s compute the first-order partial derivatives of the function \( u = \arctan\left(\frac{x^2 + y^2}{x + y}\right) \) with respect to \( x \) and \( y \).


To compute the first-order partial derivatives of the function \( u = \arctan\left(\frac{x^2 + y^2}{x + y}\right) \) with respect to \( x \) and \( y \), we follow the steps below:

### Function:
\[
u(x, y) = \arctan\left(\frac{x^2 + y^2}{x + y}\right)
\]

We'll compute the partial derivatives \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \).

### Step 1: Differentiate \( u(x, y) \) with respect to \( x \)
We use the chain rule, noting that the derivative of \( \arctan(z) \) is \( \frac{1}{1 + z^2} \). Let
\[
z = \frac{x^2 + y^2}{x + y}.
\]
Then,
\[
\frac{\partial u}{\partial x} = \frac{1}{1 + z^2} \cdot \frac{\partial z}{\partial x}.
\]

Now, differentiate \( z \) with respect to \( x \). Using the quotient rule, the derivative of \( z = \frac{x^2 + y^2}{x + y} \) is:
\[
\frac{\partial z}{\partial x} = \frac{(2x)(x + y) - (x^2 + y^2)(1)}{(x + y)^2}.
\]
Simplifying the numerator:
\[
\frac{\partial z}{\partial x} = \frac{2x(x + y) - (x^2 + y^2)}{(x + y)^2} = \frac{2x^2 + 2xy - x^2 - y^2}{(x + y)^2} = \frac{x^2 + 2xy - y^2}{(x + y)^2}.
\]

Thus,
\[
\frac{\partial u}{\partial x} = \frac{1}{1 + z^2} \cdot \frac{x^2 + 2xy - y^2}{(x + y)^2}.
\]

### Step 2: Differentiate \( u(x, y) \) with respect to \( y \)
Again, we apply the chain rule:
\[
\frac{\partial u}{\partial y} = \frac{1}{1 + z^2} \cdot \frac{\partial z}{\partial y}.
\]

Now differentiate \( z = \frac{x^2 + y^2}{x + y} \) with respect to \( y \). Again, use the quotient rule:
\[
\frac{\partial z}{\partial y} = \frac{(2y)(x + y) - (x^2 + y^2)(1)}{(x + y)^2}.
\]
Simplifying the numerator:
\[
\frac{\partial z}{\partial y} = \frac{2y(x + y) - (x^2 + y^2)}{(x + y)^2} = \frac{2xy + 2y^2 - x^2 - y^2}{(x + y)^2} = \frac{-x^2 + 2xy + y^2}{(x + y)^2}.
\]

Thus,
\[
\frac{\partial u}{\partial y} = \frac{1}{1 + z^2} \cdot \frac{-x^2 + 2xy + y^2}{(x + y)^2}.
\]

### Final Partial Derivatives
\[
\frac{\partial u}{\partial x} = \frac{1}{1 + \left(\frac{x^2 + y^2}{x + y}\right)^2} \cdot \frac{x^2 + 2xy - y^2}{(x + y)^2}
\]
and
\[
\frac{\partial u}{\partial y} = \frac{1}{1 + \left(\frac{x^2 + y^2}{x + y}\right)^2} \cdot \frac{-x^2 + 2xy + y^2}{(x + y)^2}.
\]

Would you like more details on any of the steps or further explanation?

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### Related Questions:
1. How do you apply the quotient rule in partial derivatives?
2. What is the chain rule for functions of multiple variables?
3. How do you compute second-order partial derivatives for this function?
4. How would you simplify the expression for \( z \) in this context?
5. What are the applications of the arctangent function in multivariable calculus?

### Tip:
When differentiating functions involving fractions, the quotient rule is essential, but sometimes expressions can be simplified before differentiation to reduce complexity.

Compute the first-order partial derivatives of the function u = arctan((x^2 + y^2) / (x + y)) with respect to x and y.

Learn how to compute the first-order partial derivatives of the function u = arctan((x^2 + y^2) / (x + y)) with respect to x and y. This detailed solution covers key concepts in multivariable calculus, including the chain rule and quotient rule. Suitable for undergraduate students studying calculus.

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