Since, $f: R \rightarrow R$ satisfies the equation $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}), \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}, \mathrm{f}(\mathrm{x}) \neq 0$. $\& f(x)$ is differentiable at $x=0$ and $f^{\prime}(0)=2$.
$$
\begin{aligned}
&\Rightarrow \quad f^{\prime}(0)=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} \\
&\Rightarrow \quad 2=\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x} \\
&\Rightarrow \quad 2=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{0+h} \\
&\Rightarrow \quad 2=\lim _{h \rightarrow 0} \frac{f(0) \cdot f(h)-f(0)}{h} \\
&\Rightarrow \quad 2=\lim _{h \rightarrow 0} \frac{f(0)[f(h)-1]}{h} \\
&\text { Since, } \\
&\Rightarrow \quad f(0+h)=f(0) \cdot f(h) \\
&\Rightarrow \quad f(h)=f(h) \cdot f(0) \\
&\therefore \quad f(0)=1
\end{aligned}
$$
$$
2=\lim _{h \rightarrow 0} \frac{f(h-1)}{h}
$$
Also, $f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$$
\begin{array}{lr}
=\lim _{h \rightarrow 0} \frac{f(x) \cdot f(h)-f(x)}{h} & {[\because f(x+y)=f(x) \cdot f(y)]} \\
=\lim _{h \rightarrow 0} \frac{f(x)[f(h)-1]}{h}=2 f(x) & \quad \text { [using Eq. (ii)] } \\
\therefore \quad f^{\prime}(x)=2 f(x) &
\end{array}
$$