Answer
$-\frac{1}{x(\ln x)^2}$
Work Step by Step
$\frac{d}{dx}\frac{1}{\ln x}$
$=\frac{d}{dx}(\ln x)^{-1}$
$=-(\ln x)^{-2}\frac{d}{dx}\ln x$
$=-(\ln x)^{-2}*\frac{1}{x}$
$=-\frac{1}{x(\ln x)^2}$
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