To calculate the derivative of the function y = (6t−1)(4t−2) - 1 with respect to t, you'll need to use the product rule, which says that the derivative of two functions multiplied together (in this case (6t-1) and (4t-2)) is the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function.
Let's start by considering the two functions as u = (6t-1) and v = (4t-2).
First, we find the derivatives of u and v with respect to t:
- The derivative of u = 6t - 1 is du/dt = 6,
- The derivative of v = 4t - 2 is dv/dt = 4.
Then, we apply the product rule, which gives us: dy/dt = (u)'v + u(v)'
Substituting our functions and their derivatives into this formula, we get: dy/dt = 6(4t-2) + (6t-1)4
Simplify this equation which results in: dy/dt = 24t - 12 + 24t -4.
Combine like terms to get the final solution: dy/dt = 48t - 16.
Therefore, the derivative of the function y = (6t−1)(4t−2) - 1 with respect to t is dy/dt = 48t - 16.