Question

If $x^y=\log x$ , then $\frac{dy}{dx}$ at the point where the curve cuts the $x-axis$ is

• A. $e$
• B. $\frac{1}{e}$
• C. $1$
• D. 0

Answer 1

Answer: (B)

We have, $x_y=\log x$ ...(i)
Taking log on both sides of (i), we get
$y\log x=\log \left(\log x\right)\Rightarrow y=\frac{\log \left(\log x\right)}{\log x}$ ....(ii) $\therefore \frac{dy}{dx}=\frac{\log x\left(\frac{1}{\log x}\right)\left(\frac{1}{x}\right)-\log \left(\log x\right)\left(\frac{1}{x}\right)-\log \left(\log x\right)\left(\frac{1}{x}\right)}{{\left(\log x\right)}^2}$
The point where the curve cuts the x-axis is (e, 0).
$\therefore \frac{dy}{dx}{|}_{at(e,0)}=\frac{\mathrm{1.1.}\frac{1}{e}-0}{(1{)}^2}=\frac{1}{e}$