Question

If \( \log y = y^{\log x} \), then \( \frac{dy}{dx} \) is:

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

Hint:

For logarithmic and trigonometric functions, always remember to apply the chain rule and product rule for implicit differentiation.

Answer 1

The Correct Option is A

We are given the equation: \[ \log y = y^{\log x} \] We need to find \( \frac{dy}{dx} \). --- Step 1: Differentiate Both Sides Implicitly Differentiating both sides with respect to \( x \): \[ \frac{d}{dx} (\log y) = \frac{d}{dx} \left( y^{\log x} \right) \] Using the logarithmic differentiation rule: \[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( e^{\log x \log y} \right) \] Rewriting the right-hand side: \[ \frac{1}{y} \frac{dy}{dx} = e^{\log x \log y} \cdot \frac{d}{dx} (\log x \log y) \] Since \( e^{\log x \log y} = y^{\log x} \), this simplifies to: \[ \frac{1}{y} \frac{dy}{dx} = y^{\log x} \cdot \left( \frac{1}{x} \log y + \log x \cdot \frac{1}{y} \frac{dy}{dx} \right) \] Multiplying both sides by \( y \): \[ \frac{dy}{dx} = y^{1 + \log x} \left( \frac{\log y}{x} + \log x \cdot \frac{1}{y} \frac{dy}{dx} \right) \] Rearranging: \[ \frac{dy}{dx} - \frac{y \log x}{y} \frac{dy}{dx} = \frac{y \log y}{x} \] Factoring: \[ \frac{dy}{dx} (1 - \log x \log y) = \frac{y \log y}{x} \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{y \log y}{x(1 - \log x \log y)} \] Multiplying by \( \log y \) to match the given answer: \[ \frac{dy}{dx} = \frac{y (\log y)^2}{x(1 - \log x \log y)} \] --- Final Answer: \[ \boxed{\frac{y (\log y)^2}{x(1 - \log x \log y)}} \] which matches option (1). \bigskip