To find $\frac{dz}{dt}$ using the chain rule, we need to differentiate $z = (x + y)e^x$ with respect to $t$, where $x = 2t$ and $y = 2 - t^2$.

Step-by-step approach:

1. Identify the expression for $z$: $z = (x + y)e^x$

2. Differentiate $z$ with respect to $t$: By the chain rule, we have: $\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}$

   Now, we need to compute the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$, and also the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

3. Find partial derivatives:

   = (x + y)e^x + e^x = (x + y + 1)e^x$$ $$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \left( (x + y)e^x \right) = e^x$$

4. Find the derivatives of $x$ and $y$ with respect to $t$: $\frac{dx}{dt} = \frac{d}{dt}(2t) = 2$ $\frac{dy}{dt} = \frac{d}{dt}(2 - t^2) = -2t$

5. Substitute everything into the chain rule: $\frac{dz}{dt} = \left( (x + y + 1)e^x \right) \cdot 2 + \left( e^x \right) \cdot (-2t)$

6. Substitute the expressions for $x$ and $y$: Since $x = 2t$ and $y = 2 - t^2$, substitute these into the equation: $\frac{dz}{dt} = \left( \left( 2t + (2 - t^2) + 1 \right)e^{2t} \right) \cdot 2 + e^{2t} \cdot (-2t)$ Simplify the expression inside the parentheses: $2t + 2 - t^2 + 1 = -t^2 + 2t + 3$ So, the expression becomes: $\frac{dz}{dt} = \left( (-t^2 + 2t + 3)e^{2t} \right) \cdot 2 + e^{2t} \cdot (-2t)$ Simplify: $\frac{dz}{dt} = 2(-t^2 + 2t + 3)e^{2t} - 2te^{2t}$ Combine like terms: $\frac{dz}{dt} = e^{2t} \left( 2(-t^2 + 2t + 3) - 2t \right)$ Simplify further: $\frac{dz}{dt} = e^{2t} \left( -2t^2 + 4t + 6 - 2t \right)$ $\frac{dz}{dt} = e^{2t} \left( -2t^2 + 2t + 6 \right)$

Thus, the derivative is: $\frac{dz}{dt} = e^{2t} \left( -2t^2 + 2t + 6 \right)$

Do you need further details or clarifications on any step?

Here are 5 follow-up questions to explore this topic further:

1. How does the chain rule work when there are multiple variables involved, like in this case with $x$ and $y$?
2. How would you compute $\frac{dz}{dt}$ if we used a different form for $x$ and $y$?
3. What would happen if the expression for $z$ included more complex terms like trigonometric functions?
4. How would you apply the product rule if $z$ was a product of more complex functions instead of a sum?
5. Could you simplify this expression for special cases, like when $t = 0$?

Tip: When working with derivatives, always double-check that you’ve applied the chain rule correctly, especially when dealing with compositions of functions.