﻿ Let f(x + y) = f(x) f(y) for all x, y Ïµ R and suppose that f is differentiable at 0 and f’(0) = 4. If f(x0) = 8 then f’(x0) is equal to : Kaysons Education

# Let f(x + y) = f(x) f(y) For All x, y Ïµ R and Suppose That f is Differentiable At 0 And f’(0) = 4. If f(x0) = 8 Then f’(x0) Is Equal To

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Let y = 0. Then (x) = (xf (0), i.e., (x) [(0) – 1] = 0. So (x) = 0 or f(0) = 1. If (x) = 0 for all x, the f is clearly differentiable. Suppose therefore, that f is a non-zero function, so that (0) = 1. Since f is differentiable at x = 0, we have

Let x0 Ïµ R. Then

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