To find the derivative of the function $f(x) = (-3x^3 + 7x + 11)(2x^2 - 4)$ using the product rule, we proceed as follows:

Product Rule

The product rule states that if $f(x) = g(x) \cdot h(x)$, then the derivative is: $f'(x) = g'(x)h(x) + g(x)h'(x)$ Here, $g(x) = -3x^3 + 7x + 11$ and $h(x) = 2x^2 - 4$.

Step 1: Differentiate $g(x)$ and $h(x)$

• $g(x) = -3x^3 + 7x + 11$ $g'(x) = -9x^2 + 7$
• $h(x) = 2x^2 - 4$ $h'(x) = 4x$

Step 2: Apply the product rule

Using the product rule: $f'(x) = g'(x)h(x) + g(x)h'(x)$ Substitute the values of $g(x)$, $g'(x)$, $h(x)$, and $h'(x)$: $f'(x) = (-9x^2 + 7)(2x^2 - 4) + (-3x^3 + 7x + 11)(4x)$

Step 3: Expand both terms

1. Expand the first term: $(-9x^2 + 7)(2x^2 - 4)$ $(-9x^2)(2x^2) + (-9x^2)(-4) + 7(2x^2) + 7(-4) = -18x^4 + 36x^2 + 14x^2 - 28 = -18x^4 + 50x^2 - 28$

2. Expand the second term: $(-3x^3 + 7x + 11)(4x)$ $(-3x^3)(4x) + (7x)(4x) + (11)(4x) = -12x^4 + 28x^2 + 44x$

Step 4: Combine all terms

Now combine the two results: $f'(x) = (-18x^4 + 50x^2 - 28) + (-12x^4 + 28x^2 + 44x)$ $f'(x) = -18x^4 - 12x^4 + 50x^2 + 28x^2 + 44x - 28$ $f'(x) = -30x^4 + 78x^2 + 44x - 28$

Thus, the derivative of the function is: $f'(x) = -30x^4 + 78x^2 + 44x - 28$

Would you like further details or explanations on any step?

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Here are 5 related questions:

1. How does the chain rule differ from the product rule?
2. What is the derivative of a product of three functions?
3. How would the derivative change if the function had a quotient?
4. Can you apply the product rule to trigonometric functions?
5. How do you find second derivatives using the product rule?

Tip: Always simplify the expressions before differentiating to make the process easier!