If f(x) = frac{x}{log x},then f(x) is increas | vedclass

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(B) Given $f(x) = \frac{x}{\log x}$.To find the interval where $f(x)$ is increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{(\log x)(1) - (

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$(e, \infty)$

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(B) Given $f(x) = \frac{x}{\log x}$.To find the interval where $f(x)$ is increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{(\log x)(1) - (x)(\frac{1}{x})}{(\log x)^2} = \frac{\log x - 1}{(\log x)^2}$.For $f(x)$ to be increasing,we must have $f'(x) > 0$.Since $(\log x)^2 > 0$ for all $x > 0$ (and $x 
eq 1$),the sign of $f'(x)$ depends on the numerator $\log x - 1$.$\log x - 1 > 0 \Rightarrow \log x > 1 \Rightarrow x > e$.Thus,$f(x)$ is increasing in the interval $(e, \infty)$.Comparing this with the given options,the function is increasing in $(e, \infty)$.

If $f(x) = \frac{x}{\log x}$,then $f(x)$ is increasing in

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