Question

If \( x^y = e^{x - y} \), prove that \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \):

Hint:

For functions involving both \( x \) and \( y \), logarithmic differentiation often simplifies the process.

Answer 1

Step 1: Apply logarithms to both sides of the equation \( x^y = e^{x - y} \). This yields \( \log(x^y) = \log(e^{x - y}) \).
Step 2: Utilize logarithmic properties to simplify the equation to \( y \log x = x - y \). Rearrange to isolate \( y \), resulting in \( y (1 + \log x) = x \), which gives \( y = \frac{x}{1 + \log x} \).
Step 3: Differentiate \( y \) with respect to \( x \) using the quotient rule on \( y = \frac{x}{1 + \log x} \). The derivative is calculated as \( \frac{dy}{dx} = \frac{(1 + \log x) \cdot 1 - x \cdot \frac{1}{x}}{(1 + \log x)^2} \).
Step 4: Simplify the derivative. The numerator becomes \( (1 + \log x) - 1 \). Thus, \( \frac{dy}{dx} = \frac{1 + \log x - 1}{(1 + \log x)^2} \). This simplifies further to \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \).
Conclusion:
The derived expression for the derivative is \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \).