# Thread: find the first derivative of ln function

1. ## find the first derivative of ln function

$\displaystyle ln(\frac{x}{e^xln(x)})$

2. ## Re: find the first derivative of ln function

Originally Posted by rabert1
$\displaystyle ln(\frac{x}{e^xln(x)})$
It would probably help if you simplified using logarithm laws first...

\displaystyle \displaystyle \begin{align*} \ln{\left(\frac{x}{e^x\ln{x}}\right)} = \ln{x} - \ln{\left(e^x\ln{x}\right)} = \ln{x} - \left[\ln{\left(e^x\right)} + \ln{\left(\ln{x}\right)}\right] = \ln{x} - x - \ln{\left(\ln{x}\right)} \end{align*}

The derivative should be much easier to find now.

3. ## Re: find the first derivative of ln function

Originally Posted by rabert1
$\displaystyle ln(\frac{x}{e^xln(x)})$
without simplification of expression :

$\displaystyle f'(x)=\frac{e^x \ln x}{x} \cdot \left(\frac{x}{e^x \ln x}\right)'=\frac{e^x \ln x}{x} \cdot \frac{e^x \ln x -x \cdot (e^x \ln x)'}{e^{2x} \ln^2 x}$

$\displaystyle f'(x)=\frac{1}{x} \cdot \frac{e^x \ln x-xe^x\ln x-e^x}{e^x \ln x}=\frac{\ln x -x\ln x-1}{x\ln x}$