Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 116

Differentiate $\displaystyle f(t) = (t^5 + 3) \cdot \left( \frac{t^3 - 1}{t^3 + 1} \right)$

If we simplify the function first before we take the derivative, we get

$
\begin{equation}
\begin{aligned}
f(t) &= \frac{t^8 - t^5 + 3t^3 - 3}{t^3 + 1}
\end{aligned}
\end{equation}
$


Now, by applying Quotient Rule,

$
\begin{equation}
\begin{aligned}
f'(t) &= \frac{(t^3 + 1) \cdot \frac{d}{dt} \left( t^8 - t^5 + 3t^3 - 3 \right) - \left( t^8 - t^5 + 3t^3 - 3 \right) \cdot \frac{d}{dt} (t^3 + 1)}{(t^3 + 1)^2}\\
\\
f'(t) &= \frac{(t^3 + 1)\left( 8t^7 - 5t^4 + 9t^2 \right) - \left( t^8 - t^5 + 3t^3 - 3 \right) (3t^2) }{(t^3 + 1)^2}\\
\\
f'(t) &= \frac{8t^{10} - 5t^7 + 9t^5 + 8t^7 - 5t^4 + 9t^2 - 3t^{10} + 3t^7 - 9t^5 + 9t^2 }{(t^3 + 1)^2}\\
\\
f'(t) &= \frac{5t^{10} + 6t^7 - 5t^4 + 18t^2}{(t^3 + 1)^2}
\end{aligned}
\end{equation}
$