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The problem asks us to find the slope of the surface \( z = f(x, y) = \ln(xy^2 + 2xy) + xy \) in the \( x \)-direction at the point \( (2, -1) \).

### Step 1: Partial derivative with respect to \( x \)
To find the slope in the \( x \)-direction, we need to compute the partial derivative of the function \( z = f(x, y) \) with respect to \( x \).

The function is:
\[
z = \ln(xy^2 + 2xy) + xy
\]

#### First term: \( \ln(xy^2 + 2xy) \)
Using the chain rule:
\[
\frac{\partial}{\partial x} \ln(xy^2 + 2xy) = \frac{1}{xy^2 + 2xy} \cdot \frac{\partial}{\partial x}(xy^2 + 2xy)
\]
We now compute \( \frac{\partial}{\partial x}(xy^2 + 2xy) \):
\[
\frac{\partial}{\partial x}(xy^2 + 2xy) = y^2 + 2y
\]
So,
\[
\frac{\partial}{\partial x} \ln(xy^2 + 2xy) = \frac{y^2 + 2y}{xy^2 + 2xy}
\]

#### Second term: \( xy \)
The partial derivative of \( xy \) with respect to \( x \) is:
\[
\frac{\partial}{\partial x}(xy) = y
\]

### Step 2: Total partial derivative with respect to \( x \)
Thus, the partial derivative of \( z \) with respect to \( x \) is:
\[
\frac{\partial z}{\partial x} = \frac{y^2 + 2y}{xy^2 + 2xy} + y
\]

### Step 3: Evaluate at the point \( (2, -1) \)
Now, we substitute \( x = 2 \) and \( y = -1 \) into the partial derivative:
\[
\frac{\partial z}{\partial x} = \frac{(-1)^2 + 2(-1)}{2(-1)^2 + 2(2)(-1)} + (-1)
\]

Simplifying the terms:
\[
\frac{\partial z}{\partial x} = \frac{1 - 2}{2(1) + 2(-2)} + (-1) = \frac{-1}{2 - 4} - 1 = \frac{-1}{-2} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}
\]

### Final Answer:
The slope of the surface in the \( x \)-direction at the point \( (2, -1) \) is:
\[
-\frac{1}{2}
\]

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Would you like further clarification or details on this process? Here are some related questions:

1. How do you find partial derivatives of more complex functions involving logarithms?
2. What is the physical interpretation of the partial derivative in the \( x \)-direction?
3. How does the chain rule apply to functions inside a logarithm?
4. Can we extend this to find the slope in the \( y \)-direction?
5. How do partial derivatives relate to the gradient of a surface?

**Tip:** For multi-variable functions, partial derivatives help you analyze how the function changes with respect to one variable while keeping others constant.

Find the slope of the surface z = f(x, y) = ln(xy^2 + 2xy) + xy in the x direction at the point (2, -1).

This problem involves calculating the partial derivative of the function z = ln(xy^2 + 2xy) + xy with respect to x at the point (2, -1). Learn step-by-step how to apply the chain rule and logarithmic differentiation to solve this university-level calculus problem.

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