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If $f (x) = |\log_{e}| x | |$ then $f^{'} (x)$ equals

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$\frac{1}{| x |}, where x \neq 0$

$\frac{1}{x} for | x | > 1 and - \frac{1}{x} for | x | < 1$

$- \frac{1}{x} for | x | > 1 and \frac{1}{x} for | x | < 1$

$\frac{1}{x} for | x | > 0 and - \frac{1}{x} for | x | < 1$

answer is B.

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Detailed Solution

$For x > 1, we have f (x) = | \log | x | | = \log x$ $or f^{'} (x) = \frac{1}{x}$

$For x < - 1, we have f (x) = | \log | x | | = \log (- x)$ $or f^{'} (x) = \frac{1}{x}$

$For 0 < x < 1, we have f (x) = | \log | x | | = - \log x$ $or f^{'} (x) = \frac{- 1}{x}$

$For - 1 < x < 0, we have f (x) = - \log (- x)$

$or f^{'} (x) = \frac{- 1}{x}; Hence, f^{'} (x) = \{\begin{cases} \frac{1}{x}, | x | > 1 \\ - \frac{1}{x} | x | < 1 \end{cases}$

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