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Find the derivative of x^x for x > 0.
Ax^x (1 - ln x)
Bx^x (1 + ln x)
Cx^x ln x
Dx * x^(x-1)
Answer & Solution
Correct answer: B. x^x (1 + ln x)
Taking logs gives ln y = x ln x, so y'/y = ln x + 1 and y' = x^x (1 + ln x).
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