# Simple Derivation of Functional Equation Question (L'Hospital's Rule)

First, the question is:

$f$ is a differentiable function and $f : R \rightarrow R$

$xf(x)-yf(y)=(x-y)f(x+y)$

$f'(2x)=?$

My approach for problem is using L'Hospital's rule: $$\frac{xf(x)-yf(y)}{x-y}=f(x+y)$$ Assuming $y=x$ $$\lim_{y\to{x}} \frac{xf(x)-yf(y)}{x-y} = f(2x)$$ Taking the limit using l'hospital's rule: $$(yf'(y)+f(y))|_{y=x} = f(2x)$$ Taking the derivative of both sides: $$f'(x)+xf''(x)+f'(x)=2f'(2x)$$ But the answer is $f'(2x)=f'(x)$

How is this possible?

• Could it be that $f$ must have the form $f(x)=ax+b$? (All such functions satisfy the functional equation.) May 18 '15 at 20:10
• @mickep thanks, but what's wrong with my solution? Is there a way not using functional equation techniques ie. finding the function? Maybe a calculus route? May 18 '15 at 20:21
• I'm not sure, but if $f$ is as I wrote above, then $f''(x)=0$, and thus $2f'(x)=2f'(2x)$, so the statement follows. I have to sleep, but I'll have a look at this problem tomorrow again... May 18 '15 at 20:25

In equation $$xf(x)+yf(y)=(x-y)f(x+y),$$ differentiate with respect to $x$ to get $$xf'(x)+f(x)=(x-y)f'(x+y)+f(x+y).$$ Now, set $y$ to $x$ and to $-x$ to deduce that $$xf'(x)+f(x)=f(2x)=2xf'(0)+f(0),$$ which proves that $f(x)=xf'(0)+f(0).$