Question:
Suppose a differentiable function $f(x)$ satisfies the identity $f(x+y)=f(x)+f(y)+x y^{2}+x^{2} y$, for all real $x$ and $y$. If $\lim _{x \rightarrow 0} \frac{f(x)}{x}=1$, then $f^{\prime}(3)$ is equal to_______.
Solution:
$f(x+y)=f(x)+f(y)+x y^{2}+x^{2} y$
Differentiate w.r.t. $x$ :
$f^{\prime}(x+y)=f^{\prime}(x)+0+y^{2}+2 x y$
Put $y=-x$
$f^{\prime}(0)=f^{\prime}(x)+x^{2}-2 x^{2}$....(i)
$\because \lim _{x \rightarrow 0} \frac{f(x)}{x}=1 \Rightarrow f(0)=0$
$\therefore f^{\prime}(0)=1$.....(ii),
From equations (i) and (ii),
$f^{\prime}(x)=\left(x^{2}+1\right) \Rightarrow f^{\prime}(3)=10$