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Q. A function $f: R \rightarrow R$ satisfies the equation $f(x +y)=f(x) .f(y), x, y \in R$ and $f(x) \neq 0$ for any $x \forall R$. If $f$ is differentiable at ' 0 ' $f'(0)=3$ and $f(5)=2$ then $f'(5)=$

Solution:

$f(x)=f(x+0)=f(x). f(0)$
$\Rightarrow f(0)=1$
$f'(5)=\displaystyle\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}$
$\displaystyle\lim _{h \rightarrow 0} \frac{f(5) f(h)-f(5)}{h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{f(5)[f(h)-1]}{h}$
$=f(5).\displaystyle\lim _{h \rightarrow 0} \frac{f(h)-f'(0)}{h}=f(5) f(0)$
$=2 \times 3=6$