Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 33

Differentiate $\displaystyle y = \frac{t^2 + 2}{t^4 - 3t^2 + 1}$



$
\begin{equation}
\begin{aligned}

y' =& \frac{(t^4 -3t^2 + 1) \displaystyle \frac{d}{dx} (t^2 + 2) - \left[ (t^2 + 2) \frac{d}{dx} (t^4 - 3t^2 + 1) \right]}{(t^4 - 3t^2 + 1)^2}
&& \text{Apply Quotient Rule}
\\
\\
y' =& \frac{(t^4 - 3t^2 + 1)(2t) - [(t^2 + 2)(4t^3 - 6t)]}{(t^4 - 3t^2 + 1)^2}
&& \text{Expand the equation}
\\
\\
y' =& \frac{2t^5 - \cancel{6t^3} + 2t - 4t^5 + \cancel{6t^3} - 8t^3 + 12t}{(t^4 - 3t^2 + 1)^2}
&& \text{Combine like terms}
\\
\\
y' =& \frac{-2t^5 - 8t^3 + 14t}{(t^4 - 3t^2 + 1)^2}
&& \text{}
\\
\\

\end{aligned}
\end{equation}
$