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For x > 0, logarithmic differentiation shows that the derivative of x^x is
Ax^x (1 - log x)
Bx^x (1 + log x)
Cx^(x-1) (1 + x)
Dx^x (log x - 1)
Answer & Solution
Correct answer: B. x^x (1 + log x)
With log y = x log x, differentiating gives y'/y = log x + 1, so y' = x^x (1 + log x).
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