What is $r^{\text{'}}(x)$ when $r(x)=ln(\frac{9f^{9}1}{4x^{2}x-1})?$

Select the correct answer below:

$r^{\text{'}}(x)=-\frac{81x^8}{9x^{5}1}\frac{8x1}{4x^{2}x-1}$

$r^{\text{'}}(x)=\frac{9x^{3}1}{81x^4}-\frac{4x^{2}x-1}{8x1}$

$r^{\text{'}}(x)=\frac{81x^8}{9x^{8}1}\frac{8x1}{4x^{2}x-1}$

$r^{\text{'}}(x)=\frac{81x^5}{9x^{5}1}-\frac{8x1}{4x^{2}x-1}$