The proof of derivative of lnx?

1. Nov 11, 2009

coki2000

The proof of derivative of lnx??

Hi all,
In proof of derivative of lnx,

$$\frac{d}{dx}(lnx)=\lim_{\Delta x\rightarrow 0}\frac{\ln(x+\Delta x)-lnx }{\Delta x}= \lim_{\Delta x\rightarrow 0}\frac{1}{\Delta x}\ln(1+\frac{\Delta x}{x})=\lim_{\Delta x\rightarrow 0}\frac{x}{\Delta x}\frac{1}{x}\ln(1+\frac{\Delta x}{x})=\lim_{\Delta x\rightarrow 0}\frac{1}{x}ln(1+\frac{\Delta x}{x})^{(\frac{x}{\Delta x})}$$

For $$\frac{x}{\Delta x}=n$$
$$\lim_{\Delta x\rightarrow 0}ln(1+\frac{\Delta x}{x})^{(\frac{x}{\Delta x})}\Rightarrow \lim_{n\rightarrow\infty}\ln (1+\frac{1}{n})^{n}$$

So
$$(lnx)'=\frac{1}{x}ln(\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n})=\frac{1}{x}$$

so we take the $$\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}=e$$

But in proof of $$\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}$$ we take the derivative of lnx is $$\frac{1}{x}$$
(https://www.physicsforums.com/showthread.php?t=348071)

Please explain to me this contradiction.
Thanks.

Last edited: Nov 11, 2009
2. Nov 11, 2009

l'Hôpital

Re: The proof of derivative of lnx??

Do you mean the circular logic problem? e is defined as the number that makes the base log such that (log x)' = 1/x.

Alternatively, you could do a proof by the binomial theorem, and take the limit, you'll see the definition for e in series form.

Last edited: Nov 11, 2009
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