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The proof of derivative of lnx?

  1. Nov 11, 2009 #1
    The proof of derivative of lnx??

    Hi all,
    In proof of derivative of lnx,

    [tex]\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})}[/tex]

    For [tex]\frac{x}{\Delta x}=n[/tex]
    [tex]\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}[/tex]

    So
    [tex](lnx)'=\frac{1}{x}ln(\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n})=\frac{1}{x}[/tex]

    so we take the [tex]\lim_{n\rightarrow\infty}(1+\frac{1}{n})^{n}=e[/tex]

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

    Please explain to me this contradiction.
    Thanks.
     
    Last edited: Nov 11, 2009
  2. jcsd
  3. Nov 11, 2009 #2
    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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