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Q. The set of real values of x for which $ f(x) = \frac {x}{log\, x}$ increasing, is

KCETKCET 2010Application of Derivatives

Solution:

$f(x)= \frac{x}{\log x}$
$f'(x)=\frac{\log x \cdot 1-x \cdot \frac{1}{x}}{(\log x)^{2}}=\frac{(\log x-1)}{(\log x)^{2}}$
We know that, $f(x)$ is increasing (strictly) When $f^{\prime}(x)>0$
$\Rightarrow \frac{(\log x-1)}{(\log x)^{2}} >0 $
$\Rightarrow (\log x-1)>0$
$\Rightarrow \log x>1$
$\Rightarrow \log _{e} x>\log _{e} e$
$\Rightarrow x >e$
Hence, $x: x \geq e$