Question

If \( \frac{x}{x-y} = \log \left( \frac{a}{x-y} \right) \), then \( \frac{dy}{dx} = \)

• A. \( 2 + \frac{1}{y} \)
• B. \( \frac{2y - x}{y} \)
• C. \( \frac{2x - y}{x} \)
• D. \( \frac{x - 2y}{y} \)

Hint:

When differentiating implicit functions, remember to apply the quotient rule and chain rule carefully.

Answer 1

The Correct Option is B

Step 1: Differentiate implicitly.
We are given: \[ \frac{x}{x-y} = \log \left( \frac{a}{x-y} \right) \] Differentiate both sides with respect to \( x \). Step 2: Apply the quotient rule and chain rule.
The left-hand side is \( \frac{x}{x - y} \). Using the quotient rule: \[ \frac{d}{dx}\left( \frac{x}{x - y} \right) = \frac{(x - y) - x(1 - \frac{dy}{dx})}{(x - y)^2} \] The right-hand side is \( \log \left( \frac{a}{x - y} \right) \). Differentiating: \[ \frac{d}{dx} \left( \log \left( \frac{a}{x - y} \right) \right) = \frac{1}{x - y} \cdot (-1) \cdot \frac{dy}{dx} \] Equating both sides and solving for \( \frac{dy}{dx} \), we find: \[ \frac{dy}{dx} = \frac{2y - x}{y} \] Step 3: Conclusion.
Thus, the value of \( \frac{dy}{dx} \) is \( \boxed{\frac{2y - x}{y}} \).