Discussion
For \( y = x^{x} \) with \( x > 0 \), the derivative is
Answer: C
Take logarithms, then differentiate both sides:
\[ \ln y = x \ln x \quad \Longrightarrow \quad \frac{1}{y}\frac{dy}{dx} \;=\; \ln x + 1. \]
So \( dy/dx = x^{x}(1 + \ln x) \), which vanishes at \( x = 1/e \), the minimum of the function. Applying the power rule as though the exponent were constant gives \\( x\\,x^{x-1} \\) instead
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