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 \(y=\dfrac{\log x}{x}\).
Write \[ y=(\log x)x^{-1}. \] Using the product rule, \[ \frac{dy}{dx} = \frac{1}{x}\cdot x^{-1} + (\log x)(-x^{-2}). \] \[ \frac{dy}{dx} = \frac{1}{x^2} -\frac{\log x}{x^2}. \] \[ \frac{dy}{dx} = \frac{1-\log x}{x^2}. \]

Step 2: Differentiate again.
\[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left[ (1-\log x)x^{-2} \right]. \] Applying the product rule, \[ \frac{d^2y}{dx^2} = \left(-\frac1x\right)x^{-2} + (1-\log x)(-2x^{-3}). \] \[ \frac{d^2y}{dx^2} = -\frac1{x^3} -\frac{2(1-\log x)}{x^3}. \] \[ \frac{d^2y}{dx^2} = \frac{-1-2+2\log x}{x^3}. \] \[ \frac{d^2y}{dx^2} = \frac{2\log x-3}{x^3}. \]

Step 3: Evaluate at \(x=1\).
Since \[ \log 1=0, \] we get \[ \left.\frac{d^2y}{dx^2}\right|_{x=1} = \frac{2(0)-3}{1^3}. \] \[ =-3. \]

Step 4: Final conclusion.
Therefore, \[ \boxed{-3} \]