Following is from Olver's book on Lie groups and Differential Equations:
Define the variational problem: \begin{eqnarray*} \mathcal{L}= \int_\Omega L(x,u^{(n)}) \end{eqnarray*} where $u^{(n)}$ represents partial derivatives upto n'th order. L is smooth.
Given $f,\eta: \mathbb{R}^m \rightarrow \mathbb{R}^q$. $\eta$ be a function with compact support in $\Omega$. We want to calculate the formula for variational derivative: \begin{eqnarray*} \frac{d}{d\epsilon}\Big|_{\epsilon=0}\mathcal{L}[f + \epsilon\eta] &=& \int_{\Omega} \frac{d}{d\epsilon} \Big|_{\epsilon = 0} L(x,pr^{(n)}(f+\epsilon\eta)(x))dx \\ &=& \int_{\Omega} \Big\{ \sum_{\alpha,J}\frac{\partial L}{\partial u^{\alpha}_{J}}(x,pr^{(n)}f(x)) . \partial_{J}\eta^{\alpha}(x)\Big\} dx \end{eqnarray*}
where $\frac{\partial L}{\partial u^{\alpha}_J}$ is derivative of $L$ with respect to $u^{\alpha}_J$, $J$ being the multi-index representing the ``$J^{th}$'' derivative of $u^{\alpha}, 1 \leq \alpha \leq q, |J| \leq n$ and $pr^{(n)}$ is used for prolongation.
Now this is the part I don't understand: Since $\eta$ has compact support, we can use the divergence theorem to integrate the latter expression by parts, with the boundary terms on $\partial\Omega$ vanishing. Each partial derivative $\partial/\partial x^j$, when applied to the derivatives $\partial L/\partial U^{\alpha}_{J}$ of the Lagrangian, becomes the total derivative $D_j$ since L depends on $x$ through the function $u=f(x)$
\begin{eqnarray*} \frac{d}{d\epsilon}\Big|_{\epsilon=0}\mathcal{L}[f + \epsilon\eta] &=& \int_{\Omega}\Big\{\sum_{\alpha}\Big[\sum_{J}(-D)_J\frac{\partial L}{\partial u^{\alpha}_J}(x,pr^{(n)}f(x))\Big]\eta^{\alpha}(x) \Big\}dx \end{eqnarray*}
where, for $J=(j_1,...,j_k)$, \begin{equation*} (-D)_{J} = (-1)^kD_{J} = (-D_{j_1})(-D_{j_2})...(-D_{j_k}) \end{equation*}
I understand that the integration by parts formula for higher dimension https://en.wikipedia.org/wiki/Green's_identities has to be used, but unable to fit this equation into the formula to get the desired result. Need help filling this gap