# Find the derivative of the following function: y(x)=(ln(2x))^(5x),x>1/2

Find the derivative of the following function: $y\left(x\right)=\left(\mathrm{ln}\left(2x\right){\right)}^{5x},\phantom{\rule{1em}{0ex}}x>\frac{1}{2}$
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Whenever you have a function of the form $y=f\left(x{\right)}^{g\left(x\right)}$, it is usually a good idea to use logarithmic differentiation:
$\mathrm{log}y=g\left(x\right)\mathrm{log}f\left(x\right),$
hence
$\frac{d}{dx}\left[\mathrm{log}y\right]=\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left[g\left(x\right)\mathrm{log}f\left(x\right)\right],$
and
$\frac{dy}{dx}=f\left(x{\right)}^{g\left(x\right)}\left(\frac{g\left(x\right)}{f\left(x\right)}\frac{df}{dx}+\frac{dg}{dx}\mathrm{log}f\left(x\right)\right).$
So, in your case, $f\left(x\right)=\mathrm{log}2x$ and $g\left(x\right)=5x$, so we have
$\frac{df}{dx}=\frac{1}{2x}\cdot 2=\frac{1}{x},$
and
$\frac{dg}{dx}=5,$
therefore after minor simplifications,
$\frac{dy}{dx}=\left(\mathrm{log}2x{\right)}^{5x}\left(\frac{5}{\mathrm{log}2x}+5\mathrm{log}\left(\mathrm{log}2x\right)\right).$
The only additional step that was done was to put the fractions over a common denominator of $\mathrm{log}2x$ and factoring:
$\frac{dy}{dx}=5\left(\mathrm{log}2x{\right)}^{5x-1}\left(1+\left(\mathrm{log}2x\right)\mathrm{log}\left(\mathrm{log}2x\right)\right).$