Question

question 8. 1 point what is r(x) when r(x) = log_8(\frac{5x^{2}+5x + 1}{x^{6}})? select the correct answer below: r(x)=\frac{10x + 5}{(5x^{2}+5x + 1)\ln 8}-\frac{6}{x\ln 8} r(x)=\frac{10x + 5}{5x^{2}+5x + 1}-\frac{6}{x} r(x)=\frac{x^{6}(1)\ln 8}{5x^{2}+5x + 1} r(x)=\frac{(5x^{2}+5x + 1)\ln 8}{x\ln 8}-\frac{6}{1}

Explanation:

Step1: Recall log - derivative formula

If $y = \log_a(u)$, then $y'=\frac{u'}{u\ln a}$. Here $a = 8$ and $u=\frac{5x^{2}+5x + 1}{x^{6}}$.

Step2: Find derivative of $u$

Using quotient - rule $(\frac{v}{w})'=\frac{v'w - vw'}{w^{2}}$, where $v = 5x^{2}+5x + 1$, $v'=10x + 5$, $w=x^{6}$, $w' = 6x^{5}$. Then $u'=\frac{(10x + 5)x^{6}-(5x^{2}+5x + 1)\times6x^{5}}{x^{12}}=\frac{(10x + 5)x-(5x^{2}+5x + 1)\times6}{x^{7}}=\frac{10x^{2}+5x-30x^{2}-30x - 6}{x^{7}}=\frac{- 20x^{2}-25x - 6}{x^{7}}$. Also, $u=\frac{5x^{2}+5x + 1}{x^{6}}$. So $r'(x)=\frac{(10x + 5)x^{6}-6x^{5}(5x^{2}+5x + 1)}{(5x^{2}+5x + 1)x^{6}\ln8}=\frac{10x + 5}{(5x^{2}+5x + 1)\ln8}-\frac{6}{x\ln8}$.

Answer:

$r'(x)=\frac{10x + 5}{(5x^{2}+5x + 1)\ln8}-\frac{6}{x\ln8}$