A function` f:R -> R` satisfies the equation` f(x+y)=f(x)f(y)` for all `x,y in R`.`f(x)!=0` Suppose that the function is differentiable at `x=0` and `f^prime(0)=2`. Prove that `f^prime(x)=2f(x)`.

Explanation

We have given,

To proof: `f^prime(x)=2f(x)`

`f(x+y)=f(x).f(y)` for all` x,y in R.......(1)`

Then `f(0+0)=f(0).f(0)`

or, `f(0)=1` [ Since `f(x) in
0` for all `x in R].......(2)`.

Also given,

`f′(0)=2`

...