find y\ny = (3x - 2)/(2x + 1)

Explanation:

Step1: Use the quotient - rule for first derivative

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 3x - 2$, so $u'=3$, and $v = 2x + 1$, so $v'=2$. $y'=\frac{3(2x + 1)-2(3x - 2)}{(2x + 1)^{2}}=\frac{6x+3-(6x - 4)}{(2x + 1)^{2}}=\frac{6x + 3-6x + 4}{(2x + 1)^{2}}=\frac{7}{(2x + 1)^{2}}=7(2x + 1)^{-2}$

Step2: Use the chain - rule for second derivative

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Let $u = 2x+1$, so $y = 7u^{-2}$. Then $y'=-14u^{-3}\cdot2$. Substitute $u = 2x + 1$ back in: $y''=-14(2x + 1)^{-3}\cdot2=-\frac{28}{(2x + 1)^{3}}$

Answer:

$y''=-\frac{28}{(2x + 1)^{3}}$