# Derivative of ( ln &#x2061;<!-- ⁡ --> x ) <mrow class="MJX-TeXAtom-ORD">

Derivative of $\left(\mathrm{ln}x{\right)}^{\mathrm{ln}x}$
How can I differentiate the following function?
$f\left(x\right)=\left(\mathrm{ln}x{\right)}^{\mathrm{ln}x}.$
Is it a composition of functions? And if so, which functions?
Thank you.
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rideonthebussp
The function $f$ can be expressed as
$f\left(x\right)=\left(\mathrm{ln}x{\right)}^{\mathrm{ln}x}={\mathrm{e}}^{\left(\mathrm{ln}x\right)\mathrm{ln}\left(\mathrm{ln}x\right)}.$
Hence $f\left(x\right)=\mathrm{exp}\left(g\left(x\right)\right)$, where $g\left(x\right)=\left(\mathrm{ln}x\right)\mathrm{ln}\left(\mathrm{ln}x\right)$, and thus
${f}^{\prime }\left(x\right)={\left({\mathrm{e}}^{\left(\mathrm{ln}x\right)\mathrm{ln}\left(\mathrm{ln}x\right)}\right)}^{\prime }={\mathrm{e}}^{\left(\mathrm{ln}x\right)\mathrm{ln}\left(\mathrm{ln}x\right)}{\left(\left(\mathrm{ln}x\right)\mathrm{ln}\left(\mathrm{ln}x\right)\right)}^{\prime }=\left(\mathrm{ln}x{\right)}^{\mathrm{ln}x}\left(\frac{1}{x}\mathrm{ln}\left(\mathrm{ln}x\right)+\mathrm{ln}x\frac{1}{x\mathrm{ln}x}\right).$
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Anthony Kramer
Note that
${a}^{b}={\left({e}^{\mathrm{ln}a}\right)}^{b}={e}^{b\mathrm{ln}a}$
so
${\left(\mathrm{ln}x\right)}^{\mathrm{ln}x}={\left({e}^{\mathrm{ln}\mathrm{ln}x}\right)}^{\mathrm{ln}x}\phantom{\rule{0ex}{0ex}}={e}^{\mathrm{ln}x\cdot \mathrm{ln}\mathrm{ln}x}={x}^{\mathrm{ln}\mathrm{ln}x}$
either of which which you should be able to do with methods you already know.
We can apply this technique generally to calculate the derivative of $f\left(x{\right)}^{g\left(x\right)}$:
$f\left(x{\right)}^{g\left(x\right)}={e}^{g\mathrm{ln}f}$
so
$\begin{array}{rl}\frac{d}{dx}{f}^{g}=\frac{d}{dx}{e}^{g\cdot \mathrm{ln}f}& =\left(\frac{d}{dx}\left(g\cdot \mathrm{ln}f\right)\right){e}^{g\cdot \mathrm{ln}f}\phantom{\rule{2em}{0ex}}\text{(chain rule)}\\ & =\left(\frac{d}{dx}\left(g\cdot \mathrm{ln}f\right)\right){f}^{g}\\ & =\left({g}^{\prime }\mathrm{ln}f+\frac{g{f}^{\prime }}{f}\right){f}^{g}\phantom{\rule{2em}{0ex}}\text{(product rule)}\end{array}$
where ${f}^{\prime }=\frac{df}{dx}$ and ${g}^{\prime }=\frac{dg}{dx}$