If $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x$and $y$ and $f^{|}\left(0\right)$ exists, then the value of $f^{|}\left(2\right)$ if   $f\left(2\right)=5$ and $f^{|}\left(0\right)=3$ is equal to

$f(x+y)=f\left(x\right)f\left(y\right);f\left(2\right)=5;f^{|}\left(0\right)=3$

$f(2+0)=f\left(2\right).f\left(0\right)$

$\Rightarrow f\left(0\right)=1$

$f^{|}\left(2\right)=\underset{h\to 0}{Lt}\frac{f(2+h)-f\left(2\right)}{h}$

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

$=\underset{h\to 0}{Lt}\frac{f\left(2\right)(f\left(h\right)-1)}{h}$

$=f\left(2\right).\underset{h\to 0}{Lt}\frac{f\left(h\right)-f\left(0\right)}{h}$

$=f\left(2\right).f^{|}\left(0\right)$

$=5\times 3$

$=15$