If \(y = \log_{x} e\), \(x \gt 1\), then \(\displaystyle\frac{dy}{dx}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\displaystyle\frac{1}{x\left(\log_e x\right)^{2}}\)
By the change of base rule, the variable base is moved out of the logarithm:
\( \displaystyle y = \log_{x} e = \frac{\log_e e}{\log_e x} = \frac{1}{\log_e x} = \left(\log_e x\right)^{-1} \)
By the chain rule:
\( \displaystyle \frac{dy}{dx} = -\left(\log_e x\right)^{-2} \cdot \frac{1}{x} \)
\( \displaystyle = -\frac{1}{x\left(\log_e x\right)^{2}} \)