Question

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

Hint:

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

Answer 1

Step 1: Apply natural logarithm 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 becomes \( y \log x = x - y \). Rearrange to isolate \( y \): \( y (1 + \log x) = x \), which simplifies to \( y = \frac{x}{1 + \log x} \).
Step 3: Differentiate \( y \) with respect to \( x \) using the quotient rule for \( y = \frac{x}{1 + \log x} \). The derivative is \( \frac{dy}{dx} = \frac{(1 + \log x) \cdot 1 - x \cdot \frac{1}{x}}{(1 + \log x)^2} \).
Step 4: Simplify the derivative. The numerator becomes \( (1 + \log x) - 1 \), resulting in \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \).
Conclusion: The final derivative is \( \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2} \).