Answer :

Given f(x) is differentiable at x = 0 and f(x) ≠ 0


And f(x + y) = f(x)f(y) also f’(0) = 2


To prove: f’(x) = 2f(x)


As we know that,


f’(x) =


as f(x+h) = f(x)f(h)


f’(x) =


f’(x) = …(1)


As


f(x + y) = f(x)f(y)


put x = y = 0


f(0+0) = f(0)f(0)


f(0) = {f(0)}2


f(0) = 1 { f(x) ≠ 0 ….given}


equation 1 is deduced as-


f’(x) =


f’(x) =


f’(x) = f(x)f’(0) {using formula of derivative}


f’(x) = 2 f(x) …proved { it is given that f’(0) = 2}


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