Functional derivative of $\int \left( \frac{df^2 }{d^2 x} \right)^2 dx$ According to page 7 of the PDF document 
$$
\frac{\delta}{\delta f} \int \left( \frac{df^2 }{d^2 x} \right)^2 dx = \int \frac{df^4}{d^4 x} dx
$$
I would like help proving this statement.
Although I can show that
$$
\int \left( \frac{df^2 }{d^2 x} \right)^2 dx  = \int \frac{df^4}{d^4 x} f dx
$$
My attempts at "constructing" the functional derivative of this expression isn't dropping the term $f$. I'm not even sure this is the right way to go about solving the problem.
 A: Well, probably this is a case of confusing notation. Especially I am not sure where you want to have your exponents.
In my understanding, both equations you stated are just wrong. However, the first one may carry some truth: Consider the functional $J(f) = \int (\frac{d^2 f}{dx^2})^2 dx$. Then the functional derivative of $J$ (or first variation) of $J$ is (by integration by parts)
$$
\delta J(f)(h) = \int 2\frac{d^2f}{dx^2}\frac{d^2h}{dx^2}dx = \int 2\frac{d^4 f}{dx^4} h dx.
$$
Hence, one may say (if put in a proper framework of function spaces) that the derivative of $J$ is
$$
J'(f) =  2\frac{d^4 f}{dx^4}.
$$
To conclude: The pdf document you linked is very sloppy with the notation and probably you may consult a book one the calculus of variations to get more information here.
A: Consider the functional $ F \left( u \right) = \int { \left( \frac{{d}^{2} u}{d{x}^{2}} \right) }^{2} dx $.
defining $ v = {u}_{x} $ the following is given:
$$ F \left( u \right) = \int_{\Omega} {{u}_{xx}}^{2} dx = \int_{\Omega} {{v}_{x}}^{2} dx = G \left( v \right) $$
Now, using the Gateaux derivative definition and $ L \left( x, v, {v}_{x} \right) = {{v}_{x}}^{2} $:
$$ {G}' \left( v \right) = \int_{\Omega} \left( \frac{\partial}{\partial v} L \left( x, v, {v}_{x} \right) - \frac{d}{dx} \frac{\partial}{\partial {v}_{x}} L \left( x, v, {v}_{x} \right) \right) h dx $$
Hence the critical point happens at $ \frac{d}{dx} \frac{\partial}{\partial {v}_{x}} L \left( x, v, {v}_{x} \right) = 0 $ since $ \frac{\partial}{\partial v} L \left( x, v, {v}_{x} \right) $ vanishes.
This implies the E-L is given by $ \frac{d}{dx} \frac{\partial}{\partial {v}_{x}} L \left( x, v, {v}_{x} \right) = \frac{d}{dx} 2 {v}_{x} = 2 {v}_{xx} = 0 $ which means $ {u}_{xxx} = 0 $.
This is probably a wrong answer.
Yet it rose from a discussion with @Mark Peletier, hence I mark it as Community Wiki so people will be able to see why this property of the E-L can not be used here.
Thank You.
