CA Foundation · Quantitative Aptitude · Differential and Integral Calculus
If y = x^x for x > 0, then dy/dx at x = 1 is:
Using logarithmic differentiation, dy/dx = x^x(1 + log x). At x = 1 the value is 1 × (1 + 0) = 1.
- A0
- B1Correct
- Ce
- D2
Explanation
Take logs: log y = x log x. Differentiating, (1/y)(dy/dx) = log x + 1, so dy/dx = x^x (1 + log x). At x = 1, this is 1 × (1 + 0) = 1. Treating it as a power rule x·x^(x−1) = x^x would give 1 here by coincidence but is not valid in general.
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