MHB How Is the Derivative of ln(x) Proven Using Limits?

  • Thread starter Thread starter Jameson
  • Start date Start date
AI Thread Summary
The derivative of the natural logarithm, defined as d/dx ln(x) = 1/x, is proven using the limit definition of a derivative. The proof involves evaluating the limit of the difference quotient as Δx approaches zero, transforming it into a form that utilizes the exponential function's limit definition. By substituting and manipulating the expression, both one-sided limits converge to 1/x. This confirms that the derivative of ln(x) is indeed 1/x. The continuity of ln(x) allows for the limit to be moved inside the logarithmic function during the proof.
Jameson
Insights Author
Gold Member
MHB
Messages
4,533
Reaction score
13
Define the derivative of the natural logarithm to be: [math]\frac{d}{dx} \ln(x) = \frac{1}{x}[/math]

Demonstrate this rule is valid by using the limit definition of a derivative.

Hint:
[sp]
This definition of the exponential function is necessary to calculate the limit.
[math]e^x = \lim_{n \to \infty} \left({1 + \frac x n}\right)^n[/math]
[/sp]

Remember to read the http://www.mathhelpboards.com/threads/773-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
Congratulations to the following members for their correct solutions:

1) Sudharaka
2) checkittwice
3) Reckoner
4) veronica1999

Solution (from Reckoner):

[sp]\[\frac d{dx}\ln(x) = \lim_{\Delta x\to 0}\frac{\ln\left(x+\Delta x\right) - \ln(x)}{\Delta x}\]

\[=\lim_{\Delta x\to0}\frac1{\Delta x}\ln\left(\frac{x+\Delta x}x\right)\]

\[=\lim_{\Delta x\to0}\frac1{\Delta x}\ln\left(1+\frac{\Delta x}x\right)\]

\[=\lim_{\Delta x\to0}\ln\left(\left[1+\frac{\Delta x}x\right]^{1/\Delta x}\right)\]

Let \(n = \frac1{\Delta x}\). Then for \(\Delta x > 0, n\to\infty\) as \(\Delta x\to0\) and for \(\Delta x < 0, n\to-\infty\) as \(\Delta x\to0\). So we have

\[\lim_{\Delta x\to0^+}\ln\left(\left[1+\frac{\Delta x}x\right]^{1/\Delta x}\right) = \lim_{n\to\infty}\ln\left(\left[1+\frac{1/x}n\right]^n\right)\]

\[=\ln\left(e^{1/x}\right) = \frac1x.\]

Note that the continuity of \(\ln(x)\) allowed us to move the limit inside the function. Similarly,

\[\lim_{\Delta x\to0^-}\ln\left(\left[1+\frac{\Delta x}x\right]^{1/\Delta x}\right) = \lim_{n\to-\infty}\ln\left(\left[1+\frac{1/x}n\right]^n\right)\]

\[ = \lim_{n\to\infty}\ln\left(\left[1+\frac{1/x}{-n}\right]^{-n}\right)\]

\[ = \lim_{n\to\infty}-\ln\left(\left[1+\frac{-1/x}n\right]^n\right)\]

\[ = -\ln\left(e^{-1/x}\right) = \frac1x.\]

Since both one-sided limits exist and are equal to \(\frac1x\),

\[\lim_{\Delta x\to0}\ln\left(\left[1+\frac{\Delta x}x\right]^{1/\Delta x}\right) = \frac1x\]

and thus,

\[\frac d{dx}\ln(x) = \frac1x.\][/sp]
 
Back
Top