How can I find $f'(2x)$ given that $xf(x)-yf(y)=(x-y)f(x+y)$? it might seem a little bit elementary. $f$ is defined on $\Bbb R$ and it is differantiable. and is not equal to zero.
if $xf(x)-yf(y)=(x-y)f(x+y)$ then find what is $f'(2x)$ equal to?. $(f'(x),2f'(x),xf'(x),4f'(x)$ or $8f'(x)?)$
from that equation
$\dfrac {f\left( x+y\right) -f\left( y\right) } {x}=\dfrac {f\left( x+y\right) -f\left( x\right) } {y}$
How to continue?
 A: The Identity function satisfies problem conditions and you sure that one of these 5 solutions are correct so the only possible one is $f'(2x)=f'(x)$ :) this is not a proof but answer your question if it is correct problem.
A: One can easily show that if $0\neq x\neq y\neq0$,
$$\frac{f(x+y)-f(y)}x=\frac{f(x)-f(y)}{x-y}=\frac{f(y)-f(x)}{y-x}=\frac{f(x+y)-f(x)}y.$$
(The first equality is a straightforward application of the functional equation and you get the rest for free by symmetry.) Letting $y\to0$, this implies $\frac{f(x)-f(0)}x=f'(x)$. The left hand side is differentiable with respect to $x$, so the right hand side is too and thus
$$f''(x)=\frac{xf'(x)-[f(x)-f(0)]}{x^2}=0$$
for all $x\neq0$. From this we find that $f'$ is constant except perhaps at zero, and so $f(x)=ax+b$ for all $x\neq0$. Since $f$ is continuous this extends to $x=0$ and hence $f'(x)=a=f'(2x)$ for all $x\in\mathbb R$.
A: It's obvious that every function of the form $f(x)=ax+b$ satisfies the equation:
$$xf(x)-yf(y)=(x-y)f(x+y)\tag0\label0$$
It can be shown that those are the only solutions indeed. To show that, let $a=f(1)-f(0)$ and $b=f(0)$, and define $g(x)=f(x)-ax-b$. It's easy to see that by \eqref{0}, $g$ satisfies
$$xg(x)-yg(y)=(x-y)g(x+y)\tag1\label1$$
and we have $g(0)=0$ and $g(1)=0$. Letting $x=1$ and $y=-1$ in \eqref{1} we get $g(-1)=0$. Now, letting $y=1$ and $y=-1$ in \eqref{1}, we respectively get:
$$xg(x)=(x-1)g(x+1)\tag2\label2$$
$$xg(x)=(x+1)g(x-1)\tag3\label3$$
Substituting $x+1$ for $x$ in \eqref{3} we have:
$$(x+2)g(x)=(x+1)g(x+1)\tag4\label4$$
Subtracting \eqref{4} and \eqref{2} we get $2g(x)=2g(x+1)$ and thus $g(x)=g(x+1)$. Hence by \eqref{4} we have $(x+2)g(x)=(x+1)g(x)$ and therefore $g$ is the constant zero function. So $f(x)=ax+b$ and $f$ is differentiable and for every $x$, $f^\prime(x)=a$ which yields $f^\prime(2x)=f^\prime(x)$.
A: This is a tempting thought, not necessarily correct though!
The fact that
$\dfrac {f\left( x+y\right) -f\left( y\right) } {x}=\dfrac {f\left( x+y\right) -f\left( x\right) } {y}$
Is tempting me to apply the limit as $y$ approaches zero for both sides and end with:
$\dfrac {f\left( x\right) -f\left( 0\right) } {x}=f'(x)$
Now you have a clear relationship between the function and its derivative and can be solved as a differential equation. Alternatively, you may want to expand the function in terms of Taylor's.
The above expression is a bit scary, what would $f(0)$ be? Not sure, but this is just a thought.
