find the indicated derivative and simplify.\ny for y = \\frac{6x - 8}{x^{2}+2x}\ny = square

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 6x - 8$, so $u'=6$, and $v=x^{2}+2x$, so $v' = 2x + 2$.

Step2: Substitute into quotient - rule formula

$y'=\frac{6(x^{2}+2x)-(6x - 8)(2x + 2)}{(x^{2}+2x)^{2}}$.

Step3: Expand the numerator

Expand $6(x^{2}+2x)=6x^{2}+12x$ and $(6x - 8)(2x + 2)=12x^{2}+12x-16x - 16=12x^{2}-4x - 16$. Then the numerator is $6x^{2}+12x-(12x^{2}-4x - 16)=6x^{2}+12x - 12x^{2}+4x + 16=-6x^{2}+16x + 16$.

Step4: Simplify the result

$y'=\frac{-6x^{2}+16x + 16}{(x^{2}+2x)^{2}}$.

Answer:

$\frac{-6x^{2}+16x + 16}{(x^{2}+2x)^{2}}$