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Question Number 158671 by mahdipoor last updated on 07/Nov/21

\forall x \in R; f (x) = f^{'} (x)

p r o v e f (x + y) = f (x) f (y)

Commented by mr W last updated on 07/Nov/21

g e n e r a l l y

f o r f (x + y) = f (x) f (y) w e h a v e

f (x) = a^{x} (a > 0)

f^{'} (x) = (\ln a) a^{x} \neq f (x) (e x c e p t a = e)

t h e r e f o r e

f^{'} (x) = f (x) {d o e s n}^{'} t m e a n

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

a n d f (x + y) = f (x) f (y) {d o e s n}^{'} t m e a n

f^{'} (x) = f (x) .

Commented by mr W last updated on 07/Nov/21

y o u {c a n}^{'} t p r o v e b e c a u s e {i t}^{'} s n o t t r u e .

e x a m p l e f (x) = 10 e^{x} .

y o u h a v e f^{'} (x) = 10 e^{x} = f (x)

b u t f (x + y) = 10 e^{x + y}

f (x) f (y) = 10 e^{x} \times 10 e^{y} = 100 e^{x + y}

f (x + y) \neq f (x) f (y)

Commented by mahdipoor last updated on 07/Nov/21

t h a n k s