The given function is:

$f(x) = (-6x - 4)e^{9x - 8}$

We are tasked with finding its derivative. To compute this, we will apply the product rule, as $f(x)$ is the product of two functions:

• $u(x) = -6x - 4$
• $v(x) = e^{9x - 8}$.

The product rule states:

$f'(x) = u'(x)v(x) + u(x)v'(x).$

Step 1: Compute $u'(x)$ and $v'(x)$

• $u(x) = -6x - 4 \implies u'(x) = -6$,
• $v(x) = e^{9x - 8} \implies v'(x) = 9e^{9x - 8}$ (using the chain rule).

Step 2: Apply the product rule

Now substitute into the product rule:

$f'(x) = (-6)(e^{9x - 8}) + (-6x - 4)(9e^{9x - 8}).$

Step 3: Simplify

Factor out $e^{9x - 8}$:

$f'(x) = e^{9x - 8} \big[ -6 + 9(-6x - 4) \big].$

Simplify the terms inside the brackets:

$-6 + 9(-6x - 4) = -6 - 54x - 36 = -54x - 42.$

Thus, the derivative is:

$f'(x) = e^{9x - 8}(-54x - 42).$

Or equivalently:

$f'(x) = -e^{9x - 8}(54x + 42).$

Let me know if you'd like further clarification or additional details on this solution!

Relative Questions:

1. How can the product rule be applied to other composite functions?
2. What is the importance of the chain rule when differentiating exponential functions?
3. How does factoring simplify derivative expressions in general?
4. Can the derivative $f'(x)$ be further simplified for specific values of $x$?
5. How would the derivative change if the exponential's base were not $e$?

Tip:

Always double-check intermediate steps when applying the product or chain rules to ensure accuracy in computations!