Question

If \( y = \log x \), then the value of \( x^2 \frac{d^2y}{dx^2} + 3x \frac{dy}{dx} + y \) at the point \( \left( \sqrt{e}, \sqrt{e} \right) \) is:

• A. \( 0 \)
• B. \( e \)
• C. \( 2e \)
• D. \( 2e^2 \)

Hint:

When dealing with logarithmic functions, differentiate step by step, ensuring that you calculate both the first and second derivatives accurately.

Answer 1

The Correct Option is A

We are given that \( y = \log x \), so we first calculate the first and second derivatives of \( y \) with respect to \( x \). First derivative: \[ \frac{dy}{dx} = \frac{1}{x} \] Second derivative: \[ \frac{d^2y}{dx^2} = -\frac{1}{x^2} \] Now, substitute these values into the expression \( x^2 \frac{d^2y}{dx^2} + 3x \frac{dy}{dx} + y \). At the point \( \left( \sqrt{e}, \sqrt{e} \right) \), we have: \[ x = \sqrt{e}, \quad y = \log x = \log \sqrt{e} = \frac{1}{2}, \quad \frac{dy}{dx} = \frac{1}{\sqrt{e}}, \quad \frac{d^2y}{dx^2} = -\frac{1}{e} \] Substitute these into the expression: \[ x^2 \frac{d^2y}{dx^2} + 3x \frac{dy}{dx} + y = (\sqrt{e})^2 \cdot \left( -\frac{1}{e} \right) + 3 \cdot \sqrt{e} \cdot \frac{1}{\sqrt{e}} + \frac{1}{2} \] \[ = e \cdot \left( -\frac{1}{e} \right) + 3 \cdot 1 + \frac{1}{2} \] \[ = -1 + 3 + \frac{1}{2} = 0 \] Thus, the value of the expression is \( 0 \), and the correct answer is option (1).