# Problem of the week #14 - July 2nd, 2012

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#### Jameson

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Define the derivative of the natural logarithm to be: $$\displaystyle \frac{d}{dx} \ln(x) = \frac{1}{x}$$

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

Hint:

This definition of the exponential function is necessary to calculate the limit.
$$\displaystyle e^x = \lim_{n \to \infty} \left({1 + \frac x n}\right)^n$$

#### Jameson

Staff member
Congratulations to the following members for their correct solutions:

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

Solution (from Reckoner):

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

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

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

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

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,

$\frac d{dx}\ln(x) = \frac1x.$

• Reckoner
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