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The function provided is:

\[
P = 4 + 3\cos(t)
\]

To differentiate \( P \) with respect to \( t \), we'll apply basic differentiation rules:

1. The derivative of a constant, like \( 4 \), is \( 0 \).
2. The derivative of \( 3\cos(t) \) with respect to \( t \) is \( -3\sin(t) \), using the chain rule since the derivative of \( \cos(t) \) is \( -\sin(t) \).

Thus, the derivative of \( P \) is:

\[
\frac{dP}{dt} = -3\sin(t)
\]

Would you like more details or further explanations on this process?

Here are 5 related questions you could explore:

1. What is the derivative of \( P = 4 + 3\sin(t) \)?
2. How would the second derivative of \( P = 4 + 3\cos(t) \) be calculated?
3. What happens when you differentiate \( P = 4 + 3e^t \)?
4. How do you apply the chain rule in differentiation more generally?
5. What would the derivative be if the function were \( P = 4 + 3\cos(2t) \)?

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

Differentiate the function P = 4 + 3cos(t).

Learn how to differentiate the function P = 4 + 3cos(t) with respect to t. This problem involves basic trigonometric differentiation, where the derivative of cos(t) is -sin(t) and the derivative of a constant is 0. Suitable for students in Grades 10-12.

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