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I need a hint to recover the definition of the multiplicative derivative as $f^*(x)=e^{f'(x)/f(x)}$ starting from the definition with the limit $$f^*(x)=\lim_{h\to0}\Big(\frac{f(x+h)}{f(x)}\Big)^{\frac{1}{h}}.$$

I've tried to add and subtract $f(x)$ finding $$f^*(x)=\lim_{h\to0}\Big(1+\frac{f(x+h)-f(x)}{f(x)}\Big)^{\frac{1}{h}}.$$ The next trick would be to rewrite $1/h$ as $$\frac{f(x)}{f(x+h)-f(x)}\frac{f(x+h)-f(x)}{h}\frac{1}{f(x)}$$ and calling $A(h)=\frac{f(x+h)-f(x)}{f(x)}$ one have $$f^*(x)=\lim_{h\to0}\Big(1+A(h)\Big)^{\frac{1}{A(h)}\frac{f(x+h)-f(x)}{h}\frac{1}{f(x)}}.$$ I suspect that $$\lim_{h\to0}(1+h)^\frac{1}{h}=e,$$ $$\lim_{h\to0}A(h)=0$$ and $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ must be used but I fail to see how. What step one can do to recover the relation $f^*(x)=e^{f'(x)/f(x)}$?

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\begin{align} f^*(x) & = \lim_{h\to0}\left(\frac{f(x + h)}{f(x)}\right)^{1/h} = \exp\log \lim_{h\to0}\left(\frac{f(x + h)}{f(x)}\right)^{1/h} \\[10pt] & = \exp \lim_{h\to0} \log \left(\frac{f(x + h)}{f(x)}\right)^{1/h} = \exp \lim_{h\to0} \frac {\log f(x+h) - \log f(x)} h \\[10pt] & = \exp \left( \frac d {dx} \log f(x) \right) = \exp \frac{f'(x)}{f(x)}. \end{align}

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    $\begingroup$ "padre del figlio di Iorio" works always.. $\endgroup$ – yngabl Sep 13 '16 at 16:09
  • $\begingroup$ @yngabl It was "La figlia di Iorio", but it's funny as well. ;-) $\endgroup$ – egreg Sep 13 '16 at 16:10

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