The problem asks us to find the derivative $f'(x)$ of the given function $f(x) = (-9 + 6x)(9 + 5x^3 - 9x)$.

This requires the product rule for differentiation. The product rule states: $(fg)' = f'g + fg'$

Step 1: Define the terms

Let: $f(x) = (-9 + 6x) \quad \text{and} \quad g(x) = (9 + 5x^3 - 9x)$

Step 2: Compute $f'(x)$ and $g'(x)$

1. Differentiate $f(x) = -9 + 6x$: $f'(x) = 6$

2. Differentiate $g(x) = 9 + 5x^3 - 9x$: $g'(x) = 15x^2 - 9$

Step 3: Apply the product rule

Using the product rule: $f'(x) = f'(x)g(x) + f(x)g'(x)$

Substitute the expressions for $f(x)$, $f'(x)$, $g(x)$, and $g'(x)$: $f'(x) = (6)(9 + 5x^3 - 9x) + (-9 + 6x)(15x^2 - 9)$

Step 4: Simplify

1. Expand the first term: $(6)(9 + 5x^3 - 9x) = 54 + 30x^3 - 54x$

2. Expand the second term: $(-9 + 6x)(15x^2 - 9) = (-9)(15x^2) + (-9)(-9) + (6x)(15x^2) + (6x)(-9)$ Simplify each term: $(-9)(15x^2) = -135x^2, \quad (-9)(-9) = 81, \quad (6x)(15x^2) = 90x^3, \quad (6x)(-9) = -54x$ Combine: $(-9 + 6x)(15x^2 - 9) = -135x^2 + 81 + 90x^3 - 54x$

3. Combine all terms: $f'(x) = 54 + 30x^3 - 54x - 135x^2 + 81 + 90x^3 - 54x$

4. Group and simplify:

   • Cubic terms: $30x^3 + 90x^3 = 120x^3$
   • Quadratic terms: $-135x^2$
   • Linear terms: $-54x - 54x = -108x$
   • Constant terms: $54 + 81 = 135$

Final simplified form: $f'(x) = 120x^3 - 135x^2 - 108x + 135$

Would you like a detailed explanation of any step or have further questions?

Related Questions:

1. How is the product rule derived from the limit definition of derivatives?
2. Can you use the chain rule on $g(x) = 9 + 5x^3 - 9x$ to verify $g'(x)$?
3. What is the geometric interpretation of the derivative $f'(x)$?
4. How would the solution change if one of the coefficients were different?
5. Can this method be generalized to more than two factors in a product?

Tip:

Always simplify terms systematically to avoid errors in combining coefficients during differentiation.