attempt 1: 10 attempts remaining. find the derivative. $y = \\frac{2x - 1}{3x + 1}$; $y=$
attempt 1: 10 attempts remaining. find the derivative. $y = \\frac{2x - 1}{3x + 1}$; $y=$
Answer
Explanation:
Step1: Recall quotient - rule
The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 2x - 1$, $u^\prime=2$, $v = 3x + 1$, and $v^\prime=3$.
Step2: Apply quotient - rule
Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the quotient - rule formula: [ \begin{align*} y^\prime&=\frac{(2)(3x + 1)-(2x - 1)(3)}{(3x + 1)^{2}}\ &=\frac{6x+2-(6x - 3)}{(3x + 1)^{2}}\ &=\frac{6x + 2-6x + 3}{(3x + 1)^{2}}\ &=\frac{5}{(3x + 1)^{2}} \end{align*} ]
Answer:
$\frac{5}{(3x + 1)^{2}}$