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What is the derivative of $x^x$ for $x>0$?
A$x^x\log x$
B$x^x(1+\log x)$
C$x^{x-1}$
D$x^x$
Answer & Solution
Correct answer: B. $x^x(1+\log x)$
Use logarithmic differentiation. Let $y=x^x$. Then $\log y=x\log x$. Differentiating gives $\frac{1}{y}\frac{dy}{dx}=1+\log x$, so $\frac{dy}{dx}=x^x(1+\log x)$.
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