for $g(y)=\frac{y - 4}{y^{2}-2y + 8}$, we have $g(y)=square$.
for $g(y)=\frac{y - 4}{y^{2}-2y + 8}$, we have $g(y)=square$.
Answer
Explanation:
Step1: Apply quotient - rule
The quotient - rule states that if $g(y)=\frac{u(y)}{v(y)}$, then $g^{\prime}(y)=\frac{u^{\prime}(y)v(y)-u(y)v^{\prime}(y)}{v(y)^2}$. Here, $u(y)=y - 4$, so $u^{\prime}(y)=1$, and $v(y)=y^{2}-2y + 8$, so $v^{\prime}(y)=2y-2$.
Step2: Substitute into quotient - rule formula
$g^{\prime}(y)=\frac{1\times(y^{2}-2y + 8)-(y - 4)\times(2y-2)}{(y^{2}-2y + 8)^{2}}$.
Step3: Expand the numerator
Expand $(y - 4)(2y-2)=2y^{2}-2y-8y + 8=2y^{2}-10y + 8$. Then the numerator is $y^{2}-2y + 8-(2y^{2}-10y + 8)=y^{2}-2y + 8-2y^{2}+10y - 8=-y^{2}+8y$.
Answer:
$\frac{-y^{2}+8y}{(y^{2}-2y + 8)^{2}}$