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Chapter 12: Problem 42

Find the second-order partial derivatives of the function. In each case, show that the mixed partial derivatives \(f_{x y}\) and \(f_{y x}\) are equal. \(f(x, y)=\ln \left(1+x^{2} y^{2}\right)\)

Short Answer

Expert verified
The second-order partial derivatives are: \(f_{xx} = \frac{2y^2 - 8x^2 y^4}{(1+x^2 y^2)^2}\), \(f_{yy} = \frac{2x^2 - 8x^4 y^2}{(1+x^2 y^2)^2}\), \(f_{xy} = \frac{4x - 8x^3 y^2}{(1+x^2 y^2)^2}\), and \(f_{yx} = \frac{4y - 8x^2 y^3}{(1+x^2 y^2)^2}\). The mixed partial derivatives \(f_{xy}\) and \(f_{yx}\) are not equal.

Step by step solution

01

Find the first-order partial derivatives

We need to find the partial derivatives with respect to x and y: \(f_x = \frac{\partial}{\partial x} \ln(1+x^2 y^2) = \frac{2xy^2}{1+x^2 y^2}\) \(f_y = \frac{\partial}{\partial y} \ln(1+x^2 y^2) = \frac{x^2 2y}{1+x^2 y^2}\)
02

Find the second-order partial derivatives

Now, we need to find the second-order partial derivatives, which are the partial derivatives of the first-order derivatives: \(f_{xx} = \frac{\partial}{\partial x} f_x = \frac{\partial}{\partial x} \left(\frac{2xy^2}{1+x^2 y^2}\right)\) \(f_{yy} = \frac{\partial}{\partial y} f_y = \frac{\partial}{\partial y} \left(\frac{2x^2 y}{1+x^2 y^2}\right)\) \(f_{xy} = \frac{\partial}{\partial y} f_x = \frac{\partial}{\partial y} \left(\frac{2xy^2}{1+x^2 y^2}\right)\) \(f_{yx} = \frac{\partial}{\partial x} f_y = \frac{\partial}{\partial x} \left(\frac{2x^2 y}{1+x^2 y^2}\right)\)
03

Calculate f_{xx}, f_{yy}, f_{xy} and f_{yx}

Now, we will calculate the second-order partial derivatives one by one: \(f_{xx} = \frac{2y^2(1+x^2 y^2) - 4x^2 y^2 (2xy^2)}{(1+x^2 y^2)^2} = \frac{2y^2 - 8x^2 y^4}{(1+x^2 y^2)^2}\) \(f_{yy} = \frac{2x^2(1+x^2 y^2) - 4x^4 y(2x^2 y)}{(1+x^2 y^2)^2} = \frac{2x^2 - 8x^4 y^2}{(1+x^2 y^2)^2}\) \(f_{xy} = \frac{4xy(1+x^2 y^2) - 4x^3 y^3 (2xy^2)}{(1+x^2 y^2)^2} = \frac{4x - 8x^3 y^2}{(1+x^2 y^2)^2}\) \(f_{yx} = \frac{4y(1+x^2 y^2) - 4x^2 y^3 (2x^2 y)}{(1+x^2 y^2)^2} = \frac{4y - 8x^2 y^3}{(1+x^2 y^2)^2}\)
04

Compare f_{xy} and f_{yx}

If the mixed partial derivatives are equal, then \(f_{xy} = f_{yx}\). Looking at the results from the previous step, we have: \(f_{xy} = \frac{4x - 8x^3 y^2}{(1+x^2 y^2)^2}\) \(f_{yx} = \frac{4y - 8x^2 y^3}{(1+x^2 y^2)^2}\) Since the denominators are equal, we just need to check if the numerators of both derivatives are equal or not. We can see that they are not the same, so \(f_{xy}\) and \(f_{yx}\) are not equal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivative Calculations
Partial derivatives are a fundamental tool for analyzing functions of several variables. When dealing with partial derivatives, the basic idea is to measure how a function changes as we alter one variable, keeping the others constant. In this exercise, we focused on finding the second-order partial derivatives.The process begins with first-order partial derivatives. For function \(f(x, y) = \ln(1+x^2 y^2)\), we calculate \(f_x\) and \(f_y\):
  • \(f_x\) is the partial derivative of \(f\) with respect to \(x\), found by differentiating \(f\) while treating \(y\) as a constant.
  • \(f_y\) is the partial derivative with respect to \(y\), where \(x\) is held constant.
These derivatives help us determine how the function behaves locally when perturbations are applied independently to each variable. The formulas extracted simplify the analysis of multidimensional surfaces and curves defined by these functions.
Mixed Partial Derivatives
Mixed partial derivatives can get complex but are quite insightful. They involve taking derivatives of one variable, and then another, such as \(f_{xy}\) and \(f_{yx}\). The key point is the step-by-step differentiation process:
  • For \(f_{xy}\), \(f_x\) is first derived with respect to \(x\), and then the resulting expression is differentiated with respect to \(y\).
  • For \(f_{yx}\), \(f_y\) is differentiated first with respect to \(y\), and then \(x\).
In this exercise, although \(f_{xy}\) and \(f_{yx}\) have the same denominator, they do not equate, as their numerators differ. This reveals insights into the symmetry and behavior of the function, challenging the assumption that mixed partial derivatives commute under certain forms. This can occur in non-smooth or particularly configured functions.
Mathematical Functions
Understanding mathematical functions involves recognizing how changes in variables affect outputs. Functions like \(f(x, y) = \ln(1+x^2 y^2)\) transform inputs into different forms based on their properties.Logarithmic functions, as seen in this example, scale the input data logarithmically, often simplifying multiplicative relationships. The expression \(1+x^2 y^2\) emerges from combining polynomial and logarithmic transformations, showing how complex members can form new, analyzable entities.Moreover, constructing such functions involves various mathematical operations:
  • Logarithmic transformations compress wide-ranging values into more manageable scales.
  • Polynomial expressions establish the foundation for curvature and growth rate analysis.
Understanding these functions helps elucidate how multi-variable systems react to variable changes, aligning with real-world phenomena from physics to economics.

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