While solving functional equations I came across questions where the functional equation is not symmetric in $x,y $
Here's an example $f (x+y)=f (x)f (y)+2x^2y+0.5xy^2$ for all $x,y$ belonging to the set of real numbers and $f (x) $ is differentiable everywhere. If $f'(0)=1$, then Find $f'(2)-f (2)$
Here's my attempt to the solution- Partial differentiating the equation w.r.t x $f'(x+y)=f'(x)f (y)+4xy+0.5y^2$
Now put $x=0$ $f'(y)=f'(0)f (y)+0.5y^2$
$f'(y)=f (y) + 0.5y^2$ Putting$ y=2$
Now if I had solved the equation by differentiating w.r.t. y, the answer comes out to be 8. $f'(x+y)=f (x)f'(y)+2x^2+xy $
$f'(x)=f (x)+ 2x^2$
How is this true?
Also, consider another unsymmetrical functional equation $f(x+y)=f (x)+f (y)+2x $ Now if I interchange $x$ and $y$ $f (x+y)=f (x)+f (y)+2y $
This implies $x=y $ What is its meaning in context of the problem? I think it means that this equation is not a two variable functional equation but rather a single variable functional equation. Am I right?