Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 18

Determine the derivative of the function $h(t) = (t^4-1)^3(t^3+1)^4$


$
\begin{equation}
\begin{aligned}
h'(t) &= \left[ (t^4-1)^3 \cdot \frac{d}{dt} (t^3+1)^4 \right] + \left[ (t^3+1)^4 \cdot \frac{d}{dt} (t^4 - 1)^3 \right]\\
\\
h'(t) &= \left[ (t^4-1)^3 \cdot 4(t^3+1)^3 \frac{d}{dt} (t^3 + 1) \right] + \left[ (t^3+1)^4 \cdot 3(t^4 - 1)^2 \frac{d}{dt} (t^4-1)\right]\\
\\
h'(t) &= \left[ (t^4 - 1)^3 \cdot 4(t^3+1)^3)(3t^2) \right] + \left[ (t^3+1)^4 \cdot 3(t^4-1)^2(4t^3)\right]\\
\\
h'(t) &= \left[ (t^4 - 1)^3 (t^3+1)^3(12t^2)\right] + \left[ (t^3+1)^4(t^4-1)^2(12t^3)\right]\\
\\
h'(t) &= (t^4-1)^2(t^3+1)^2(12t^2) \left[ (t^4-1)+(t^3+1)(t) \right]\\
\\
h'(t) &= (t^4-1)^2(t^3+1)^3(12t^2)(t^4-1+t^4+t)\\
\\
h'(t) &= (t^4-1)^2(t^3+1)^3(12t^2) (2t^4+t-1)
\end{aligned}
\end{equation}
$