Question

what is the first derivative of $r(x)=log_6left(\frac{x^{2}+x + 4}{x^{3}6^{x}}
ight)? select the correct answer below: $r(x)=\frac{x^{3}(6)ln6}{x^{2}+x + 4}$ $r(x)=\frac{1}{(x^{2}+x + 4)ln6}-\frac{3}{xln6}-1$ $r(x)=\frac{2x + 1}{(x^{2}+x + 4)ln6}-\frac{3}{xln6}-1$ $r(x)=\frac{2x + 1}{x^{2}+x + 4}-\frac{3}{x}-1$

Explanation:

Step1: Use log - property

$r(x)=\log_6(x^{2}+x + 4)-\log_6(x^{3})-\log_6(6^{x})$.

Step2: Differentiate each term

$r'(x)=\frac{2x + 1}{(x^{2}+x + 4)\ln6}-\frac{3}{x\ln6}-1$.

Answer:

$r'(x)=\frac{2x + 1}{(x^{2}+x + 4)\ln6}-\frac{3}{x\ln6}-1$ (third option)