Derivative of $n^{\log n}$?

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What would be the derivative of $n^{\log n}$? I have to prove that $(\log n)^n$ = $\omega$($n^{\log n}$). I am trying to implement L'Hopital rule.

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2 Answers

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Note that $f(x)=x^{\log x}=e^{(\log x)^2}$. By the chain rule, the derivative is

$$f^\prime(x)=\frac{2\log x}{x} e^{(\log x)^2}=\frac{2\log x}{x} x^{\log x}$$

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Put $y = n^{log n} $ $$\implies \log y = \log n \log n \implies \frac{y'}{y} = \frac{\log n}{n} + \frac{ \log n}{n} = \frac{2 \log n}{n} $$

$$ y' = \frac{y 2 \log n}{n} \implies y' = \frac{2 n^{\log n} \log n}{n} $$

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