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Problem is that Suppose f is differentiable everywhere and $f(x+y)=f(x)f(y)$ for all $x,y$ . Show that $f'(x)=f'(0)f(x)$ and determine the value of $f'(0)$.

I can show $f'(x)=f'(0)f(x)$

but i don't know how to determine the value of $f'(0)$.

Please help!

nonuser
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fn5six
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  • See https://math.stackexchange.com/questions/1944229/does-a-function-that-satisfies-the-equality-fab-fafb-have-to-be-expon?noredirect=1&lq=1. – StubbornAtom Sep 22 '18 at 11:11
  • In the two equations,$f(x+y)=f(x)f(y)$ and $f'(x)=f'(0)f(x)$ substitute $x=y=0$. – Soham Sep 22 '18 at 11:27
  • @tatan If I do what you say I get first $f(0)=1$ and then $f'(0)=f'(0)$. I don't think that counts as determining the value of $f'(0)$. – David C. Ullrich Sep 22 '18 at 12:45
  • @tatan "substitute $x=y=0$" Right, done, and now what? – Did Sep 22 '18 at 12:56
  • The value of $f'(0)$ is a function of $f(1)$ and it can't be determined unless $f(1)$ is known. It can be proved with some effort that either $f'(0)=0$ (when $f$ is constant) or $f'(0)=\log f(1)$. – Paramanand Singh Sep 22 '18 at 15:17

1 Answers1

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It's impossible to determine the value of $f'(0)$ from the information given - perhaps you left something out, perhaps the question was stated somewhat differently, or perhaps it's a bad question.

If $a$ is any real number and $f(t)=e^{at}$ then $f$ is differentiable, $f(x+y)=f(x)f(y)$, and $f'(0)=a$. So the information given literally says nothing at all about the value of $f'(0)$; that value can be anything.

David C. Ullrich
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