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To determine \(\frac{dy}{dt}\) for the function \(y = 2\cos(t) - 4t^3 - 10\), we need to differentiate the function with respect to \(t\). Here's how we can proceed:

Given:
\[
y = 2\cos(t) - 4t^3 - 10
\]

Differentiate each term with respect to \(t\):

1. The derivative of \(2\cos(t)\) with respect to \(t\) is:
   \[
   \frac{d}{dt}[2\cos(t)] = 2(-\sin(t)) = -2\sin(t)
   \]

2. The derivative of \(-4t^3\) with respect to \(t\) is:
   \[
   \frac{d}{dt}[-4t^3] = -4 \times 3t^2 = -12t^2
   \]

3. The derivative of the constant \(-10\) with respect to \(t\) is:
   \[
   \frac{d}{dt}[-10] = 0
   \]

Now, combine these results to get \(\frac{dy}{dt}\):
\[
\frac{dy}{dt} = -2\sin(t) - 12t^2
\]

So, the derivative \(\frac{dy}{dt}\) is:
\[
\boxed{-2\sin(t) - 12t^2}
\]

Would you like further details or have any questions about the process?

Here are five related questions you might find helpful:

1. How do you differentiate \(y = \sin(t)\) with respect to \(t\)?
2. What is the second derivative of \(y = 2\cos(t) - 4t^3 - 10\)?
3. How would the derivative change if the function was \(y = 2\cos(2t) - 4t^3 - 10\)?
4. Can you find the critical points of \(y = 2\cos(t) - 4t^3 - 10\) by setting \(\frac{dy}{dt} = 0\)?
5. How do the trigonometric derivatives influence the overall behavior of the function?

**Tip:** When differentiating trigonometric functions, remember the basic derivatives: \(\frac{d}{dt}[\sin(t)] = \cos(t)\) and \(\frac{d}{dt}[\cos(t)] = -\sin(t)\).

Determine dy/dt for y = 2cos(t) - 4t^3 - 10.

Explore the step-by-step process of differentiating y = 2cos(t) - 4t^3 - 10 with respect to t. This solution covers key concepts in differentiation, including trigonometric derivatives and the power rule, suitable for students in Grades 11-12.

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