Given the function $f(t) = (2t^4 - 7t^3 + 2t - 1)^2$, we can find the derivative of this function using the chain rule.

The chain rule for a function of the form $g(t)^2$ states: $\frac{d}{dt} \left[ g(t)^2 \right] = 2g(t) \cdot g'(t)$ where $g(t) = 2t^4 - 7t^3 + 2t - 1$.

Step 1: Define $g(t)$

Let $g(t) = 2t^4 - 7t^3 + 2t - 1$.

Step 2: Differentiate $g(t)$

We now find the derivative $g'(t)$: $g'(t) = \frac{d}{dt} \left[ 2t^4 - 7t^3 + 2t - 1 \right]$ Differentiating term by term: $g'(t) = 8t^3 - 21t^2 + 2$

Step 3: Apply the Chain Rule

Using the chain rule: $f'(t) = 2g(t) \cdot g'(t)$ Substitute $g(t)$ and $g'(t)$: $f'(t) = 2(2t^4 - 7t^3 + 2t - 1)(8t^3 - 21t^2 + 2)$

This is the derivative of $f(t)$.

Would you like further details on simplifying this result or any related explanation?

Here are 5 related questions:

1. How would you expand $f(t)$ and find its derivative directly?
2. What is the derivative of a cubic function?
3. Can we apply the product rule in addition to the chain rule here?
4. What are common applications of the chain rule in calculus?
5. How does the degree of a polynomial affect its derivative?

Tip: Always remember to differentiate constants as zero and apply the power rule carefully for polynomial functions!