Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 46

Differentiate $\displaystyle f(x) = \frac{3x^2 - 5x}{x^2 - 1}$
By applying Long Division, we have



Thus,
$\displaystyle f(x) = \frac{3x^2 - 5x}{x^2 - 1} = 3 + \frac{-5x + 3}{x^2 - 1}$

Then, by taking the derivative using Quotient Rule, we obtain

$\begin{equation} \begin{aligned} f'(x) &= \frac{d}{dx} (3) + \frac{d}{dx} \left[ \frac{-5x + 3}{x^2 - 1} \right]\\ \\ f'(x) &= 0 + \frac{(x^2 - 1) \cdot \frac{d}{dx} (-5x + 3) - (-5x + 3) \cdot \frac{d}{dx} (x^2 - 1) }{(x^2 - 1)^2}\\ \\ f'(x) &= \frac{(x^2 -1)(-5)-(-5x +3)(2x)}{(x^2 - 1)^2}\\ \\ f'(x) &= \frac{-5x^2 + 5 + 10x^2 - 6x}{(x^2 - 1)^2}\\ \\ &= \frac{5x^2 - 6x + 5}{(x^2 - 1)^2} \end{aligned} \end{equation}$