Applying Leibniz to an integral equation The Integral equation
$$\frac{1}{2y} \int_{x-y}^{x+y} f(t) dt=f(x), \forall x,y \in R ~~~~(1)$$
relates to a question at NSE:
If $\frac{1}{2y} \int_{x-y}^{x+y} f(t) \space \mathrm{d}t = f(x)$, then $f$ is linear text*
By the Leibniz' rule, the D. w. r. t. $x$ on both sides leads to
$$\frac{1}{2y}[ f(x+y)- f(x-y)]= f'(x)~~~~(2).$$
It may be checked that $f(x)=x^2$ satisfies (2) but it does not satisfy (1).
The question is to pin point  why (1) and (2) are not equivalent.

$f(x)=g(x)$ implies $f'(x)=g'(x)$ but the converse is not true. ($g(x)$ could be $c+f(x)$ where $c$ is a constant).
In this case since you are differentiating w.r.t. $x$ treating $y$ as  a constant you can only conclude that when  the second one is assumed the first equation holds when you add a function of $y$ ( i.e. something  not depending on $x$) to one side.
There is indeed a function of the form $x^{2}+\phi (y)$ which satisfies the first equation. Can  you  find it?
