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To find the derivative of the expression \((4x^2 - 5) \cdot (4x^4 - 7x)\) using the product rule, we follow the formula for the derivative of a product of two functions \( u(x) \) and \( v(x) \), which is given by:

\[
(uv)' = u'v + uv'
\]

Here, let:
\[ u = 4x^2 - 5 \]
\[ v = 4x^4 - 7x \]

First, compute the derivatives \( u' \) and \( v' \):
\[ u' = \frac{d}{dx}(4x^2 - 5) = 8x \]
\[ v' = \frac{d}{dx}(4x^4 - 7x) = 16x^3 - 7 \]

Now, apply the product rule:
\[ (uv)' = u'v + uv' \]
\[ (uv)' = (8x)(4x^4 - 7x) + (4x^2 - 5)(16x^3 - 7) \]

Let's expand and simplify this expression:
\[ = 32x^5 - 56x^2 + 64x^5 - 28x^2 - 80x^3 + 35 \]
\[ = 96x^5 - 84x^2 - 80x^3 + 35 \]

The derivative of the function \( (4x^2 - 5) \cdot (4x^4 - 7x) \) is:
\[ 96x^5 - 80x^3 - 84x^2 + 35 \]

If you have any further questions or need details, feel free to ask!

Related Questions:
1. What happens if you apply the quotient rule to the ratio of these two functions instead?
2. How would the derivative change if the power of \( x \) in the first function were higher?
3. Can you explain how to derive the general form of the product rule from first principles?
4. What are the practical applications of finding derivatives in real-world problems?
5. How would you use this derivative to find the local maxima and minima of the function?

Tip: When using the product rule, always ensure to distribute both \( u'v \) and \( uv' \) completely to avoid missing terms during simplification.

Find the derivative of (4x^2 - 5) · (4x^4 - 7x) using the PRODUCT RULE (Theorem G).

Learn how to compute the derivative of (4x^2 - 5)(4x^4 - 7x) by applying the product rule. This problem demonstrates key concepts in calculus and derivative applications, suitable for students in Grades 11-12.

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