Question

step 1
the function (h(t)=(t^{4}-1)^{3}(t^{3}+1)^{6}) is a product, and so we must use the product - rule to find its derivative. also, the factors of the product are compositions, so finding their derivatives will require using the chain rule.
using the chain rule, the derivative of ((t^{4}-1)^{3}) is
3(t^{4}-1)^{2}(4t^{3})
step 2
similarly, the derivative of ((t^{3}+1)^{6}) is
6(space)^{5}(3cdotspace)

Explanation:

Step1: Apply chain - rule to \((t^{4}-1)^{3}\)

The chain - rule states that if \(y = u^{n}\) and \(u = g(t)\), then \(y^\prime=nu^{n - 1}\cdot g^\prime(t)\). For \(y=(t^{4}-1)^{3}\), let \(u = t^{4}-1\), \(n = 3\), and \(g(t)=t^{4}-1\), \(g^\prime(t)=4t^{3}\). So the derivative is \(3(t^{4}-1)^{2}(4t^{3})\).

Step2: Apply chain - rule to \((t^{3}+1)^{6}\)

Let \(u=t^{3}+1\), \(n = 6\), and \(g(t)=t^{3}+1\), \(g^\prime(t)=3t^{2}\). By the chain - rule, the derivative of \((t^{3}+1)^{6}\) is \(6(t^{3}+1)^{5}(3t^{2})\).

Answer:

The derivative of \((t^{3}+1)^{6}\) is \(6(t^{3}+1)^{5}(3t^{2})\)