The given problem involves computing partial derivatives using the chain rule. Let's break down the steps:

The function is defined as:

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

We are asked to find:

1. $\frac{\partial z}{\partial u}$
2. $\frac{\partial z}{\partial v}$

To find the partial derivatives, we need to use the chain rule.

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

First, apply the chain rule:

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

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

Thus:

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

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

Next, apply the chain rule similarly:

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

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

Thus:

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

Final Answers

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

Let me know if you'd like a more detailed explanation of these steps!

Further Questions:

1. What happens to the partial derivatives when we set $u = 0$ and $v = 0$?
2. How can the chain rule be applied to more complex multivariable functions?
3. Can you identify the higher-order partial derivatives of $z$ with respect to $u$ and $v$?
4. What would happen if we switched the roles of $u$ and $v$ in the equations for $x$ and $y$?
5. How would these derivatives change if $x$ and $y$ were replaced with trigonometric functions?

Tip:

When applying the chain rule, always keep track of how each variable (in this case, $x$ and $y$) depends on the variables $u$ and $v$. This ensures that each term is correctly differentiated with respect to the relevant variable.