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A function f: R->R satisfies the equ...

A function `f: R->R` satisfies the equation `f(x+y)=f(x)f(y)` for all`x , y in R` and `f(x)!=0 ` for all ` x in Rdot` If `f(x)` is differentiable at `x=0a n df^(prime)(0)=2,` then prove that `f^(prime)(x)=2f(x)dot`

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