Question

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

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 2x-1$, $u'=2$, $v = 3x + 1$, and $v'=3$.

Step2: Substitute values into the formula

$y'=\frac{2(3x + 1)-(2x - 1)\times3}{(3x + 1)^{2}}$.

Step3: Expand the numerator

$y'=\frac{6x+2-(6x - 3)}{(3x + 1)^{2}}=\frac{6x + 2-6x + 3}{(3x + 1)^{2}}$.

Step4: Simplify the numerator

$y'=\frac{(6x-6x)+(2 + 3)}{(3x + 1)^{2}}=\frac{5}{(3x + 1)^{2}}$.

Answer:

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