Q. If $f(x) = \log_{x^{2}}\ \left( \log_{e}x \right),\ then\ f^{'}(x)\ at\ x = e$ is
A) 1
B) $\frac{1}{e}$
C) $\frac{1}{2e}$
D) 0
Explanation
(c)
Given, $f(x) = {\log_{x^{2}}\ }{(\log_{e}{x) = \frac{1}{2}\log_{x}{(\log_{e}{x)}}}}$
$\Rightarrow (x) = \frac{1}{2}\frac{\log_{e}{\log_{e}x}}{\log_{e}x}$
$\Rightarrow f^{'}(x) = \frac{1}{2}\frac{\log_{e}{x\left( \frac{1}{x\log_{e}x} \right) - \log_{e}{\log_{e}{x \times \frac{1}{x}}}}}{{(\log_{e}{x)}}^{2}}$
$\Rightarrow \ f^{'}(x) = \frac{1}{2}\frac{\frac{1}{x} - \frac{1}{x}\log_{e}{\log_{e}x}}{\left( \log_{e}x \right)^{2}}$
At $x = e,f^{'}(e) = \frac{1}{2}\frac{\frac{1}{e} - \frac{1}{e}\log_{e}1}{(1)^{2}}$
$\Rightarrow \ f^{'}(e) = \frac{1}{2e}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.