To find the derivative of the function $y = (2x^2 + 5)(2x - 5)$ using the product rule, follow these steps:

Step 1: Identify the two functions

Let:

• $u(x) = 2x^2 + 5$
• $v(x) = 2x - 5$

Step 2: Find the derivatives of $u(x)$ and $v(x)$

Differentiate each function with respect to $x$:

$u'(x) = \frac{d}{dx}(2x^2 + 5) = 4x$

$v'(x) = \frac{d}{dx}(2x - 5) = 2$

Step 3: Apply the product rule

The product rule states:

$y' = u'(x)v(x) + u(x)v'(x)$

Substitute the derivatives and the original functions:

$y' = (4x)(2x - 5) + (2x^2 + 5)(2)$

Step 4: Simplify the expression

Expand and simplify:

$y' = 4x(2x - 5) + 2(2x^2 + 5)$

First, expand $4x(2x - 5)$:

$4x(2x - 5) = 8x^2 - 20x$

Next, expand $2(2x^2 + 5)$:

$2(2x^2 + 5) = 4x^2 + 10$

Finally, combine these results:

$y' = 8x^2 - 20x + 4x^2 + 10$

Combine like terms:

$y' = 12x^2 - 20x + 10$

Final Answer:

The derivative of $y = (2x^2 + 5)(2x - 5)$ is:

$y' = 12x^2 - 20x + 10$

Would you like any further explanations or details?

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Here are 8 related questions you might find helpful:

1. How do you apply the product rule to more complicated functions?
2. What is the quotient rule, and how is it different from the product rule?
3. Can the product rule be used for more than two functions multiplied together?
4. What are common mistakes to avoid when using the product rule?
5. How does the product rule apply to trigonometric functions?
6. What is the significance of the product rule in calculus?
7. How do you find the derivative of a product of functions using the chain rule?
8. How can the product rule be visualized geometrically?

Tip: When working with the product rule, always double-check your differentiation of each individual function before applying the rule. This helps prevent errors in more complex problems.