The correct option is D 1/x(log |x|)^2 ddxlog_|x|e =ddx(logelog|x|)=ddx(1log|x|) = 1(log|x|)^2×1|x|×x|x| = 1(log|x|)^2×x|x|^2 = ...

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Logarithmic Differentiation

d/dx log|x|e=...

Question

$\frac{d}{d x} l o g_{| x |} e =$ ____________.

\frac{- 1}{x (l o g | x |)^{2}}

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\frac{1}{(l o g x)^{2}}

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\frac{1}{| x |}

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e^{x}

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Solution

The correct option is D $\frac{- 1}{x (l o g | x |)^{2}}$
$\frac{d}{d x} l o g_{| x |} e$

$= \frac{d}{d x} (\frac{l o g e}{l o g | x |}) = \frac{d}{d x} (\frac{1}{l o g | x |})$

$= \frac{- 1}{(l o g | x |)^{2}} \times \frac{1}{| x |} \times \frac{x}{| x |}$

$= \frac{- 1}{(l o g | x |)^{2}} \times \frac{x}{| x |^{2}}$

$= \frac{- 1}{(l o g | x |)^{2}} \times \frac{x}{x^{2}}$

$= \frac{- 1}{x (l o g | x |)^{2}}$

The answer is option (A).

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