Question

Suppose a differentiable function $f\left(x\right)$ satisfies the identity $f(x+y)=f\left(x\right)+f\left(y\right)+x{y}^2+x^{2}y$ for all real $x$ and $y.$ If $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=1$, then $f^{\prime }\left(3\right)$ is equal to

Answer 1

Given: $f(x+y)=f\left(x\right)+f\left(y\right)+x{y}^2+x^{2}y$
At $x=y=0$
$f\left(0\right)=2f\left(0\right)\Rightarrow f\left(0\right)=0$
Now,
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
(take $y=h$)
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}+\underset{h\to 0}{lim}\left(xh\right)+x^2$
$f^{\prime }\left(x\right)=1+0+x^2$
$f^{\prime }\left(x\right)=1+x^2$
$f^{\prime }\left(3\right)=10$