We are given the following equations:

$z = (x + y)e^y$ $x = u^2 + v^2$ $y = u^2 - v^2$

We are tasked with finding the partial derivatives $\frac{\partial z}{\partial u}$ and $\frac{\partial z}{\partial v}$ using the chain rule.

Step 1: Find $\frac{\partial z}{\partial u}$

To apply the chain rule, we'll first need to express the derivative of $z$ with respect to $u$, considering how $x$ and $y$ depend on $u$.

1. Use the chain rule:

$\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial u}$

2. Find the partial derivatives:

   • $\frac{\partial z}{\partial x} = e^y$ (since $z = (x + y)e^y$, and differentiating with respect to $x$ gives just $e^y$)
   • $\frac{\partial z}{\partial y} = (x + y)e^y + e^y$ (differentiate $z = (x + y)e^y$ with respect to $y$, applying the product rule)
   • $\frac{\partial x}{\partial u} = 2u$ (since $x = u^2 + v^2$)
   • $\frac{\partial y}{\partial u} = 2u$ (since $y = u^2 - v^2$)

3. Combine the terms:

$\frac{\partial z}{\partial u} = e^y \cdot 2u + \left( (x + y)e^y + e^y \right) \cdot 2u$

Simplify the expression:

$\frac{\partial z}{\partial u} = 2u e^y + 2u \left( (x + y)e^y + e^y \right)$ $= 2u e^y + 2u \left( (x + y + 1) e^y \right)$ $= 2u e^y \left( 1 + (x + y + 1) \right)$

Step 2: Find $\frac{\partial z}{\partial v}$

We apply a similar process for $v$:

1. Use the chain rule:

$\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial v}$

2. Find the partial derivatives:

   • $\frac{\partial x}{\partial v} = 2v$
   • $\frac{\partial y}{\partial v} = -2v$

3. Combine the terms:

$\frac{\partial z}{\partial v} = e^y \cdot 2v + \left( (x + y)e^y + e^y \right) \cdot (-2v)$

Simplify the expression:

$\frac{\partial z}{\partial v} = 2v e^y - 2v \left( (x + y)e^y + e^y \right)$ $= 2v e^y - 2v \left( (x + y + 1) e^y \right)$ $= 2v e^y \left( 1 - (x + y + 1) \right)$

Final answers:

$\frac{\partial z}{\partial u} = 2u e^y \left( x + y + 2 \right)$ $\frac{\partial z}{\partial v} = 2v e^y \left( -x - y \right)$

Let me know if you need further clarification or if you want to see more steps.

Related questions:

1. How can we calculate the second-order partial derivatives for $z$?
2. What is the meaning of the chain rule in this context?
3. How can we find the total derivative $dz$ in terms of $du$ and $dv$?
4. Can we use a different method to find these derivatives more efficiently?
5. What are some common mistakes when applying the chain rule?

Tip: When using the chain rule, carefully track each term and variable that depends on the others. Writing out all steps can help prevent errors in complex problems.