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To solve this problem, we are given:

\[
u = f(r, s)
\]
with the relations:
\[
r = x^2 + y^2 \quad \text{and} \quad s = x^2 - y^2
\]

We are tasked with finding the partial derivatives \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \).

### Step 1: Chain Rule Application
Since \( u \) is a function of \( r \) and \( s \), and \( r \) and \( s \) are functions of \( x \) and \( y \), we can apply the chain rule for multivariable calculus:

\[
\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cdot \frac{\partial r}{\partial x} + \frac{\partial u}{\partial s} \cdot \frac{\partial s}{\partial x}
\]

\[
\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \cdot \frac{\partial r}{\partial y} + \frac{\partial u}{\partial s} \cdot \frac{\partial s}{\partial y}
\]

### Step 2: Compute the partial derivatives of \( r \) and \( s \)

Now, let's compute the partial derivatives of \( r = x^2 + y^2 \) and \( s = x^2 - y^2 \) with respect to \( x \) and \( y \).

- \( \frac{\partial r}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^2) = 2x \)
- \( \frac{\partial r}{\partial y} = \frac{\partial}{\partial y}(x^2 + y^2) = 2y \)
- \( \frac{\partial s}{\partial x} = \frac{\partial}{\partial x}(x^2 - y^2) = 2x \)
- \( \frac{\partial s}{\partial y} = \frac{\partial}{\partial y}(x^2 - y^2) = -2y \)

### Step 3: Substituting into the chain rule expressions

Now we substitute the partial derivatives of \( r \) and \( s \) into the chain rule equations:

\[
\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cdot 2x + \frac{\partial u}{\partial s} \cdot 2x
\]

\[
\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \cdot 2y + \frac{\partial u}{\partial s} \cdot (-2y)
\]

### Step 4: Final result

Simplifying the equations:

\[
\frac{\partial u}{\partial x} = 2x \left( \frac{\partial u}{\partial r} + \frac{\partial u}{\partial s} \right)
\]

\[
\frac{\partial u}{\partial y} = 2y \left( \frac{\partial u}{\partial r} - \frac{\partial u}{\partial s} \right)
\]

Thus, these are the expressions for the partial derivatives of \( u \) with respect to \( x \) and \( y \).

Would you like a detailed breakdown of any part of the process, or do you have additional questions? Here are some related questions that expand on this topic:

1. How would the result change if \( r \) and \( s \) had different functional forms?
2. How does the chain rule apply in higher dimensions?
3. Can this method be generalized to functions involving three or more variables?
4. What are the geometric interpretations of the functions \( r = x^2 + y^2 \) and \( s = x^2 - y^2 \)?
5. How does one solve similar partial derivative problems in polar coordinates?

**Tip:** When solving multivariable calculus problems, always identify the dependent and independent variables clearly before applying the chain rule.

If u = f(r,s); r = x^2 + y^2; s = x^2 - y^2, then find ∂u/∂x and ∂u/∂y.

This problem explores how to find the partial derivatives ∂u/∂x and ∂u/∂y when u is a function of two variables r and s, which are functions of x and y. Using the chain rule in multivariable calculus, we compute these derivatives step by step.

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