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

Explanation

Given that,
`f: R->R`
satisfies that equation

`f(x+y)=f(x)f(y)`
for all `x ,\ y in R`
, `f(x)!=0`
Then `f(0+0)=f(0).f(0)`

`=> f(0)=1`
Also given,

`f'(0)=2`
`=>lim_(h->0)(f(0+h)-f(0))/h=2`
`=>lim_(h->0)(f(0)f(h)-f(0))/h=2`
`=> lim_(h->0)(f(h)-1)/h=2`
now,
`f'(x)=lim_(h->0)(f(x+h)-f(x))/h`
`=lim_(h->0)(f(x)f(h)-f(x))/h`
`=f(x)lim_(h->0)(f(h)-1)/h`
`=2f(x)`
so,`f^(prime)(x)=2\ f(x)`