\mathcal{L}(x,u,u')dx$, where $dx$ is Lebesgue measure, $A$ is open and bounded with regular boundary, $\mathcal{L}\in\mathcal{C}^2(A\times\mathbb{R}\times\mathbb{R}^n)$ and convex $\forall x\in A$, also $u(x)\in\mathcal{C}^1(\bar A)$. Moreover we introduce $f(t)=F[\psi]=F[u+t(v-u)]$, where $v(x)\in\mathcal{C}^1(\bar A), \ t\in [0,1]$ and I want to calculate $\frac{df}{dt}$, but I am a bit confused and not sure. My attempt:
$\frac{df}{dt}=\frac{d}{dt}\int_A\mathcal{L}[x,u+t(v-u),(u+t(v-u))']dx$
Since $\mathcal{L}$ is continuous and bounded, we can put the differential under the integral sign.
$\frac{d\mathcal{L}}{dt}=\frac{d\mathcal{L}}{d\psi}\frac{d\psi}{dt}+\frac{d\mathcal{L}}{d\psi'}\frac{d\psi'}{dt}+\frac{d\mathcal{L}}{dx}\frac{dx}{dt}=\mathcal{L}_\psi\cdot(v-u)+\langle\nabla_{\psi'}\mathcal{L},\nabla v-\nabla u\rangle$
and here I am confused because professor got here
$\mathcal{L}_u\cdot(v-u)+\langle\nabla_{\xi}\mathcal{L},\nabla v-\nabla u\rangle$, where $\xi=u'$.
What I am missing? thank you