To find the rate of change of concentration, we need to compute the derivative of $C(t)$ with respect to $t$:

$C(t) = 5e^{-0.3t}$

The derivative $C'(t)$ is:

$C'(t) = 5 \cdot (-0.3) e^{-0.3t} = -1.5 e^{-0.3t}$

Now, let's evaluate $C'(t)$ at $t = 1$ and $t = 5$:

1. At $t = 1$: $C'(1) = -1.5 e^{-0.3 \cdot 1} \approx -1.5 e^{-0.3} \approx -1.5 \cdot 0.7408 \approx -1.1112$

2. At $t = 5$: $C'(5) = -1.5 e^{-0.3 \cdot 5} \approx -1.5 e^{-1.5} \approx -1.5 \cdot 0.2231 \approx -0.3347$

Thus, the rate of change of concentration is approximately:

• After 1 hour: $-1.11 \, \text{mg/mL/hr}$
• After 5 hours: $-0.33 \, \text{mg/mL/hr}$

Would you like more details on this calculation? Here are some related questions you might consider:

1. How does the concentration change at $t = 0$?
2. What does the negative rate of change indicate about the drug concentration?
3. How would the rate of change be affected if the exponent changed?
4. Can you graph the concentration over time?
5. What is the significance of the constant $5$ in the equation?

Tip: Understanding derivatives helps in analyzing how quantities change over time in real-world applications!