Question

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

Answer 1

Given $f\left(x+y\right)=f\left(x\right)+f\left(y\right)+x{y}^2+x^{2}y$ for all real $x$ and $y$. and $\underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=1$.

Step 1: To find $f^{\text{'}}\left(x\right)$

We know that,

$f^{\text{'}}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$$\begin{gathered} =\underset{h\to 0}{\lim }\frac{f\left(x\right)+f\left(h\right)+x{h}^2+x^{2}h-f\left(x\right)}{h}\left[\text{given}\right] \\ =\underset{h\to 0}{\lim }\frac{f\left(h\right)}{h}+\underset{h\to 0}{\lim }\frac{x{h}^2}{h}+\underset{h\to 0}{\lim }\frac{x^{2}h}{h} \\ =\underset{h\to 0}{\lim }\frac{f\left(h\right)}{h}+x^2 \\ =1+x^2\left[\because \underset{x\to 0}{\lim }\frac{f\left(x\right)}{x}=1\right] \end{gathered}$$

Therefore,

$f^{\text{'}}\left(x\right)=1+x^2$

Step 2: To find $f^{\text{'}}\left(3\right)$

$f^{\text{'}}\left(x\right)=1+x^2$

$$\begin{gathered} \Rightarrow f^{\text{'}}\left(3\right)=1+3^2 \\ =1+9 \\ =10 \end{gathered}$$

Hence, the value of $f^{\text{'}}\left(3\right)=10$