Question

Find the derivative of the function.
h(t) = (t4 − 1)^7(t3 + 1)^8

Answer 1

\(h(t) = (t^4 − 1)^7(t^3 + 1)^8\). You have a product of two functions,
say \(f(t) = (t^4-1)^7\) and \(g(t) = (t^3+1)^8\), then use the product rule:
\(f'g + fg' \). When you calculate \(f' \) and \(g'\), realize that you must use the chain rule to find those derivatives!

\[ \large h'(t)=\underbrace{[7(t^4-1)^64t^3]}_{f'}\underbrace{[(t^3+1)^8]}_g+\underbrace{[(t^4-1)^7]}_f\underbrace{[8(t^3+1)^7(3t^2)]}_{g'}\]

Answer 2

\[h \prime \left( t \right)=7\left( t^4-1 \right)^64 t^3\left( t^3+1 \right)^8+8\left( t^3+1 \right)^7 3 t^2\left( t^4-1 \right)^7\]
\[=4t^2\left( t^4-1 \right)^6\left( t^3+1 \right)^7\left[7t \left( t^3+1 \right)+6\left( t^4-1 \right) \right]\]