If y = x^x, what is dy/dx for x > 0? (Hard)
Answer and explanation
Correct answer: A. x^x (1 + ln x)
- Step 1: Take natural log of both sides: ln y = x ln x. Step 2. Differentiate implicitly: (1/y) dy/dx = 1 ln x + x (1/x) = ln x + 1. Thus dy/dx = y (1 + ln x) = x^x (1 + ln x). Answer: x^x (1 + ln x). Why not the others: x * x^(x-1) treats exponent as constant; x^x ln x treats base as constant; x^x (1 - ln x) has wrong sign.
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