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Question Number 158671 by mahdipoor last updated on 07/Nov/21
∀x∈R ; f(x)=f ′(x)  prove f(x+y)=f(x)f(y)
$$\forall{x}\in{R}\:;\:{f}\left({x}\right)={f}\:'\left({x}\right) \\ $$$${prove}\:{f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right) \\ $$
Commented by mr W last updated on 07/Nov/21
generally  for f(x+y)=f(x)f(y) we have   f(x)=a^x  (a>0)  f′(x)=(ln a)a^x ≠f(x) (except a=e)  therefore  f′(x)=f(x) doesn′t mean   f(x+y)=f(x)f(y)  and f(x+y)=f(x)f(y) doesn′t mean   f′(x)=f(x).
$${generally} \\ $$$${for}\:{f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)\:{we}\:{have}\: \\ $$$${f}\left({x}\right)={a}^{{x}} \:\left({a}>\mathrm{0}\right) \\ $$$${f}'\left({x}\right)=\left(\mathrm{ln}\:{a}\right){a}^{{x}} \neq{f}\left({x}\right)\:\left({except}\:{a}={e}\right) \\ $$$${therefore} \\ $$$${f}'\left({x}\right)={f}\left({x}\right)\:{doesn}'{t}\:{mean}\: \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right) \\ $$$${and}\:{f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)\:{doesn}'{t}\:{mean}\: \\ $$$${f}'\left({x}\right)={f}\left({x}\right).\: \\ $$
Commented by mr W last updated on 07/Nov/21
you can′t prove because it′s not true.  example f(x)=10e^x .  you have f′(x)=10e^x =f(x)  but f(x+y)=10e^(x+y)   f(x)f(y)=10e^x ×10e^y =100e^(x+y)   f(x+y)≠f(x)f(y)
$${you}\:{can}'{t}\:{prove}\:{because}\:{it}'{s}\:{not}\:{true}. \\ $$$${example}\:{f}\left({x}\right)=\mathrm{10}{e}^{{x}} . \\ $$$${you}\:{have}\:{f}'\left({x}\right)=\mathrm{10}{e}^{{x}} ={f}\left({x}\right) \\ $$$${but}\:{f}\left({x}+{y}\right)=\mathrm{10}{e}^{{x}+{y}} \\ $$$${f}\left({x}\right){f}\left({y}\right)=\mathrm{10}{e}^{{x}} ×\mathrm{10}{e}^{{y}} =\mathrm{100}{e}^{{x}+{y}} \\ $$$${f}\left({x}+{y}\right)\neq{f}\left({x}\right){f}\left({y}\right) \\ $$
Commented by mahdipoor last updated on 07/Nov/21
thanks
$${thanks} \\ $$

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