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A function f:R rarr R" satisfies the equ...

A function `f:R rarr R"` satisfies the equation `f(x+y)=f(x)f(y)` for all values of `x" and "y` and for any `x in R,f(x)!=0`. Suppose the function is differentiable at `x=0" and "f'(0)=2`, then for all `x in R,f(x)=`

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Let `f : R rarr R` satisfies the equation `f(x+y) = f(x) . f(y), AA x, y in R, f(x) ne 0`.
Let `f(x)` is differentiable at `x = 0` and `f'(0) = 2`
`rArr f'(0) = underset(xrarr0)(lim) (f(x) - f(0))/(x-0)`
`rArr 2 = underset(xrarr0)(lim) (f(x) - f(0))/(x)`
` rArr 2 = underset(hrarr0)(lim)(f(0+h)-f(0))/(h)`
`rArr 2 = underset(hrarr0)lim(f'(0).f(h)-f(0))/(h)`
`rArr 2 = underset(hrarr0)(lim) (f(0) [f(h) - 1])/(h)` , `[:' f(0) = f (h) "....."(i)]`
Also, `f'(x) = underset(hrarr0)(lim)(f(x+h)-f(x))/(h)`
` =underset(hrarr0)(lim)(f(x).f(h)-f(x))/(h), [:' f(x+y)=f(x).f(y)]`
`=underset(hrarr0)(lim)(f(x)[f(h)-1])/(h) = 2f(x)` , [using Eq. (i)]
`:. f'(x) = 2f(x)`
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