If f (x + y) = f(x). f( y) ∀ x.y and f (5) = 2, f'(0) = 3 , then f'(5) is
(a) 0
(b) 1
(c) 6
(d) 2
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Answer:
(c) 6
Step-by-step explanation:
use x, y=1
f(2) = f(1)^2
f(3) = f(1) f(2) = f(1)^3
f(4) = f(1) f(3) = f(1)^4
Therefore f(nx) = f(x)^n
nf'(nx) = nf(x)^n-1 f'(x)
f'(nx) = f(x)^n-1 f'(x)
putting n=0 we get
f'(0) = f'(x)/f(x)
putting x = 5
3 = f'(5)/2 . Hence f'(5) = 6
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