Not sure of derivative in derivation of Euler-Lagrange equation: is it some kind of chain rule?

Question

I am reading a derivation on the Euler-Lagrange equations and I don't see how a certain expression arises. In the following I don't see precisely how the derivative in $t$ becomes a summation of partial derivatives in $u$ and $p$: $$ \begin{align} \frac{d}{dt}J[u + tv] & = \int_a^b \frac{d}{dt}L(x, u+tv, u' + tv')dx \\ & = \int_a^b \bigg[v \frac{\partial}{\partial u}L(x, u+tv, u' + tv') + v' \frac{\partial}{\partial p}L(x, u+tv, u' + tv')] dx, \end{align} $$ where $p$ represents $u'$. What is the formula being used here, is it some version of the chain rule?

Answer 1

Yes, it's exactly the chain rule. In general if $f: \mathbb{R}^n \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}^n$ are differentiable, then $$ \frac{d}{dt} f(g(t)) = \nabla f(g(t)) \cdot g'(t). $$ This is what's being used here with $n=3$, $$ \nabla L = (\partial_x L, \partial_u L, \partial_p L), $$ and $$ g'(t) = (0,v,\partial_x v). $$