Question

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 $3(\square)^{2}(4\cdot\square)$

Explanation:

Step1: Recall chain - rule formula

The chain - rule states that if \(y = u^n\) and \(u\) is a function of \(t\), then \(\frac{dy}{dt}=n\cdot u^{n - 1}\cdot\frac{du}{dt}\). For the function \(y=(t^{4}-1)^{3}\), let \(u = t^{4}-1\) and \(n = 3\).

Step2: Identify \(u\) and \(\frac{du}{dt}\)

We have \(u=t^{4}-1\) and \(\frac{du}{dt}=4t^{3}\). Then, by the chain - rule, the derivative of \((t^{4}-1)^{3}\) is \(3(t^{4}-1)^{2}(4t^{3})\).

Answer:

The first blank is \(t^{4}-1\) and the second blank is \(t^{3}\).