alleryd/dx (xˣ) where x > 0:
Axˣ × (1 + ln x)
Bxˣ × x
Cx × xˣ⁻¹
Dxˣ
Answer & Solution
Correct answer: A. xˣ × (1 + ln x)
Let y = xˣ. Take ln: ln y = x ln x.
**Differentiate:** (1/y) y' = ln x + 1. So y' = xˣ × (ln x + 1) = xˣ × (1 + ln x).
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