To solve part (b) of this problem, we need to determine how fast the product was selling 23 months after its introduction. This requires finding the rate of change of $N(t)$ with respect to $t$, i.e., $N'(t)$, and evaluating it at $t = 23$.

The total number of units sold is given by: $N(t) = 20,000 \left( 1 - e^{-0.05t} \right)^2$

Step 1: Differentiate $N(t)$

Using the chain rule and product rule, the derivative of $N(t)$ is: $N'(t) = 20,000 \cdot 2 \cdot \left( 1 - e^{-0.05t} \right) \cdot \frac{d}{dt}\left(1 - e^{-0.05t}\right)$

The derivative of $1 - e^{-0.05t}$ is: $\frac{d}{dt}(1 - e^{-0.05t}) = 0 - (-0.05)e^{-0.05t} = 0.05e^{-0.05t}$

Thus: $N'(t) = 20,000 \cdot 2 \cdot \left( 1 - e^{-0.05t} \right) \cdot 0.05e^{-0.05t}$

Simplify: $N'(t) = 2000 \cdot \left( 1 - e^{-0.05t} \right) \cdot e^{-0.05t}$

Step 2: Evaluate $N'(t)$ at $t = 23$

Substitute $t = 23$ into the derivative: $N'(23) = 2000 \cdot \left( 1 - e^{-0.05 \cdot 23} \right) \cdot e^{-0.05 \cdot 23}$

First, calculate $e^{-0.05 \cdot 23}$: $e^{-0.05 \cdot 23} = e^{-1.15} \approx 0.316$

Now calculate $1 - e^{-0.05 \cdot 23}$: $1 - 0.316 = 0.684$

Finally, substitute into $N'(23)$: $N'(23) = 2000 \cdot 0.684 \cdot 0.316$

Perform the multiplication: $N'(23) \approx 2000 \cdot 0.216 \approx 432$

Final Answer:

The product was selling at approximately 432 units per month 23 months after its introduction.

Would you like further details or clarification on the differentiation steps? Here are 5 related questions for exploration:

1. How is the chain rule applied in more complex differentiation problems?
2. What does the derivative tell us in the context of product sales?
3. How can exponential decay functions model real-world phenomena like sales trends?
4. How do we interpret the value of $e^{-0.05t}$ in this formula?
5. Can you calculate the second derivative to understand changes in the sales rate over time?

Tip: Always double-check your calculations for exponential terms, as small errors can propagate!