This question has already been asked but the solution was not satisfactory
Suppose $f$ is a function satisfying the equation $$f(x+y)=f(x)+f(y)+x^2y+xy^2$$ for all real numbers $x$ and $y$ Suppose also that $\lim \limits_{x \to 0}\frac{f(x)}{x}=1$
Find $f'(x)$
By using the definition of derivative,I obtained $f(0)=0$ and $f'(0)=1$
How will I get $f'(x)$?
The definition of the derivative is $$ f'(x) = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h} $$ Using the property we've been given about $f$, we get $$ \frac{f(x+h) - f(x)}{h} = \frac{f(x) + f(h) + x^2h + xh^2 - f(x)}{h}\\ = \frac{f(h)}{h} + x^2 + xh $$ This gives $$ f'(x) = \lim_{h\to 0}\frac{f(x+h) - f(x)}{h}\\ = \lim_{h\to 0}\left(\frac{f(h)}h + x^2 + xh\right)\\ = 1 + x^2 $$
