I have to take the derivative of the following function with respect to $\varepsilon$:
$$\phi(\varepsilon)=\int_{a}^{b+\varepsilon C}F(x, y(x)+\varepsilon\eta(x), y'(x)+\varepsilon\eta'(x))\;dx$$
My study material says that it should be:
$$\frac{d\phi}{d\varepsilon}\rvert_{\varepsilon=0}=CF(x,y,y')\rvert_{x=b}+F_{y'}\eta\rvert_{x=b}+\int_a^b\left(F_y-\frac{d}{dx}F_{y'}\right)\;dx,$$
but I have no idea how this was calculated. I'm having a course about Calculus of Variations. $F$ is a functional. Could someone explain how the derivative was calculated?
Here is my reference:
We get the first component of the first variation by considering a problem with only one fixed end-point, and allowing $x^*$ to vary, so that $$ \begin{align*} 0 = \frac{d\phi_1(\epsilon)}{d\epsilon} &= \left.\frac{d}{d\epsilon}\right\vert_{\epsilon=0} \int_{x_0}^{\color{red}{\hat{x}^*}} F(x, y_1 + \epsilon\eta, y_1' + \epsilon\eta')\,dx \\ &= \left.\frac{d}{d\epsilon}\right\vert_{\epsilon=0} \int_{x_0}^{\color{red}{\hat{x}^* + \epsilon X}} F(x, y_1 + \epsilon\eta, y_1' + \epsilon\eta')\,dx \\ &= \color{red}{\left.X\,F(x,y_1,y_1')\right\vert_{x=x^*}} + \color{red}{\left. F_{y_1'}\eta\right\vert_{x=x^*}} + \int_{x_0}^{x^*} \left(F_{y_1} - \frac{d}{dx}\,F_{y_1'}\right)dx \end{align*} $$
In general, if you have a function $$ \phi(\varepsilon) = \int_{a(\varepsilon)}^{b(\varepsilon)} g(\varepsilon,x)\, dx, $$ you can introduce an auxiliary function $$ H(\varepsilon,u,v) = \int_{u}^{v} g(\varepsilon,x)\, dx $$ and notice that $\phi(\varepsilon) = H(\varepsilon,a(\varepsilon),b(\varepsilon))$. Now you just apply the chain rule to differentiate with respect to $\varepsilon$. The final result should be $$ \phi'(\varepsilon) = \frac{\partial H}{\partial \varepsilon} + \frac{\partial H}{\partial u}a'(\varepsilon) + \frac{\partial H}{\partial v} b'(\varepsilon). $$