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) $\frac{1}{x\log a} - \frac{\log a}{x\left( \log x \right)^{2}}$
Explanation
(d)
$\because y = \log_{a}{x + \frac{\log a}{\log x} + 1 + 1}$
$\Rightarrow \frac{dy}{dx} = \frac{1}{x}\log_{a}{e - \log a}\left( \frac{1}{\log x} \right)^{2}\frac{1}{x}$
$= \frac{1}{x\log a} - \frac{\log a}{x\left( \log x \right)^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.