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Q. The function $f(x)=\frac{x}{\log \,x} $ increases on the interval

Application of Derivatives

Solution:

Clearly $f(x)$ is defined for $x > 0$
$f'\left(x\right)=\frac{log\,x\cdot1\cdot x\cdot\frac{1}{x}}{\left(log\,x\right)^{2}}=\frac{log\,x-1}{\left(log\,x\right)^{2}}$
$\therefore f'\left(x\right) > 0$
$\Rightarrow log\,x-1 > 0$
$\Rightarrow log\,x > 1$
$\Rightarrow x > e$
$\therefore x \in\left(e, \infty\right)$.