Question

If \[ x = \log t, \quad y + 1 = \frac{1}{t}, \] then \[ e^{-x} \frac{d^2 x}{dy^2} + \frac{dx}{dy} = \]

• A. 0
• B. 2
• C. -1
• D. 1

Hint:

For problems involving logarithmic and inverse functions, differentiate with respect to the variable and use the chain rule to find higher derivatives.

Answer 1

The Correct Option is A

Step 1: Differentiate the given equations.
We are given \( x = \log t \) and \( y + 1 = \frac{1}{t} \). Differentiating \( x = \log t \) with respect to \( t \), we get: \[ \frac{dx}{dt} = \frac{1}{t}. \] Now, differentiating \( y + 1 = \frac{1}{t} \) with respect to \( t \), we get: \[ \frac{dy}{dt} = -\frac{1}{t^2}. \]
Step 2: Calculate the second derivative.
We compute \( \frac{d^2 x}{dy^2} \) using the chain rule and simplify. After calculations, we find that \( e^{-x} \frac{d^2 x}{dy^2} + \frac{dx}{dy} = 0 \). Thus, the correct answer is option (A).