Let $y$ be an extremal function of the functional $$J[y]=\int_{x_1}^{x_2} L(x, y(x), y'(x)) \mathrm{d}x$$ So it satisfies the Euler-Lagrange equation: $$\frac{\partial L}{\partial y}-\frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial L}{\partial y'}=0$$ If we then substitute this $y$ into the original functional, we get the eigonal: $$S(x_1,x_2,y_1,y_1)=\int_{x_1}^{x_2} L(x, y(x), y'(x)) \mathrm{d}x$$ Which is now the function of $x_1, x_2, y_1=y(x_1), y_2=y(x_2)$. Then we have that $$\frac{\partial S}{\partial x_2}=-E(x_2)$$ And $$\frac{\partial S}{\partial y_2}=p(x_2)$$ Where $$p=\frac{\partial L}{\partial y'}$$ And $$E=py'-L$$
Question: How to show these identities with differentiating $S$ directly?
I've seen the derivation with discretization, but I would like to do it this way. I tried to apply the Leibniz integral rule: $$\frac{\partial S}{\partial x_2}=L(x_2)+\int_{x_1}^{x_2}\frac{\partial L}{ \partial x_2} \mathrm{d}x$$ But I could not show that $$\int_{x_1}^{x_2}\frac{\partial L}{ \partial x_2} \mathrm{d}x=-p(x_2)y'(x_2)$$