What is \( \frac{dy}{dx} \) equal to, given that \( y = y_0 \) when x = 1 ?

For the next two (02) items that follow : Let \( (e^y)^x - y = 0 \), where y is a function of x whose domain is (0, 10].

  1. A. \frac{y_0}{1+e^{y_0}}
  2. B. -\frac{y_0 e^{y_0}}{1+e^{y_0}}
  3. C. \frac{y_0 e^{y_0}}{1+e^{y_0}}
  4. D. \frac{y_0 e^{y_0}}{1-e^{y_0}} ✓

Correct Answer: D. \frac{y_0 e^{y_0}}{1-e^{y_0}}

Explanation

Substitute x = 1 into \( xy = \ln y \) to get \( y_0 = \ln y_0 \), meaning \( e^{y_0} = y_0 \). Evaluating the derivative \( \frac{dy}{dx} = \frac{y^2}{1-xy} \) at x=1 gives \( \frac{y_0^2}{1-y_0} \). Replacing one \( y_0 \) with \( e^{y_0} \) yields \( \frac{y_0 e^{y_0}}{1-e^{y_0}} \).

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