Differentiate using product and quotient rule (1- 4x^2) cos x

To differentiate \((1 - 4x^2) \cos x\), we apply both the **Product Rule** and the **Quotient Rule**. 

The Product Rule formula is:

\[
\frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)
\]

Here, we have:
- \(f(x) = 1 - 4x^2\)
- \(g(x) = \cos x\)

### Step 1: Differentiate \(f(x)\) and \(g(x)\)
- \(f'(x) = -8x\)
- \(g'(x) = -\sin x\)

### Step 2: Apply the Product Rule
\[
\frac{d}{dx} \left[ (1 - 4x^2) \cos x \right] = f'(x)g(x) + f(x)g'(x)
\]

Substitute the values of \(f(x)\), \(f'(x)\), \(g(x)\), and \(g'(x)\):
\[
= (-8x) \cos x + (1 - 4x^2) (-\sin x)
\]
\[
= -8x \cos x - (1 - 4x^2) \sin x
\]

Thus, the derivative of \((1 - 4x^2) \cos x\) is:
\[
\frac{d}{dx} \left[ (1 - 4x^2) \cos x \right] = -8x \cos x - (1 - 4x^2) \sin x
\]

Would you like more details or further explanations?

Here are five questions related to this:
1. How does the product rule work in general?
2. Can you explain why we use \(-8x\) for \(f'(x)\)?
3. What is the significance of \(\cos x\) and \(\sin x\) in trigonometric differentiation?
4. How would the answer change if the trigonometric function was \(\sin x\) instead of \(\cos x\)?
5. What are other examples where the product rule is useful?

**Tip**: Always remember to simplify the final expression after applying the product rule for clearer results.

Differentiate using product and quotient rule (1 - 4x^2) cos x

Learn how to differentiate the expression (1 - 4x^2) cos x using the product rule. This detailed step-by-step solution involves calculus concepts such as the product rule and trigonometric differentiation, making it suitable for students in advanced high school calculus or first-year university courses.

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