Answer
$r^{\prime}(x)=\displaystyle \frac{4\ln(x^{2})}{x}$
Work Step by Step
With $u=\displaystyle \ln(x-1),\quad \frac{du}{dx}=\frac{1}{x-1}$,
the generalized power rule (chain rule) gives
$r^{\prime}(x)=\displaystyle \frac{d}{dx}[u^{n}]=nu^{n-1}\frac{du}{dx}$
$r^{\prime}(x)=2\displaystyle \ln(x^{2})\cdot\frac{du}{dx}$
With $v=x^{2}, \displaystyle \quad\frac{dv}{dx}=2x$
$u=\ln v$, so, using the "logarithms of functions"
$\displaystyle \frac{d}{dx}[\ln v]=\frac{1}{v}\frac{dv}{dx}$,
$\displaystyle \frac{du}{dx}=\frac{1}{x^{2}}\cdot 2x$
So,
$r^{\prime}(x)=2\displaystyle \ln(x^{2})\cdot\frac{1}{x^{2}}\cdot 2x=\frac{4x\ln(x^{2})}{x^{2}}$
$r^{\prime}(x)=\displaystyle \frac{4\ln(x^{2})}{x}$
