F (x + y) = f (x) f (y), and the derivative of F (0) exists. It is proved that f '(x) = f (x) f' (0)

According to the definition of derivative, f '(x) = [f (x + T) - f (x)] / T (where t is a small partition of X tending to 0), then f' (x) = [f (x) * f (T) - f (x)] / T = f (x) * [f (T) - 1] / T. from the meaning of the title, we can see: F (x) = f (x + 0) = f (x) * f (0), then f '(x) = f (x) * [f (0 + T) - f (0)] / T = f (x) * f' (0)

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