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A function f: RvecR satisfies the equati...

A function `f: RvecR` satisfies the equation `f(x+y)=f(x)f(y)` for all`x , y in Ra n df(x)!=0fora l lx 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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We have f(x+y) =f(x)f(y) for all `x, y in R`
`therefore" "f(0)=f(0)f(0) or f(0) {f(0)-1}=0`
`or" "f(0)=1" "[becausef(0)ne0]`
Now, f'(0)=2
`or" "underset(hrarr0)lim(f(0+h)-f(0))/(h)=2`
`or" "underset(hrarr0)lim(f(h)-1)/(h)=2" "(becausef(0)=1)" (1)"`
`"Now, "f'(x)=underset(hrarr0)lim(f(x+h)-f(x))/(h)`
`=underset(hrarr0)lim(f(x)f(h)-f(x))/(h)" "[becausef(x+y)=f(x)f(y)]`
`=f(x)(underset(hrarr0)lim(f(h)-1)/(h))=2f(x)" [Using (1)]"`
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