Question

question 5 · 1 point
what is $r(x)$ when $r(x)=lnleft(\frac{9x^{9}+1}{4x^{2}+x - 1}
ight)$?
select the correct answer below:
$r(x)=-\frac{81x^{8}}{9x^{9}+1}+\frac{8x + 1}{4x^{2}+x - 1}$
$r(x)=\frac{9x^{9}+1}{81x^{8}}-\frac{4x^{2}+x - 1}{8x + 1}$
$r(x)=\frac{81x^{8}}{9x^{9}+1}+\frac{8x + 1}{4x^{2}+x - 1}$
$r(x)=\frac{81x^{8}}{9x^{9}+1}-\frac{8x + 1}{4x^{2}+x - 1}$

Explanation:

Step1: Use log - quotient rule

$r(x)=\ln(9x^{9}+1)-\ln(4x^{2}+x - 1)$

Step2: Differentiate each term

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

Answer:

$r'(x)=\frac{81x^{8}}{9x^{9}+1}-\frac{8x + 1}{4x^{2}+x - 1}$ (last option)