Real Analysis, differentiation

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Solved: Real Analysis, differentiation

Homework Statement


If g is differentiable and g(x+y)=g(x)(g(y) find g(0) and show g'(x)=g'(0)g(x)

The Attempt at a Solution


I solved g(0)=1

and

I got as far as

<br /> g&#039;(x)=\lim_{\substack{x\rightarrow 0}}g(x) \frac{g(h)-1}{h}<br />

but now I am stuck.

Thank you in advance
 
Last edited:
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Never mind... Just did g'(0) and plug in...
 
Your formula for g' is incorrect- though it may be a typo.
g&#039;(x)= \lim_{\substack{h\rightarrow 0}}g(x)\frac{g(h)- 1}{h}
where the limit is taken as h goes to 0, not x. Since "g(x)" does not depend on h, you can factor that out:
g&#039;(x)= g(x)\left(\lim_{\substack{h\rightarrow 0}}\frac{g(h)-1}{h}\right)
and you should be able to see that the limit is simply the definition of g'(0).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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