Given y = x^x for x > 0, find dy/dx using logarithmic differentiation.
Answer and explanation
Correct answer: C. x^x * (1 + ln(x))
- Step 1: Take the natural logarithm of both sides: ln(y) = ln(x^x) = x * ln(x).
- Step 2: Differentiate implicitly with respect to x: (1/y) dy/dx = (1) ln(x) + x * (1/x) = ln(x) + 1.
- Step 3: Multiply both sides by y: 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)' improperly treats the exponent as a constant using the power rule; 'x^x ln(x)' improperly treats the base as a constant using the exponential rule; 'x^x + ln(x)' adds terms instead of multiplying by the derivative.
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