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}
Given the equation:\[x^y = e^{x-y}.\]Taking the natural logarithm of both sides yields:\[\log(x^y) = \log(e^{x-y}).\]Step 2: {Use logarithmic properties and rearrange for y}
Applying logarithmic properties:\[y \log x = x - y.\]Rearranging to solve for \( y \):\[y (1 + \log x) = x.\]This gives:\[y = \frac{x}{1 + \log x}.\]Step 3: {Differentiate y with respect to x}
Differentiating \( y = \frac{x}{1 + \log x} \) 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}
Simplifying the numerator:\[\frac{dy}{dx} = \frac{(1 + \log x) - 1}{(1 + \log x)^2}.\]This simplifies to:\[\frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}.\]Conclusion: The derivative is:\[\frac{dy}{dx} = \frac{\log x}{(1 + \log x)^2}.\]