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A function f: R->R satisfies that equati...

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)` .

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