Find

∂u

∂x and

∂u

∂y

when u = tan−1 x2−y2

We are given the function \( u = \tan^{-1}(x^2 - y^2) \), and we are asked to find the partial derivatives \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \).

### Step 1: Differentiate \( u \) with respect to \( x \)

We know that the derivative of \( \tan^{-1}(z) \) with respect to \( z \) is \( \frac{1}{1+z^2} \). Applying the chain rule, we have:

\[
\frac{\partial u}{\partial x} = \frac{1}{1 + (x^2 - y^2)^2} \cdot \frac{\partial}{\partial x}(x^2 - y^2)
\]

The partial derivative of \( x^2 - y^2 \) with respect to \( x \) is:

\[
\frac{\partial}{\partial x}(x^2 - y^2) = 2x
\]

Thus,

\[
\frac{\partial u}{\partial x} = \frac{2x}{1 + (x^2 - y^2)^2}
\]

### Step 2: Differentiate \( u \) with respect to \( y \)

Similarly, using the chain rule again for the derivative of \( \tan^{-1}(z) \) with respect to \( y \), we get:

\[
\frac{\partial u}{\partial y} = \frac{1}{1 + (x^2 - y^2)^2} \cdot \frac{\partial}{\partial y}(x^2 - y^2)
\]

The partial derivative of \( x^2 - y^2 \) with respect to \( y \) is:

\[
\frac{\partial}{\partial y}(x^2 - y^2) = -2y
\]

Thus,

\[
\frac{\partial u}{\partial y} = \frac{-2y}{1 + (x^2 - y^2)^2}
\]

### Final Answer:

- \( \frac{\partial u}{\partial x} = \frac{2x}{1 + (x^2 - y^2)^2} \)
- \( \frac{\partial u}{\partial y} = \frac{-2y}{1 + (x^2 - y^2)^2} \)

Do you want further details on any of the steps or have any questions?

Here are 5 related questions to deepen understanding:

1. How do you apply the chain rule for multivariable functions?
2. What are the properties of the inverse tangent function, \( \tan^{-1}(z) \)?
3. What is the geometric interpretation of partial derivatives in 3D space?
4. How does the form of the derivative change when using a different function like \( \sin^{-1} \) instead of \( \tan^{-1} \)?
5. Can you apply a similar method to find the second-order partial derivatives?

**Tip:** When taking partial derivatives, always treat the other variables as constants while differentiating with respect to one specific variable.

Find ∂u/∂x and ∂u/∂y when u = tan⁻¹(x² - y²).

This solution explains how to find the partial derivatives of u = tan⁻¹(x² - y²) with respect to x and y. It covers concepts in multivariable calculus, including partial derivatives and the application of the chain rule. Detailed steps are provided for students at the undergraduate calculus level.

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