Question

If \(y^x = e^{y - x}\), then \(\frac{dy}{dx}\) is equal to:

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

Hint:

For equations where variables appear both in the base and exponent, apply logarithm first to simplify the expression before differentiating.

Answer 1

The Correct Option is D

Given, \[ y^x = e^{y-x} \] Taking logarithm on both sides: \[ \log(y^x) = \log(e^{y-x}) \] Using logarithmic properties: \[ x \log y = y - x \] Rearranging: \[ x \log y + x = y \] \[ x(1+\log y)=y \] Thus, \[ x=\frac{y}{1+\log y} \] Differentiating both sides with respect to \(y\): \[ \frac{dx}{dy} = \frac{(1+\log y)-y\left(\frac1y\right)}{(1+\log y)^2} \] \[ \frac{dx}{dy} = \frac{\log y}{(1+\log y)^2} \] Therefore, \[ \frac{dy}{dx} = \frac{1}{\dfrac{\log y}{(1+\log y)^2}} \] \[ \boxed{ \frac{dy}{dx} = \frac{(1+\log y)^2}{\log y} } \]