The slope of normal to the curve y= log (logx) at x = e is

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-e

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\frac{1}{e}

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\frac{1}{e}

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Solution

The correct option is C $\frac{1}{e}$
$\begin{matrix} \frac{d}{d x} log (log x) S u b s t i t u t e u = log x = \frac{d}{d u} log u \frac{d}{d x} log x = \frac{1}{u} (\frac{1}{x}) = \frac{1}{log x} (\frac{1}{x}) S u b s t i t u t e x = e = \frac{1}{log e} (\frac{1}{e}) = \frac{1}{e} \end{matrix}$

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