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A variable base and exponent give y = x^x for x > 0. Its derivative is
Ax^x (1 + ln x)
Bx^x + ln x
Cx^x ln x
Dx^(x-1) + 1
Answer & Solution
Correct answer: A. x^x (1 + ln x)
Taking ln y = x ln x and differentiating gives y'/y = ln x + 1, so y' = x^x (1 + ln x).
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