Proof of Euler Lagrange equation in calculus of variation I am learning calculus of variations from the  article page 9 
There it says $J[u]=\int_a^bL(x,u,u')\, dx$ and $u'$ is represented by $p$.  
$$\langle\nabla J[u],v\rangle={dJ[u+tv] \over dt} \text{ when }t=0.$$
$$h(t)=J[u+tv]=\int_a^bL(x,u+tv,u'+tv') \, dx.$$
My question is how is $h'(t)$ computed.
$$h'(t)=\int_a^bv \frac{\partial L(x,u+tv,u'+tv')}{\partial u}+v'\frac{\partial L(x,u+tv,u'+tv')}{\partial p} \, dx.$$
Also can some one please explain to me what $\langle \nabla J[u],v\rangle={dJ[u+tv] \over dt} $ means?
 A: The term $h'(t)$ is computed as follows. We have defined
\begin{equation}
h(t) = \int_{a}^{b} L(x, u + tv, u' + tv') dx,
\end{equation}
and differentiating both sides gives
\begin{equation}
h'(t) = \frac{d}{dt}\int_{a}^{b} L(x, u + tv, u' + tv') dx.
\end{equation}
One of the crucial steps here is moving the derivative with respect to $t$ inside of the integral with respect to $x$. This is allowed since $x$ and $t$ are independent and because we are assuming sufficient smoothness of the function $L$. 
Moving the derivative inside the integral, we use the chain rule:
\begin{align}
h'(t) &= \int_{a}^{b} \frac{d}{dt}L(x, u + tv, u' + tv') dx\\ &= \int_{a}^{b} \frac{\partial}{\partial t}L(x, u + tv, u' + tv') + \frac{\partial}{\partial u}L(x, u + tv, u' + tv')\frac{du}{dt} + \frac{\partial}{\partial u'}L(x, u + tv, u' + tv')\frac{du'}{dt}dx,
\end{align}
where the steps above just reflect applying the chain rule. Now the $\frac{\partial}{\partial t}$ term goes away since $L$ does not explicitly depend upon $t$. 
For the next two terms we have to be careful. When we write $\frac{du}{dt}$, we are really writing $\frac{d}{dt}\big(\text{the second argument of $L$}\big)$, not $\dot{u}$. The notation here is overloaded and $u$ can refer either to the variable $u$ or the second argument of $J$ and here it refers to the second argument of $J$. 
Then we see that $\frac{du}{dt} = \frac{d}{dt}(u + tv) = v$. Similarly, we have
$\frac{du'}{dt} = v'$. Then we have
\begin{equation}
h'(t) = \int_{a}^{b} \frac{\partial L(x, u + tv, u' + tv')}{\partial u}v + \frac{\partial L(x, u + tv, u' + tv')}{\partial u'}v'dx. 
\end{equation}
As for the term $\langle \nabla J[u], v\rangle$, it is the inner product of $\nabla J[u]$ and $v$. Here, this gives us the directional derivative of $J$ in the direction $v$ evaluated at the point $u$. To see why, think of $u + tv$ as defining a line between $u$ and $v$. Changing the parameter $t$ lets us slide along this line. Picking any $t$, we can evaluate $J$ for that value of $t$ to get $J[u + tv]$. If we take the derivative of this with respect to $t$, we have
\begin{equation}
\frac{d}{dt} J[u + tv] = \frac{\partial}{\partial u}J[u + tv]\frac{du}{dt}
\end{equation}
for the same reason as above. Also as above, the term $\frac{du}{dt}$ really means "the time derivative of the argument of $J$" and this is simply $v$. If we call $\frac{\partial J}{\partial u}$ by the name $\nabla J$, the above equation gives us 
\begin{equation}
\frac{d}{dt} J[u + tv] = \nabla J[u + tv]\cdot v.
\end{equation}
Let's now evaluate this at $u$. We can do this by setting $t = 0$, which gives
\begin{equation}
\frac{d}{dt} J[u + tv] \bigg|_{t = 0} = \nabla J \cdot v = \langle \nabla J, v\rangle.
\end{equation}
