Answer
$\displaystyle \frac{1}{x\ln x}$
Work Step by Step
By theorem 5.3/2:$\ \ \ \displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$
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Here,$ u(x)=\ln x^{2}=2\ln x$ (property of ln(x), Th.5.2.3)
$u^{\prime}(x)=2\displaystyle \cdot\frac{1}{x}=\frac{2}{x}$
So, $\ \ \ \displaystyle \frac{d}{dx}[\ln$($\ln x^{2}$)]$=\displaystyle \frac{1}{\ln x^{2}}\cdot\frac{2}{x}=\frac{2}{2x\ln x}=\frac{1}{x\ln x}$
