Find the first order partial derivatives of the following:
1.
(i) u = arctan((x ^ 2 + y ^ 2)/(x + y))

Let'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 \( y \).

### 1. Partial derivative of \( u \) with respect to \( x \)
The formula for the derivative of the arctangent function is:

\[
\frac{d}{dx} \arctan(f(x)) = \frac{1}{1 + f(x)^2} \cdot f'(x)
\]

Let \( f(x, y) = \frac{x^2 + y^2}{x + y} \), then:

\[
u = \arctan\left(f(x, y)\right)
\]

#### Step 1: Compute \( f_x \), the partial derivative of \( f(x, y) \) with respect to \( x \)

We use the quotient rule to differentiate \( f(x, y) = \frac{x^2 + y^2}{x + y} \). The quotient rule is:

\[
\frac{d}{dx} \left(\frac{g(x)}{h(x)}\right) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}
\]

Here, \( g(x) = x^2 + y^2 \) and \( h(x) = x + y \).

\[
g'(x) = 2x, \quad h'(x) = 1
\]

Using the quotient rule:

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

Simplifying:

\[
f_x = \frac{2x(x + y) - (x^2 + y^2)}{(x + y)^2}
\]
\[
f_x = \frac{2x^2 + 2xy - x^2 - y^2}{(x + y)^2}
\]
\[
f_x = \frac{x^2 + 2xy - y^2}{(x + y)^2}
\]

#### Step 2: Apply the chain rule for the arctangent function

Now, using the chain rule:

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

Substituting \( f_x \):

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

---

### 2. Partial derivative of \( u \) with respect to \( y \)
Similarly, we now compute the partial derivative with respect to \( y \).

#### Step 1: Compute \( f_y \), the partial derivative of \( f(x, y) \) with respect to \( y \)

We again use the quotient rule with \( g(y) = x^2 + y^2 \) and \( h(y) = x + y \).

\[
g'(y) = 2y, \quad h'(y) = 1
\]

Applying the quotient rule:

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

Simplifying:

\[
f_y = \frac{2y(x + y) - (x^2 + y^2)}{(x + y)^2}
\]
\[
f_y = \frac{2xy + 2y^2 - x^2 - y^2}{(x + y)^2}
\]
\[
f_y = \frac{-x^2 + 2xy + y^2}{(x + y)^2}
\]

#### Step 2: Apply the chain rule for the arctangent function

Now, using the chain rule:

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

Substituting \( f_y \):

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

---

### Final Results:

- The partial derivative with respect to \( x \) is:

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

- The partial derivative with respect to \( y \) is:

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

---

Would you like any further clarification or additional details?

Here are some related questions to further your understanding:
1. What is the geometric interpretation of partial derivatives?
2. How do the quotient and chain rules work in partial differentiation?
3. Can we simplify the partial derivatives using symmetry properties of the function?
4. How does the arctangent function behave for large or small values of \( x \) and \( y \)?
5. What are the second-order partial derivatives of \( u \)?

**Tip**: Always double-check the application of the quotient rule in complex expressions, as errors often occur in simplifying terms.

Find the first order partial derivatives of the following: 1. (i) u = arctan((x ^ 2 + y ^ 2)/(x + y))

Explore the first-order partial derivatives of the function u = arctan((x^2 + y^2)/(x + y)). This detailed solution uses the quotient rule and chain rule to compute the derivatives with respect to x and y. Ideal for undergraduate calculus students.

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