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We are tasked with finding the derivative of the function:

\[
y = (6t - 1)(2t - 4)^{-1}
\]

To differentiate this, we need to apply the **product rule**, which states:

\[
\frac{d}{dt}(u \cdot v) = u' \cdot v + u \cdot v'
\]

Here, we let:
- \( u = 6t - 1 \)
- \( v = (2t - 4)^{-1} \)

Now, let's compute the derivatives of \( u \) and \( v \):
1. \( u' = \frac{d}{dt}(6t - 1) = 6 \)
2. \( v' \) requires the chain rule. We know \( v = (2t - 4)^{-1} \), so:

\[
v' = -1 \cdot (2t - 4)^{-2} \cdot \frac{d}{dt}(2t - 4) = -\frac{2}{(2t - 4)^2}
\]

Now, apply the product rule:

\[
\frac{dy}{dt} = u' \cdot v + u \cdot v'
\]

Substitute \( u = 6t - 1 \), \( u' = 6 \), \( v = (2t - 4)^{-1} \), and \( v' = -\frac{2}{(2t - 4)^2} \):

\[
\frac{dy}{dt} = 6 \cdot (2t - 4)^{-1} + (6t - 1) \cdot \left(-\frac{2}{(2t - 4)^2}\right)
\]

Let me simplify this expression for you.The derivative of the function \( y = (6t - 1)(2t - 4)^{-1} \) with respect to \( t \) is:

\[
\frac{dy}{dt} = -\frac{11}{2(t - 2)^2}
\]

Would you like more details or have any questions about this? Here are 5 questions that might expand on this concept:

1. How do you apply the product rule in general?
2. What role does the chain rule play when differentiating functions of functions?
3. Can you explain why \( v' \) requires the chain rule in this example?
4. What happens to the function behavior as \( t \) approaches 2?
5. How would you integrate the resulting derivative?

**Tip:** When dealing with rational functions, simplifying before applying differentiation rules can often make the process easier.

Find the derivative of the function y = (6t - 1)(2t - 4)^{-1}.

Learn how to find the derivative of the function y = (6t - 1)(2t - 4)^{-1} using the product and chain rules. This problem involves applying the product rule for differentiating a product of functions and the chain rule for nested functions.

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