Let's first work through the given equation step by step to simplify it and find the derivative.

You gave the function:

\[
y = (u^2 - 1^3)\cos(3u + 2)
\]

### Step 1: Simplify the expression
Note that \(1^3 = 1\), so the equation becomes:

\[
y = (u^2 - 1)\cos(3u + 2)
\]

### Step 2: Find the derivative \(y'\)
To differentiate this, we will use the **product rule**, which states that if \(y = f(u)g(u)\), then:

\[
y' = f'(u)g(u) + f(u)g'(u)
\]

Here, let:
- \(f(u) = u^2 - 1\)
- \(g(u) = \cos(3u + 2)\)

Now, differentiate both parts:

1. \(f'(u) = \frac{d}{du}(u^2 - 1) = 2u\)
2. \(g'(u)\) requires the **chain rule**. We have:
   \[
   g'(u) = \frac{d}{du}\left(\cos(3u + 2)\right) = -\sin(3u + 2) \cdot 3 = -3\sin(3u + 2)
   \]

### Step 3: Apply the product rule
Now substitute into the product rule formula:

\[
y' = f'(u)g(u) + f(u)g'(u)
\]

\[
y' = (2u)\cos(3u + 2) + (u^2 - 1)(-3\sin(3u + 2))
\]

### Step 4: Simplify the derivative
This gives:

\[
y' = 2u\cos(3u + 2) - 3(u^2 - 1)\sin(3u + 2)
\]

Thus, the derivative of the given function is:

\[
y' = 2u\cos(3u + 2) - 3(u^2 - 1)\sin(3u + 2)
\]

Would you like more details or further explanation? 

Here are 5 related questions:
1. How does the product rule work in different situations?
2. What is the chain rule, and how do we apply it here?
3. What are the basic steps for differentiating trigonometric functions?
4. How can we handle more complex chain rule examples?
5. What happens if we differentiate a product of three or more functions?

**Tip**: Always simplify constants (like \(1^3\)) early in the process to make the differentiation easier!

y = (u² - 1³)cos(3u + 2), find the derivative y'.

Learn how to find the derivative of the function y = (u² - 1)cos(3u + 2) step by step using the product rule and chain rule. This solution covers key calculus concepts, including trigonometric functions, and is suitable for students in Grades 11-12 and early university-level calculus.

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