Question

Let $f\left(x\right)$ is a differentiable function on $x\in R$ , such that $f\left(x+y\right)=f\left(x\right)f\left(y\right)$ for all $x,y\in R$ where $f\left(0\right)\ne 0.$ If $f\left(5\right)=10,f^{{}^{\prime }}\left(0\right)=6$ , then the value of $f^{{}^{\prime }}\left(5\right)$ is equal to

Answer 1

Answer: 60

$f^{{}^{\prime }}\left(5\right)=\underset{h\to 0}{\lim }\frac{f\left(5+h\right)-f\left(5\right)}{h}$
$=\underset{h\to 0}{\lim }\frac{f\left(5+h\right)-f\left(5+0\right)}{h}$
$=\underset{h\to 0}{\lim }\frac{f\left(5\right).f\left(h\right)-f\left(5\right).f\left(0\right)}{h}$
$[\because f\left(x+y\right)=f\left(x\right).f\left(y\right)forallx,y]$
$=\left(\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h}\right).f\left(5\right)$
$=f^{{}^{\prime }}\left(0\right).f\left(5\right)=6\times 10$
$=60$