Chapter 14: Problem 59
Verify that the conclusion of Clairaut's Theorem holds, that is, \(u_{x y}=u_{y x}\) $$ u=x^{4} y^{3}-y^{4} $$
Short Answer
Expert verified
The equality \(u_{xy} = u_{yx} = 12x^3 y^2\) holds.
Step by step solution
01
Calculate Partial Derivative \(u_x\)
To find the partial derivative with respect to \(x\), differentiate \(u = x^{4} y^{3} - y^{4}\) treating \(y\) as a constant.\[ u_x = \frac{\partial}{\partial x}(x^4 y^3 - y^4) = 4x^3 y^3 \]
02
Calculate Partial Derivative \(u_{xy}\)
Now, differentiate \(u_x = 4x^3 y^3\) with respect to \(y\).\[ u_{xy} = \frac{\partial}{\partial y}(4x^3 y^3) = 4x^3 \cdot 3y^2 = 12x^3 y^2 \]
03
Calculate Partial Derivative \(u_y\)
Now calculate the partial derivative with respect to \(y\), treating \(x\) as a constant.\[ u_y = \frac{\partial}{\partial y}(x^4 y^3 - y^4) = 3x^4 y^2 - 4y^3 \]
04
Calculate Partial Derivative \(u_{yx}\)
Differentiate \(u_y = 3x^4 y^2 - 4y^3\) with respect to \(x\).\[ u_{yx} = \frac{\partial}{\partial x}(3x^4 y^2 - 4y^3) = 3 \cdot 4x^3 y^2 = 12x^3 y^2 \]
05
Verify Equality \(u_{xy} = u_{yx}\)
The final step is to verify if \(u_{xy}\) is equal to \(u_{yx}\).Since both \(u_{xy} = 12x^3 y^2\) and \(u_{yx} = 12x^3 y^2\) are indeed equal,\[ u_{xy} = u_{yx} \] This confirms that the conclusion of Clairaut's Theorem holds true.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a fundamental concept in calculus, particularly when dealing with functions that have more than one variable. When we talk about partial derivatives, we are essentially concerned with how a function changes as one specific variable is varied while keeping all other variables constant. This is different from regular derivatives, which consider the change of a function in terms of a single variable.
- For example, consider a function \( u = f(x, y) \), which depends on two variables \( x \) and \( y \).
- To find the partial derivative with respect to \( x \) (denoted as \( u_x \)), you treat \( y \) as a constant and differentiate the function with respect to \( x \).
- Similarly, to find the partial derivative concerning \( y \) (denoted as \( u_y \)), you treat \( x \) as a constant and differentiate the function concerning \( y \).
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions of several variables. These functions are often represented as \( f(x, y, z) \) or more generally \( f(x_1, x_2, \, ... \, ,x_n) \). In this branch of mathematics, we explore concepts such as gradients, partial derivatives, double integrals, and more.
- Functions in multivariable calculus can describe surfaces and regions in higher dimensions, providing powerful models for real-world phenomena.
- Many physical systems, such as temperature distribution over an area, can be analyzed with multivariable calculus, enabling the modeling of complex biological, chemical, and physical systems.
- Even simple functions can lead to fascinating geometric and analytic insights in this realm.
Mixed Partial Derivatives
Mixed partial derivatives involve taking partial derivatives of a function with respect to two or more different variables in succession. For instance, when you have a function \( u = f(x, y) \), you might seek to calculate \( \frac{\partial^2 u}{\partial x \partial y} \) or \( \frac{\partial^2 u}{\partial y \partial x} \).
- Mathematically, we first differentiate with respect to one variable, and then differentiate the resulting function with respect to another variable.
- A significant result of mixed partials is Clairaut's Theorem, which states that if the mixed partial derivatives are continuous, their order can be switched, meaning \( u_{xy} = u_{yx} \).