Single Variable Calculus, Chapter 3, Review Exercises, Section Review Exercises, Problem 38

Find $y'$ of $\displaystyle y = \frac{(x-1)(x-4)}{(x-2)(x-3)}$


$
\begin{equation}
\begin{aligned}
y' &= \frac{d}{dx} \left[\frac{(x-1)(x-4)}{(x-2)(x-3)} \right]\\
\\
y' &= \frac{(x-2)(x-3)\frac{d}{dx}\left[ (x-1)(x-4)\right]-(x-1)(x-4)\frac{d}{dx}\left[ (x-2)(x-3)\right] }{\left[(x-2)(x-3) \right]^2}\\
\\
y' &= \frac{(x-2)(x-3)\left[ (x-1) \frac{d}{dx} (x-4) + (x-4)\frac{d}{dx}(x-1)\right] - (x-1)(x-4) \left[ (x-2) \frac{d}{dx} ( x-3)+(x-3)\frac{d}{dx}(x-2)\right]}{(x-2)^2(x-3)^2}\\
\\
y' &= \frac{(x-2)(x-3)\left[(x-1)(1)+(x-4)(1) \right]-(x-1)(x-4)\left[ (x-2)(1)+(x-3)(1)\right]}{(x-2)^2(x-3)^2}\\
\\
y' &= \frac{(x-2)(x-3)(x-1+x-4)-(x-1)(x-4)(x-2+x-3)}{(x-2)^2(x-3)^2}\\
\\
y' &= \frac{(x-2)(x-3)(2x-5)-(x-1)(x-4)(2x-5)}{(x-2)^2(x-3)^2}\\
\\
y' &= \frac{(2x-5)\left[ (x-2)(x-3)-(x-1)(x-4) \right]}{(x-2)^2(x-3)^2}\\
\\
y' &= \frac{(2x-5)(\cancel{x^2} - \cancel{3x} - \cancel{2x} + 6 - \cancel{x^2} + \cancel{4x} + \cancel{x} - 4 )}{(x-2)^2(x-3)^2}\\
\\
y' &= \frac{(2x-5)(2)}{(x-2)^2(x-3)^2}\\
\\
y' &= \frac{2(2x-5)}{(x-2)^2(x-3)^2}
\end{aligned}
\end{equation}
$