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By logarithmic differentiation, d/dx of x^x (x > 0) equals what?
Ax x^(x-1)
Bx^x log x
Cx^x (1 - log x)
Dx^x (1 + log x)
Answer & Solution
Correct answer: D. x^x (1 + log x)
Taking logs gives log y = x log x, so y'/y = log x + 1 and y' = x^x (1 + log x).
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