Math Problem Statement

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

Solution

The problem asks us to find the slope of the surface z=f(x,y)=ln(xy2+2xy)+xyz = f(x, y) = \ln(xy^2 + 2xy) + xy in the xx-direction at the point (2,1)(2, -1).

Step 1: Partial derivative with respect to xx

To find the slope in the xx-direction, we need to compute the partial derivative of the function z=f(x,y)z = f(x, y) with respect to xx.

The function is: z=ln(xy2+2xy)+xyz = \ln(xy^2 + 2xy) + xy

First term: ln(xy2+2xy)\ln(xy^2 + 2xy)

Using the chain rule: xln(xy2+2xy)=1xy2+2xyx(xy2+2xy)\frac{\partial}{\partial x} \ln(xy^2 + 2xy) = \frac{1}{xy^2 + 2xy} \cdot \frac{\partial}{\partial x}(xy^2 + 2xy) We now compute x(xy2+2xy)\frac{\partial}{\partial x}(xy^2 + 2xy): x(xy2+2xy)=y2+2y\frac{\partial}{\partial x}(xy^2 + 2xy) = y^2 + 2y So, xln(xy2+2xy)=y2+2yxy2+2xy\frac{\partial}{\partial x} \ln(xy^2 + 2xy) = \frac{y^2 + 2y}{xy^2 + 2xy}

Second term: xyxy

The partial derivative of xyxy with respect to xx is: x(xy)=y\frac{\partial}{\partial x}(xy) = y

Step 2: Total partial derivative with respect to xx

Thus, the partial derivative of zz with respect to xx is: zx=y2+2yxy2+2xy+y\frac{\partial z}{\partial x} = \frac{y^2 + 2y}{xy^2 + 2xy} + y

Step 3: Evaluate at the point (2,1)(2, -1)

Now, we substitute x=2x = 2 and y=1y = -1 into the partial derivative: zx=(1)2+2(1)2(1)2+2(2)(1)+(1)\frac{\partial z}{\partial x} = \frac{(-1)^2 + 2(-1)}{2(-1)^2 + 2(2)(-1)} + (-1)

Simplifying the terms: zx=122(1)+2(2)+(1)=1241=121=121=12\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 xx-direction at the point (2,1)(2, -1) is: 12-\frac{1}{2}


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 xx-direction?
  3. How does the chain rule apply to functions inside a logarithm?
  4. Can we extend this to find the slope in the yy-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.

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Multivariable Calculus
Logarithmic Functions

Formulas

Partial derivative formula: ∂f/∂x
Chain rule: d/dx[ln(g(x))] = 1/g(x) * g'(x)

Theorems

Chain Rule for differentiation
Logarithmic differentiation

Suitable Grade Level

University Level Calculus