To find the derivative of the function $f(x) = (3x^2 - 7)(2x + 1)$, we can use the product rule for derivatives. The product rule states that for two functions $u(x)$ and $v(x)$, the derivative of their product is:

$(f \cdot g)' = f' \cdot g + f \cdot g'$

Step-by-Step Solution

Let:

• $u(x) = 3x^2 - 7$
• $v(x) = 2x + 1$

Step 1: Differentiate $u(x)$ and $v(x)$ separately

1. $u'(x) = \frac{d}{dx}(3x^2 - 7) = 6x$
2. $v'(x) = \frac{d}{dx}(2x + 1) = 2$

Step 2: Apply the product rule

Using $f'(x) = u'(x)v(x) + u(x)v'(x)$, we get:

$f'(x) = (6x)(2x + 1) + (3x^2 - 7)(2)$

Step 3: Simplify each term

1. $(6x)(2x + 1) = 12x^2 + 6x$
2. $(3x^2 - 7)(2) = 6x^2 - 14$

Step 4: Combine terms

$f'(x) = 12x^2 + 6x + 6x^2 - 14$ $f'(x) = 18x^2 + 6x - 14$

Final Answer

$f'(x) = 18x^2 + 6x - 14$

Would you like more details or have any questions about the steps?

Here are some related questions for further exploration:

1. How would you find the second derivative of this function?
2. Can we verify this derivative by expanding $f(x)$ before differentiating?
3. How does the product rule apply if one of the factors was a constant?
4. What would happen if the function had three factors instead of two?
5. How does the chain rule differ from the product rule?

Tip: Always simplify each term in differentiation before combining for a clear answer.