differentiate. y(x) = (3x - 3)/(x² - 2x + 5)

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, $u = 3x - 3$, $v=x^{2}-2x + 5$.

Step2: Find $u^\prime$ and $v^\prime$

Differentiate $u$ with respect to $x$: $u^\prime=\frac{d}{dx}(3x - 3)=3$. Differentiate $v$ with respect to $x$: $v^\prime=\frac{d}{dx}(x^{2}-2x + 5)=2x-2$.

Step3: Apply quotient - rule

$y^\prime=\frac{3(x^{2}-2x + 5)-(3x - 3)(2x - 2)}{(x^{2}-2x + 5)^{2}}$. Expand the numerator: [ \begin{align*} &3(x^{2}-2x + 5)-(3x - 3)(2x - 2)\ =&3x^{2}-6x + 15-(6x^{2}-6x-6x + 6)\ =&3x^{2}-6x + 15-(6x^{2}-12x + 6)\ =&3x^{2}-6x + 15 - 6x^{2}+12x - 6\ =&-3x^{2}+6x + 9 \end{align*} ] So, $y^\prime=\frac{-3x^{2}+6x + 9}{(x^{2}-2x + 5)^{2}}$.

Answer:

$y^\prime=\frac{-3x^{2}+6x + 9}{(x^{2}-2x + 5)^{2}}$