https://file.aiquickdraw.com/mathbot/file/1726663130555cwpzb5vi.jpg

To solve the problem, we need to find the derivative of the function:

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

We will apply the **product rule** for differentiation, as this function is a product of two functions: \( u(x) = 3x^2 \) and \( v(x) = \cos(2x - 1) \).

The product rule states:

\[
\frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x)
\]

### Step 1: Differentiate \( u(x) = 3x^2 \)
\[
u'(x) = 6x
\]

### Step 2: Differentiate \( v(x) = \cos(2x - 1) \) using the chain rule
\[
v'(x) = -\sin(2x - 1) \cdot \frac{d}{dx} (2x - 1) = -\sin(2x - 1) \cdot 2 = -2 \sin(2x - 1)
\]

### Step 3: Apply the product rule
\[
y' = u'(x) v(x) + u(x) v'(x)
\]
Substituting the values:
\[
y' = 6x \cos(2x - 1) + 3x^2 (-2 \sin(2x - 1))
\]
Simplifying:
\[
y' = 6x \cos(2x - 1) - 6x^2 \sin(2x - 1)
\]

Thus, the derivative is:

\[
y' = 6x \cos(2x - 1) - 6x^2 \sin(2x - 1)
\]

This corresponds to the third option in the image.

Do you need more details on this solution or have any further questions?

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Here are 5 related questions for further exploration:
1. What is the derivative of \( y = x^2 \sin(3x + 2) \)?
2. How does the product rule work when both terms involve trigonometric functions?
3. Can you derive the second derivative of \( y = 3x^2 \cos(2x - 1) \)?
4. What happens if you apply the quotient rule instead of the product rule in similar problems?
5. How can the chain rule be applied to functions involving higher-order trigonometric expressions?

**Tip:** Always simplify your derivative step by step to avoid missing any terms, especially when applying the product and chain rules together.

What is the derivative of y = 3x^2 cos(2x - 1)?

Learn how to find the derivative of y = 3x^2 cos(2x - 1) using the product rule and chain rule. This problem involves differentiating a product of a quadratic function and a trigonometric function, suitable for students in Grades 11-12 or college-level calculus.

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