To solve for $g'(s)$, we need to differentiate the function:

$g(s) = \frac{9s^2 - 5s + 9}{e^{13s}}$

This expression is a quotient of two functions, so we can apply the quotient rule, which states:

$\left( \frac{f}{h} \right)' = \frac{f' h - f h'}{h^2}$

Here:

• $f(s) = 9s^2 - 5s + 9$
• $h(s) = e^{13s}$

Let's differentiate both the numerator $f(s)$ and the denominator $h(s)$ separately.

1. Differentiate $f(s) = 9s^2 - 5s + 9$: $f'(s) = 18s - 5$

2. Differentiate $h(s) = e^{13s}$: $h'(s) = 13e^{13s}$

Now, we can apply the quotient rule:

$g'(s) = \frac{(18s - 5)e^{13s} - (9s^2 - 5s + 9)(13e^{13s})}{(e^{13s})^2}$

Simplify this expression:

$g'(s) = \frac{e^{13s} \left(18s - 5 - 13(9s^2 - 5s + 9)\right)}{e^{26s}}$

Since $e^{13s}$ in the numerator and $e^{26s}$ in the denominator simplify to $e^{13s}$ in the denominator, we have:

$g'(s) = \frac{18s - 5 - 13(9s^2 - 5s + 9)}{e^{13s}}$

Expanding the terms inside the numerator:

$g'(s) = \frac{18s - 5 - 117s^2 + 65s - 117}{e^{13s}}$

Combine like terms:

$g'(s) = \frac{-117s^2 + 83s - 122}{e^{13s}}$

Thus, the derivative is:

$g'(s) = \frac{-117s^2 + 83s - 122}{e^{13s}}$

Would you like further details on this solution, or do you have any questions?

Here are five related questions you might find helpful:

1. How does the quotient rule differ from the product rule in calculus?
2. What are some real-world applications of exponential functions in derivatives?
3. How would the result change if we had a different exponent on $e$ in the denominator?
4. How can we verify the correctness of this derivative?
5. What would happen if we differentiated this function a second time?

Tip: In problems involving exponential functions, keep an eye out for opportunities to simplify by canceling terms like $e^{kx}$ between the numerator and denominator.