Q. If $y = \log_{a}\ x + \log_{x}{a + \log_{x}{x + \log_{a}{a,\ then\frac{dy}{dx}\ }}\ }$is equal to
A) $\frac{1}{x} + x\log a$
B) $\frac{\log a}{x} + \frac{x}{\log a}$
C) $\frac{1}{x\log a} + x\log a$
D) None of these
Explanation
(d)
Given, $y = \frac{\log x}{\log a} + \frac{\log a}{\log x} + 1 + 1$
$\Rightarrow \frac{dy}{dx} = \frac{1}{x\log a} - \frac{\log a}{x{(\log{x)}}^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.