Derivatives and Functional Derivatives!

We want to extremize $I[y] \; = \; \int_0^1 \sqrt{1 + y(x) + y'(x)}\,dx$ subject to suitable continuity/differentiability conditions and $y(0) = y(1) = 0$ . Standard Calculus of Variations theory tells us that we need to solve the equation $\frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right) - \frac{\partial F}{\partial y} \; = \; 0$ where $F(x,y,y') = \sqrt{1 + y + y'}$ . Since $F$ is not explicity dependent on $x$ , a first integral of this equation is $\begin{aligned} F - y'\frac{\partial F}{\partial y'} & = \; c \\ \frac{2 + 2y + y'}{2\sqrt{1 + y + y'}} & = \; c \\ (2 + 2y + y')^2 & = \; 4c^2(1 + y + y') \\ \big[2 + 2y + y' - 2c^2\big]^2 & = \; 4c^2(c^2 - 1 - y^2) \end{aligned}$ If we put $c^2 - 1 - y \; = \; z^2$ , this becomes $\begin{aligned} (-2zz' - 2z^2)^2 & = \; 4c^2z^2 \\ (z'+z)^2 & = \; c^2 \\ z' + z & = \; \pm c \\ z & = \; \pm c + de^{-x} \end{aligned}$ The case $d=0$ means that $z$ is constant, and hence $y$ is constant, and hence $y \equiv 0$ , in which case $I[y] = 0$ . Otherwise $z$ is monotonic, and for $y(0)$ to be equal to $y(1)$ , we have to have $z(1) = -z(0)$ , which means that $d = \mp\tfrac{2e}{e+1}c$ and hence $z\; = \; \pm c\left(1 - \frac{2e^{1-x}}{e+1}\right)$ Thus we require $0 \; = \; y(0) \; =\; c^2 - 1 - c^2\left(\tfrac{e-1}{e+1}\right)^2 \; = \; \tfrac{4e}{(e+1)^2}c^2 - 1$ so that $c = \cosh\tfrac12$ . We note that $z'(x)$ is never zero, and so $y'(x) = - 2z(x)z'(x)$ is only zero when $z(x) = 0$ , which occurs precisely when $x \; = \; \ln\left(\tfrac{2e}{1+e}\right) = \boxed{0.3798854930}$