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We are tasked with finding the derivative of \( y = \cos^2(4\pi t + 5) \) with respect to \( t \).

### Solution:

First, we apply the **chain rule**. Since \( y = \cos^2(4\pi t + 5) \), we can rewrite it as:

\[
y = (\cos(4\pi t + 5))^2
\]

Now, let's differentiate \( y \) using the chain rule:

\[
\frac{dy}{dt} = 2 \cdot \cos(4\pi t + 5) \cdot \frac{d}{dt}[\cos(4\pi t + 5)]
\]

Next, we need to differentiate \( \cos(4\pi t + 5) \). Using the chain rule again:

\[
\frac{d}{dt}[\cos(4\pi t + 5)] = -\sin(4\pi t + 5) \cdot \frac{d}{dt}[4\pi t + 5]
\]

Since the derivative of \( 4\pi t + 5 \) with respect to \( t \) is \( 4\pi \), we have:

\[
\frac{d}{dt}[\cos(4\pi t + 5)] = -\sin(4\pi t + 5) \cdot 4\pi
\]

Substitute this back into the expression for \( \frac{dy}{dt} \):

\[
\frac{dy}{dt} = 2 \cdot \cos(4\pi t + 5) \cdot \left(-\sin(4\pi t + 5) \cdot 4\pi\right)
\]

Simplify:

\[
\frac{dy}{dt} = -8\pi \cdot \cos(4\pi t + 5) \cdot \sin(4\pi t + 5)
\]

Using the **double-angle identity** for sine, \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we can express this as:

\[
\frac{dy}{dt} = -4\pi \cdot \sin(2(4\pi t + 5))
\]

Thus, the derivative is:

\[
\boxed{\frac{dy}{dt} = -4\pi \cdot \sin(8\pi t + 10)}
\]

Let me know if you'd like further clarification!

### Related questions:
1. How does the chain rule apply to trigonometric functions like sine and cosine?
2. What is the significance of the double-angle identity in simplifying trigonometric derivatives?
3. How would the derivative change if the function were \( \sin^2(4\pi t + 5) \) instead?
4. Can you explain how to differentiate composite functions involving more complex expressions?
5. What is the role of the constant \( 4\pi \) in this differentiation process?

### Tip:
When differentiating squared trigonometric functions, remember to apply the chain rule, and always look out for opportunities to simplify using trigonometric identities!

Learn how to find the derivative of y = cos^2(4πt + 5) with respect to t, using the chain rule and trigonometric identities. This calculus problem involves differentiating composite trigonometric functions.

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