Question

Suppose that for all $x,y\in R,f(x+y)=f\left(x\right).f\left(y\right)$ and $f^{\prime }\left(0\right)$ exist then show that $f\left(x\right)$ exists and equals to $f\left(x\right)\cdot f^{\prime }\left(0\right)$ for all $x\in R$

Answer 1

$f(x+y)=f\left(x\right).f\left(y\right)$

Let $f\left(x\right)=a^x$

$a^{x+y}=a^x.a^y$

$⟹f\left(x\right)=a^x$

$f^{\prime }\left(x\right)=a^x{\log }_{e}a{f}^{\prime }\left(0\right)=1$

$⟹f\left(x\right)=f\left(x\right).f^{\prime }\left(0\right)$