Q. If $f(x) = \log_{x}{(\log_{e}{x),\ then\ f^{'}(x)\ at\ x = e}}$ is equal to
A) 1
B) 2
C) 0
D) $\frac{1}{e}$
Explanation
(d)
Given,
$f(x) = \log_{x}{(\log_{e}{x) = \frac{\log_{e}{\log_{e}x}}{\log_{e}x}}}$
On differentiating w.r.t. $x,$ we get
$f^{'}(x) = \frac{\log_{e}{x.\frac{1}{\log_{e}x}.\frac{1}{x} - \log_{e}{\log_{e}{x.}}\frac{1}{x}}}{\left( \log_{e}x \right)^{2}}$
$\Rightarrow f^{'}(x) = \frac{1 - \log_{e}{\log_{e}x}}{x\left( \log_{e}x \right)^{2}}$
$\Rightarrow f^{'}(e) = \frac{1 - \log_{e}{\log_{e}e}}{e\left( \log_{e}e \right)^{2}} = \frac{1 - \log_{e}1}{e} = \frac{1}{e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.