Question

Let $f(x+y)=f\left(x\right)\cdot f\left(y\right)\forall x,y\in R.$ If $f\left(5\right)=2$ and $f^{\prime }\left(0\right)=3,$ then the value of $f^{\prime }\left(5\right)$ is

Answer 1

Given, $f\left(5\right)=2$ and $f^{\prime }\left(0\right)=3$
$f(x+y)=f\left(x\right)\cdot f\left(y\right)...\left(1\right)$
Put $x=5$ and $y=0,$ we get
$f\left(5\right)=f\left(5\right)\cdot f\left(0\right)$
$\Rightarrow f\left(0\right)=1$

Differentiating $\left(1\right)$ w.r.t $x$
$f^{\prime }(x+y)(1+y^{\prime })=f^{\prime }\left(x\right)\cdot f\left(y\right)+f\left(x\right)\cdot f^{\prime }\left(y\right)\cdot y^{\prime }$
Put $x=5$ and $y=0,$ we get
$\Rightarrow f^{\prime }\left(5\right)(1+y^{\prime })=f^{\prime }\left(5\right)\cdot f\left(0\right)+f\left(5\right)\cdot f^{\prime }\left(0\right)\cdot y^{\prime }$
Since, $f\left(0\right)=1,f\left(5\right)=2$ and $f^{\prime }\left(0\right)=3$
$\therefore f^{\prime }\left(5\right)+y^{\prime }{f}^{\prime }\left(5\right)=f^{\prime }\left(5\right)+2\cdot 3\cdot y^{\prime }$
$\Rightarrow y^{\prime }{f}^{\prime }\left(5\right)=6y^{\prime }$
$\Rightarrow f^{\prime }\left(5\right)=6$