Sorry for the vague title. I barely fitted it in. Here is the general context. Let $y^*(x)$ be a solution and a local minimum of the $I_0(y) \equiv \int_{x_0}^{x_1} {f_0(x,y,y')dx} \to extr; I_i(y) = \int_{x_0}^{x_1}{f_i(x,y,y')dx} = \alpha_i, i = 1,...,m; y(x_0) = y_0, y(x_1) = y_1$ isoperimetric problem, $f_i \in C^2(\mathbb{R}^3),i=0,...m$. Assign $\delta I_i(y^*,h) \equiv \lim \alpha \downarrow 0 \frac{I_i(y^* + \alpha h) - I_i(y^*)}{\alpha}$. Show that $\delta I_i (y^*,h) = \int_{x_0}^{x_1} {(-d/dx f_{iy'}' + f_{iy}')h(x)dx}, i = 0,...m$. By writing everything in integral form and multiplying the inner expression of the integral by $h(x)/h(x)$ I got the expression in the post title. That one is not coming from the textbook it's only the shortest way I found to express this problem. I tried using the definition of derivative and also Taylor formula with no success. Any hints and or suggestions? Also I can't understand how that negative sign appears in front of $d/dx$. Thanks in advance.
P.S. for further general context, this is from a theorem that states the following. If $y^*$ is a solution of the isoperimetric problem, then $\exists \lambda_0,...\lambda_m$ as the parameters of the problem such that at least one is not $0$ and for $y^*$ holds the $-d/dx L_{y'}' + L_y' = 0, y(x_0) = y_0, y(x_1) = y_1$ differential equation with edge conditions.
My problem comes as a statement in the proof of this theorem, which is considered as something obvious, though I don't see it.