frac{d}{d x}(log _{|x|} e) = . . . . . . | vedclass

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Question

(A) We know that $\log _{|x|} e = \frac{1}{\log _e |x|} = \frac{1}{\ln |x|}$.Let $y = \frac{1}{\ln |x|}$.Using the chain rule,$\frac{d y}{d x} = \frac

Vedclass · Accepted Answer

$\frac{-1}{x(\log |x|)^2}$

Vedclass · Answer

(A) We know that $\log _{|x|} e = \frac{1}{\log _e |x|} = \frac{1}{\ln |x|}$.Let $y = \frac{1}{\ln |x|}$.Using the chain rule,$\frac{d y}{d x} = \frac{d}{d x}(\ln |x|)^{-1} = -1 \cdot (\ln |x|)^{-2} \cdot \frac{d}{d x}(\ln |x|)$.Since $\frac{d}{d x}(\ln |x|) = \frac{1}{x}$,we have $\frac{d y}{d x} = -(\ln |x|)^{-2} \cdot \frac{1}{x} = \frac{-1}{x(\ln |x|)^2}$.Thus,the correct option is $A$.

$\frac{d}{d x}(\log _{|x|} e) =$ . . . . . .

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