frac{d}{d x}left(log left(frac{1}{x}right)+lo | vedclass

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Question

(A) Let $f(x) = \log \left(\frac{1}{x}\right) + \log \left(\frac{1}{x^2}\right) + \log \left(\frac{1}{x^3}\right)$.Using the property $\log(a^b) = b \

Vedclass · Accepted Answer

$-\frac{6}{x}$

Vedclass · Answer

(A) Let $f(x) = \log \left(\frac{1}{x}ight) + \log \left(\frac{1}{x^2}ight) + \log \left(\frac{1}{x^3}ight)$.Using the property $\log(a^b) = b \log(a)$ and $\log(\frac{1}{x}) = -\log(x)$,we can simplify the expression:$f(x) = -\log(x) - 2\log(x) - 3\log(x)$$f(x) = -6\log(x)$.Now,differentiate with respect to $x$:$\frac{d}{dx} f(x) = \frac{d}{dx} (-6\log(x))$$= -6 	imes \frac{1}{x} = -\frac{6}{x}$.Thus,the correct option is $A$.

$\frac{d}{d x}\left(\log \left(\frac{1}{x}\right)+\log \left(\frac{1}{x^2}\right)+\log\left(\frac{1}{x^3}\right)\right) = \text{ . . . . . . }$,$x > 1$

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