Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 48

Differentiate $\displaystyle f(t) = \frac{3t^2 + 2t - 1}{-t^2 + 4t +1}$
By applying Long Division, we have



Thus,
$\displaystyle f(t) = -3 + \frac{14t + 2}{-t^2 + 4t + 1}$
Then, by taking the derivative using Quotient Rule, we obtain

$
\begin{equation}
\begin{aligned}
f'(t) &= \frac{d}{dt} (-3) + \frac{d}{dt} \left( \frac{14t + 2}{-t^2 + 4t + 1} \right)\\
\\
f'(t) &= 0 + \frac{(-t^2 + 4t + 1) \cdot \frac{d}{dt} (14t + 2) - (14t + 2) \cdot \frac{d}{dt}(-t^2 + 4t + 1) }{(-t^2 + 4t + 1)^2}\\
\\
f'(t) &= \frac{(-t^2 + 4t + 1)(14) - (14t + 2)( -2t + 4)}{(-t^2 + 4t + 1)^2}\\
\\
f'(t) &= \frac{-14t^2 + 56t + 14 - \left[ -28t^2 + 56t - 4t + 8 \right]}{(-t^2 +4t + )^2}\\
\\
&= \frac{-14t^2 + 56t + 14 + 28t^2 - 56t + 4t - 8 }{(-t^2 + 4t + 1)^2}\
\\
&= \frac{14t^2 + 4t + 6}{(-t^2 + 4t + 1)^2} \quad \text{ or } \quad \frac{2(7t^2 + 2t + 3)}{\left[ (-1) (t^2 - 4t - 1)\right]^2}\\
\\
&= \frac{2(7t^2 + 2t + 3)}{(-1)^2(t^2 - 4t - 1)^2}\\
\\
&= \frac{2(7t^2 + 2t + 3)}{t^2 - 4t - 1}

\end{aligned}
\end{equation}
$