Question

attempt 1: 10 attempts remaining. find the derivative. $y = \frac{2x - 1}{3x + 1}$; $y=$

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:

$\frac{5}{(3x + 1)^{2}}$