Discussion
If \((xe)^{y} - e^{x} = 0\), then \(\frac{dy}{dx}\) at \(x = 1\) is
Answer: A
\(\frac{dy}{dx}\) at x = 1 is 0.
Take logarithms of \((xe)^{y} = e^{x}\): \(y(\ln x + 1) = x\), so \(y = \frac{x}{1 + \ln x}\). Differentiate by the quotient rule:
\[ \begin{aligned} \frac{dy}{dx} &= \frac{(1 + \ln x) - x \cdot \frac{1}{x}}{(1 + \ln x)^{2}} \\ &= \frac{\ln x}{(1 + \ln x)^{2}} \end{aligned} \]
At x = 1, \(\ln 1 = 0\), so the derivative is 0: y has a stationary point there. The other values come from forgetting that the derivative of \(\ln x\) is \(\frac{1}{x}\).
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