d/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).
Related questions
Rolle's theorem applies to a function $f$ on $[a, b]$ if $f$ is:The derivative of $ in(3x)$ is:A function differentiable at $x = a$ is always:A function $f$ is continuous at $x = a$ if:The derivative of $ in x$ is:The derivative of $f(x) = x^3 + 2x^2 + 1$ at $x = 1$ is:The limit $\lim_{x\to 0} in x/x$ equals:The limit $\lim_{x\to 2}(x^2 + 3x - 1)$ equals: