alleryIf $y = x^x$, then $\dfrac{dy}{dx}$ equals:
A$x^x(1 + \ln x)$
B$x^{x-1}$
C$x^x \ln x$
D$x \cdot x^{x-1}$
Answer & Solution
Correct answer: A. $x^x(1 + \ln x)$
Take $\ln$: $\ln y = x \ln x$.
**Differentiate:** $\frac{1}{y}\frac{dy}{dx} = \ln x + 1$
⇒ $dy/dx = y(1 + \ln x) = x^x(1 + \ln x)$.
Classic logarithmic differentiation.
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