Answer
$ -\displaystyle \frac{1}{x\ln^{2}|x|}$
Work Step by Step
By the quatient rule, $\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$f(x)=1,\qquad f^{\prime}(x)=0$
$g(x)=\ln|x|$
... Derivatives of logarithms of absolute values:
$g^{\prime}(x)=\displaystyle \frac{d}{dx} [\displaystyle \ln|x|]=\frac{1}{x} $
So
$\displaystyle \frac{d}{dx}[\frac{1}{\ln|x|}]= \displaystyle \frac{0-1\cdot\frac{1}{x}}{[\ln|x|]^{2}}$
$= -\displaystyle \frac{1}{x\ln^{2}|x|}$
