Answer
See explanations.
Work Step by Step
Step 1. Identify the given conditions: (i) $f(x+y)=f(x)\cdot f(y)$, (ii) $f(x)=1+x\ g(x)$ and $\lim_{x\to0}g(x)=1$
Step 2. Use the definition of the limit and the above relations: $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{f(x)(f(h)-1)}{h}=f(x)\lim_{h\to0}\frac{f(h)-1}{h}$
Step 3. As $f(h)=1+h\ g(h)$ and $\lim_{h\to0}g(h)=1$, we have $f'(x)=f(x)\lim_{h\to0}\frac{1+h\ g(h)-1}{h}=f(x)\lim_{h\to0}g(h)=f(x)$
