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Using logarithmic differentiation, find d/dx of x^x for x greater than 0.
Ax^x (1 + ln x)
Bx^x (1 - ln x)
Cx^x (x + ln x)
Dx^(x-1) (1 + x)
Answer & Solution
Correct answer: A. x^x (1 + ln x)
With ln y = x ln x, differentiating gives y'/y = ln x + 1, so y' = x^x (1 + ln x).
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