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: Take logarithms of both sides
The given equation is: \[ x^y = e^{x-y}. \] Taking the natural logarithm on both sides, we get: \[ \log(x^y) = \log(e^{x-y}). \]
Step 2: Simplify using logarithmic properties
Using the properties of logarithms: \[ y \log x = x - y. \] Rearranging the terms to express \( y \): \[ y (1 + \log x) = x. \] Thus, we have: \[ y = \frac{x}{1 + \log x}. \]
Step 3: Differentiate \( y \) with respect to \( x \)
Differentiate both sides of \( y = \frac{x}{1 + \log x} \) with respect to \( x \) using the quotient rule: \[ \frac{dy}{dx} = \frac{(1 + \log x) \cdot 1 - x \cdot \frac{1}{x}}{(1 + \log x)^2}. \]
Step 4: Simplify the derivative
Simplify the numerator: \[ \frac{dy}{dx} = \frac{(1 + \log x) - 1}{(1 + \log x)^2}. \] This reduces to: \[ \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}. \]
Conclusion: The derivative is: \[ \frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}. \]