We are asked to find the derivative of the function:

$y = (3t - 1)(6t - 6)^{-1}$

Steps to Solve:

1. Rewrite the function: $y = \frac{3t - 1}{6t - 6}$ This simplifies the expression. Now we can proceed using the quotient rule for differentiation.

2. Apply the Quotient Rule: The quotient rule states that if $y = \frac{u(t)}{v(t)}$, then: $\frac{dy}{dt} = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2}$ Here, $u(t) = 3t - 1$ and $v(t) = 6t - 6$.

3. Find $u'(t)$ and $v'(t)$:

   • $u'(t) = \frac{d}{dt}(3t - 1) = 3$
   • $v'(t) = \frac{d}{dt}(6t - 6) = 6$

4. Substitute into the quotient rule: $\frac{dy}{dt} = \frac{(3)(6t - 6) - (3t - 1)(6)}{(6t - 6)^2}$

5. Simplify the numerator:

   • First term: $3(6t - 6) = 18t - 18$
   • Second term: $(3t - 1)(6) = 18t - 6$

   The numerator becomes: $(18t - 18) - (18t - 6) = -18 + 6 = -12$

6. Final answer: $\frac{dy}{dt} = \frac{-12}{(6t - 6)^2}$

Thus, the derivative of the given function is:

$\frac{dy}{dt} = \frac{-12}{(6t - 6)^2}$

Would you like further details or clarifications? Here are some related questions to explore:

1. How does the quotient rule work with more complex functions?
2. What are the conditions for applying the chain rule in derivatives?
3. Can this derivative be simplified further under certain values of $t$?
4. How does this derivative behave as $t \to 1$?
5. What are the differences between the product and quotient rules in differentiation?

Tip: When simplifying functions, always check if you can rewrite them into a more manageable form before applying differentiation rules.