Let \(y = \log_e x\), \(x \gt 0\). Consider the following statements:
(A) \(\displaystyle\frac{dy}{dx} = \frac{1}{x}\)
(B) \(\displaystyle\frac{d^{2}y}{dx^{2}} = \frac{1}{x^{2}}\)
(C) \(\displaystyle x\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} = 0\)
(D) \(\displaystyle x^{2}\frac{d^{2}y}{dx^{2}} = -1\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (C) and (D) only
Since \(\frac{d}{dx}\log_e x = \frac{1}{x} = x^{-1}\), the power rule gives the second derivative:
\( \displaystyle \frac{dy}{dx} = \frac{1}{x} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} = -\frac{1}{x^{2}} \)
Then
\( \displaystyle x\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} = -\frac{1}{x} + \frac{1}{x} = 0 \)
\( \displaystyle x^{2}\frac{d^{2}y}{dx^{2}} = x^{2} \cdot \left(-\frac{1}{x^{2}}\right) = -1 \)
So \(y' = \frac{1}{x}\), \(y'' = -\frac{1}{x^{2}}\), \(xy'' + y' = 0\) and \(x^{2}y'' = -1\).
Checking the statements
(A) \(\frac{dy}{dx} = \frac{1}{x}\) — Correct. Standard derivative of \(\log_e x\).
(B) \(\frac{d^{2}y}{dx^{2}} = \frac{1}{x^{2}}\) — Incorrect. The second derivative is \(-\frac{1}{x^{2}}\); the sign is negative.
(C) \(x\frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} = 0\) — Correct. \(x \cdot \left(-\frac{1}{x^{2}}\right) + \frac{1}{x} = 0\).
(D) \(x^{2}\frac{d^{2}y}{dx^{2}} = -1\) — Correct. \(x^{2} \cdot \left(-\frac{1}{x^{2}}\right) = -1\).