The sales function provided in the image is:

$S(t) = \frac{180t + 1}{t^2 + 6}$

To find the rate of change of sales when $t = 0$, we need to compute the derivative of $S(t)$ with respect to $t$ and evaluate it at $t = 0$.

The function is a quotient, so we will use the quotient rule for differentiation, which is:

$\frac{d}{dt} \left( \frac{f(t)}{g(t)} \right) = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2}$

Here, $f(t) = 180t + 1$ and $g(t) = t^2 + 6$.

Steps:

1. Differentiate $f(t) = 180t + 1$ to get $f'(t)$.
2. Differentiate $g(t) = t^2 + 6$ to get $g'(t)$.
3. Apply the quotient rule to find $S'(t)$.
4. Evaluate $S'(t)$ at $t = 0$.

I'll now calculate this.The rate of change of sales in millions of dollars per year when $t = 0$ is $30$ million dollars per year.

Would you like more details on the steps involved or have any other questions?

Here are 5 related questions for further practice:

1. What is the rate of change of sales when $t = 1$?
2. How do we interpret the rate of change of a sales function?
3. What is the second derivative of the sales function, and what does it represent?
4. How would the rate of change of sales behave as $t \to \infty$?
5. Can we use any other rule besides the quotient rule for this function?

Tip: The quotient rule is a key tool for differentiating ratios of two functions—make sure to carefully apply it step by step!