To find the derivative of the function $H(t) = 0.8 \sin\left(\frac{\pi}{2}(t - 1)\right) + 1.8$, we will apply the chain rule.

1. Identify the components: The function consists of a sine function and a constant.

2. Differentiate the sine function:

   • The derivative of $\sin(x)$ is $\cos(x)$.
   • In our case, $x = \frac{\pi}{2}(t - 1)$.

3. Apply the chain rule:

   • If $u = \frac{\pi}{2}(t - 1)$, then the derivative $\frac{du}{dt} = \frac{\pi}{2}$.
   • Therefore, the derivative of $\sin(u)$ is $\cos(u) \cdot \frac{du}{dt} = \cos\left(\frac{\pi}{2}(t - 1)\right) \cdot \frac{\pi}{2}$.

4. Combine everything:

   • The derivative $H'(t)$ is: $H'(t) = 0.8 \cdot \cos\left(\frac{\pi}{2}(t - 1)\right) \cdot \frac{\pi}{2}$
   • Since the constant $1.8$ does not contribute to the derivative, it is omitted.

5. Final expression: $H'(t) = 0.4\pi \cos\left(\frac{\pi}{2}(t - 1)\right)$

Would you like more details on any step, or do you have further questions? Here are some related questions you might find interesting:

1. What is the value of $H'(t)$ at $t = 1$?
2. How would you find the second derivative $H''(t)$?
3. Can you explain the significance of the sine and cosine functions in this context?
4. What are the applications of this type of function in real-world scenarios?
5. How would you graph $H(t)$ and its derivative $H'(t)$?

Tip: Always remember to check the function's domain and behavior when working with trigonometric derivatives!