x/x^2lnx

**Boo** · January 22nd 2013, 03:21 AM

All!
Please, how do I get f'(x) and f''(x) if I have f(x)=x/(x^lnx)?

Many thanks!

**topsquark** · January 22nd 2013, 05:49 AM

Originally Posted by Boo

All!
Please, how do I get f'(x) and f''(x) if I have f(x)=x/(x^lnx)?

Many thanks!

Which do you have? $f(x) = \frac{x}{x^2 ln(x)}$

or
$f(x) = \frac{x}{x^{ln(x)}}$

-Dan

Or maybe:
$f(x) = \frac{x}{x^{2ln(x)}}$

**Soroban** · January 22nd 2013, 06:16 AM

Hello, Boo!

$\text{Given: }\:f(x) \:=\:\dfrac{x}{x^{\ln x}}. \quad \text{Find }f'(x)\text{ and }f''(x).$

You need Logarithmic Differentiation.

We have: . $y \;=\;\dfrac{x}{x^{\ln x}}$

Take logs: . $\ln(y) \;=\;\ln\left(\frac{x}{x^{\ln x}}\right)$

. . . . . . . . $\ln(y) \;=\;\ln(x) - \ln\left(x^{\ln x}\right)$

. . . . . . . . $\ln(y) \;=\;\ln(x) - \ln x\ln x$

. . . . . . . . $\ln(y) \;=\;\ln(x) - [\ln(x)]^2$

Differentiate implicitly:

. . . . . . . . $\frac{y'}{y} \;=\;\frac{1}{x} - 2\ln(x)\!\cdot\!\frac{1}{x}$

. . . . . . . . $\frac{y'}{y} \;=\;\frac{1}{x}\left[1 - 2\ln(x)\right]$

. . . . . . . . $y' \;=\;y\cdot \frac{1-2\ln(x)}{x}$

. . . . . . . . $y' \;=\;\frac{x}{x^{\ln x}}\cdot\frac{1-2\ln(x)}{x}$

Hence: . . $y' \;=\;\frac{1-2\ln(x)}{x^{\ln x}}$

Now repeat the process to find the second derivative, $y''.$

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