A function $f:R\to R$ satisfies the equation $f\left(x+y\right)=f\left(x\right)f\left(y\right)$ for all $x,y\in R$ and $f\left(x\right)\ne 0$ for all $x\in R.$ If $f\left(x\right)$ is differentiable at $x=0$and $f^1\left(0\right)=2$, then $f^1\left(x\right)$ equals__

We have $f\left(x+y\right)=f\left(x\right)f\left(y\right)$ for all $x,y\in R$

$\therefore f\left(0\right)=f\left(0\right)f\left(0\right)\Rightarrow f\left(0\right)\left\{f\left(0\right)-1\right\}=0\Rightarrow f\left(0\right)=1$

$\left[\because f\left(0\right)\ne 1\right]$

Now, $f^1\left(0\right)=2$

$\Rightarrow \underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}=2$

$\Rightarrow \underset{h\to 0}{\lim }\frac{f\left(h\right)-1}{h}=2$ $\left[\because f\left(0\right)=1\right]\to \left(1\right)$

Now $f^1\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$\underset{h\to 0}{\lim }\frac{f\left(x\right)f\left(h\right)-f\left(x\right)}{h}\left[\because f\left(x+y\right)=f\left(x\right)f\left(y\right)\right]$

$=f\left(x\right)\left(\underset{h\to 0}{\lim }\frac{f\left(h\right)-1}{h}\right)=2f\left(x\right)$(using (i))