What is the derivative of function respective to it's integral?

Question

To make the notation here more readable I transposed it all down one derivative, so the question I'm having is what is this:

$$ \frac{d \frac{d f(b)}{d b}}{d f(a)} $$

The question came up while I was trying to derive the wave equation from the Lagrangian (from what I read on the internet, this is not how one's supposed to be doing it, but I'm not a physics student but a hobbyist who's doing it for fun, so if I want to hit my head against a rock I'm going to do it), so it's actually the square (but this shouldn't change anything):

$$ \frac{d \left(\frac{d f(b)}{d b}\right)^2}{d f(a)} $$

Wolfram Alpha says both are zero and a manual derivation from the definition of the derivative says the same, but – assuming I didn't do something wrong – it has to be $\ne 0$.

Edit:

For a bit of context, I was trying to derive the wave equation $$ a \frac{d^2 V}{d^2 t} + b \frac{d^2 V}{d^2 x} = 0 $$ from – what I believe to be the appropriate – Lagrangian (on the internet the Lagrangian is usually formulated with tools of general relativity or something, which I don't fully understand): $$ L = \frac{1}{2} m \frac{d V}{d t}^2 - \frac{1}{2} k \frac{d V}{d x}^2 $$ Applying now the Euler-Lagrange equation $$ \frac{d L}{d q} = \frac{d}{d t} \frac{d L}{d \dot{q}} $$ causes me to have to evaluate the asked about expression (just with $V$ instead of $f$).

Answer 1

Let The integral be F. Then f= dF/dx and you are asking about df/dF. By the chain rule df/dF= (df/dx)(dx/dF)= (df/dx)(1/(dF/dx))= (1/f)(df/dx).

Answer 2

The expression $$\frac{d \frac{d f(b)}{d b}}{d f(a)}$$ makes absolutely no sense. It is a combination of how bad can Lebniz's notation go, intertwined with the catastrophic confusion that has people write $f(x)$ when they mean $f$ and the really bad practice of confusing a formal argument of a function with the value at which that argument is taken.

What is meant is probably the following. We have a function $f$ defined on some interval $I$, and it has a derivative $f'$ on that interval. The function is such that the set of points $\{(f(x),f'(x)):x\in I\}$ happens to be (part of) the graph of a funtion, that is, there exists a function $u$ such that $f'(x)=u(f(x))$ for each $x\in I$. The expression in the question is intended to denote the derivative $u'$.

Given this, the meaningless notation $$\frac{d \left(\frac{d f(b)}{d b}\right)^2}{d f(a)}$$ is probably trying to denote the derivative of the function $u^2$, and of course, neither $u$ nor $u^2$ need have zero derivative in general.

For example, let us take consider the function $f$ such that $f(x)=x^3$ for all $x>0$. Its derivative has $f'(x)=3x^2$ for all $x>0$, and therefore there is indeed a function whose graph contains the set $\{(f(x),f'(x)):x>0\}$: the function $u$ such that $u(t)=3t^{2/3}$ for all $t>2$ clearly is such that $$u(f(x))=3(x^3)^{2/3}=3x^2=f'(x)$$ for all $x>0$. In this case we have that neither $u$ nor $u^2$ have zero derivative, of course.