What is the derivative of f(x) = x^x for x > 0 using logarithmic differentiation?
Answer and explanation
Correct answer: C. f'(x) = x^x * (ln(x) + 1)
- Step 1: Let y = x^x. Take natural logarithms: ln(y) = ln(x^x) = x ln(x). Step 2. Differentiate implicitly: (1/y) (dy/dx) = 1 ln(x) + x (1/x) = ln(x) + 1. Multiply by y: dy/dx = x^x (ln(x) + 1). Answer: f'(x) = x^x (ln(x) + 1). Why not the others:** The power rule applies only to constant powers (x^n); the exponential rule applies only to constant bases (a^x). Both vary here.
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