If $f(x+y)=f(x)\cdot f(y)$ for all $x,y\in \mathbb{R}$ such that $f(5)=2$ and $f^{\prime }(0)=3$, then the value of $f^{\prime }(5)$ is:

For the functional equation $f(x+y)=f(x)f(y)$, the standard solution is $f(x)=a^x$.Differentiating with respect to $y$:$f^{\prime }(x+y)=f(x)\cdot f^{\prime }(y)$ Putting $y=0$:$f^{\prime }(x)=f(x)\cdot f^{\prime }(0)$ Given $f^{\prime }(0)=3$:$f^{\prime }(x)=3f(x)$ For $x=5$:$f^{\prime }(5)=3f(5)$ Given $f(5)=2$:$f^{\prime }(5)=3\cdot 2=6$.