Question

If \[ y=\frac{\log x}{x}, \] then \[ \frac{d^2y}{dx^2} at x=1 = \]

• A. \(-e^{-3}\)
• B. \(-3\)
• C. \(3\)
• D. \(e^3\)

Hint:

For functions of the form \(\dfrac{\log x}{x}\), first rewrite as \((\log x)x^{-1}\). Product rule and power rule make differentiation much simpler.

Answer 1

The Correct Option is B

Step 1: Differentiate y = (log x)/x using the product rule.
y = (log x)·x⁻¹. dy/dx = (1/x)·x⁻¹ + (log x)(-x⁻²) = 1/x² - (log x)/x² = (1 - log x)/x².

Step 2: Differentiate again to find the second derivative.
d²y/dx² = d/dx[(1 - log x)x⁻²] = (-1/x)x⁻² + (1 - log x)(-2x⁻³) = -1/x³ - 2(1 - log x)/x³ = (2 log x - 3)/x³.

Step 3: Evaluate at x = 1.
Since log 1 = 0, d²y/dx²|_{x=1} = (2·0 - 3)/1³ = -3.

Step 4: Final conclusion.
The second derivative at x = 1 is -3.