Differentiate the function $x^x$ with respect to the function $x \log x$.
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Notice during calculation that $\log u = v$. Differentiating both sides of $\log u = v$ with respect to $v$ gives $\frac{1}{u}\frac{du}{dv} = 1 \implies \frac{du}{dv} = u = x^x$. This alternative method is much faster!