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The problem asks to calculate the gradient of the function \( f(x, y) = \sqrt{x + y} \) at the point \( (3, 6) \).

The gradient of a function \( f(x, y) \) is denoted as \( \nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Let's proceed with the solution step-by-step.

### Step 1: Compute the partial derivatives.

1. **Partial derivative with respect to \( x \)**:
   \[
   \frac{\partial}{\partial x} f(x, y) = \frac{\partial}{\partial x} \left( \sqrt{x + y} \right)
   \]
   Using the chain rule:
   \[
   \frac{\partial}{\partial x} \left( \sqrt{x + y} \right) = \frac{1}{2\sqrt{x + y}} \cdot \frac{\partial}{\partial x} (x + y) = \frac{1}{2\sqrt{x + y}} \cdot 1 = \frac{1}{2\sqrt{x + y}}
   \]

2. **Partial derivative with respect to \( y \)**:
   \[
   \frac{\partial}{\partial y} f(x, y) = \frac{\partial}{\partial y} \left( \sqrt{x + y} \right)
   \]
   Similarly:
   \[
   \frac{\partial}{\partial y} \left( \sqrt{x + y} \right) = \frac{1}{2\sqrt{x + y}} \cdot \frac{\partial}{\partial y} (x + y) = \frac{1}{2\sqrt{x + y}} \cdot 1 = \frac{1}{2\sqrt{x + y}}
   \]

### Step 2: Evaluate the partial derivatives at \( (x, y) = (3, 6) \).

- **For \( \frac{\partial f}{\partial x} \) at \( (3, 6) \)**:
  \[
  \frac{\partial f}{\partial x} (3, 6) = \frac{1}{2\sqrt{3 + 6}} = \frac{1}{2\sqrt{9}} = \frac{1}{2 \times 3} = \frac{1}{6}
  \]

- **For \( \frac{\partial f}{\partial y} \) at \( (3, 6) \)**:
  \[
  \frac{\partial f}{\partial y} (3, 6) = \frac{1}{2\sqrt{3 + 6}} = \frac{1}{2\sqrt{9}} = \frac{1}{6}
  \]

### Step 3: Write the gradient.

Thus, the gradient of \( f(x, y) \) at the point \( (3, 6) \) is:
\[
\nabla f(3, 6) = \left( \frac{1}{6}, \frac{1}{6} \right)
\]

Would you like more details on any of these steps or have any further questions?

Here are some related questions:
1. What is the geometric interpretation of the gradient?
2. How can you apply the gradient to find the direction of steepest ascent?
3. How would the gradient change if the function were \( f(x, y) = \sqrt{x^2 + y^2} \)?
4. How does the chain rule apply in multivariable calculus when calculating gradients?
5. Can you calculate higher-order derivatives (such as Hessians) of this function?

**Tip:** The gradient always points in the direction of the steepest increase in a function.

Calcula el gradiente de f(x, y) = √(x + y) en el punto (3, 6).

Learn how to calculate the gradient of the function f(x, y) = √(x + y) at the point (3, 6) using partial derivatives. This problem covers multivariable calculus concepts, including the chain rule. Suitable for college-level calculus students.

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