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Find the four second-order partial derivatives. $$f(x, y)=x^{4} y^{3}-x^{2} y^{3}$$

Short Answer

Expert verified
The four second-order partial derivatives are: \(f_{xx} = 12x^2 y^3 - 2y^3\), \(f_{xy} = 12x^3 y^2 - 6xy^2\), \(f_{yy} = 6x^4 y - 6x^2 y\), \(f_{yx} = 12x^3 y^2 - 6xy^2\).

Step by step solution

01

- Find the first-order partial derivatives

Calculate the first-order partial derivatives of the function with respect to both variables, x and y.For x: \[ f_x = \frac{\partial}{\partial x} (x^4 y^3 - x^2 y^3) = 4x^3 y^3 - 2x y^3 \]For y: \[ f_y = \frac{\partial}{\partial y} (x^4 y^3 - x^2 y^3) = 3x^4 y^2 - 3x^2 y^2 \]
02

- Find the second-order partial derivative \( \frac{\partial^2 f}{\partial x^2} \)

Take the partial derivative of the first-order partial derivative with respect to x again.\[ f_{xx} = \frac{\partial}{\partial x}(4x^3 y^3 - 2x y^3) = 12x^2 y^3 - 2y^3 \]
03

- Find the second-order partial derivative \( \frac{\partial^2 f}{\partial x \partial y} \)

Take the partial derivative of the first-order partial derivative with respect to y.\[ f_{xy} = \frac{\partial}{\partial y}(4x^3 y^3 - 2x y^3) = 12x^3 y^2 - 6xy^2 \]
04

- Find the second-order partial derivative \( \frac{\partial^2 f}{\partial y^2} \)

Take the partial derivative of the first-order partial derivative with respect to y again.\[ f_{yy} = \frac{\partial}{\partial y}(3x^4 y^2 - 3x^2 y^2) = 6x^4 y - 6x^2 y \]
05

- Find the second-order partial derivative \( \frac{\partial^2 f}{\partial y \partial x} \)

Take the partial derivative of the first-order partial derivative with respect to x.\[ f_{yx} = \frac{\partial}{\partial x}(3x^4 y^2 - 3x^2 y^2) = 12x^3 y^2 - 6xy^2 \]Note that this derivative is equivalent to \( f_{xy} \), which confirms Clairaut's theorem.

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Key Concepts

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

partial derivatives
Partial derivatives help us understand how a function changes as we vary one variable at a time while keeping the others constant. For a function of two variables, like \(f(x, y) = x^4 y^3 - x^2 y^3\), we can find the partial derivatives with respect to \(x\) and \(y\). This means we treat one variable as a constant and differentiate the function with respect to the other variable. For instance, to find \(f_x\), the partial derivative with respect to \(x\), we differentiate with respect to \(x\) while treating \(y\) as constant. Similarly, to find \(f_y\), we differentiate with respect to \(y\) while treating \(x\) as constant.
These first-order partial derivatives are valuable because they tell us the rate of change of the function in one specific direction. For our function, the first-order partial derivatives are:
  • \( f_x = 4x^3 y^3 - 2x y^3 \)
  • \( f_y = 3x^4 y^2 - 3x^2 y^2 \)
Once we have these, we can explore how to get second-order partial derivatives, which explain how the rate of change itself changes.
Clairaut's theorem
Clairaut's theorem, also known as the symmetry of mixed partial derivatives, tells us that for a function and given that the mixed partial derivatives are continuous, the mixed partial derivatives are equal. This means \( \frac{\text{d}^2f}{\text{d}x\text{d}y} = \frac{\text{d}^2f}{\text{d}y\text{d}x} \). To verify this theorem, we calculate the mixed second-order partial derivatives and check if they are indeed the same.
For our function \( f(x, y) = x^4 y^3 - x^2 y^3 \), the mixed partial derivatives found are:
  • \( f_{xy} = 12x^3 y^2 - 6x y^2 \)
  • \( f_{yx} = 12x^3 y^2 - 6x y^2 \)
Since both partial derivatives \( f_{xy} \) and \( f_{yx} \) are equal, Clairaut's theorem is confirmed for this function. Understanding this property is essential when dealing with more complex multivariable functions, as it ensures consistency and predictability in calculations.
multivariable calculus
Multivariable calculus extends the principles of calculus to functions of several variables. This branch covers concepts like partial derivatives, multiple integrals, and the behavior of functions in higher-dimensional spaces. Functions of several variables can have more complex surfaces and shapes, unlike single-variable functions which are limited to curves. When working with these functions, we need to understand how changes in one variable influence the function while considering other variables' effects as well.
For instance, the function \( f(x, y) = x^4 y^3 - x^2 y^3 \) depends on both \(x\) and \(y\). Observing how it behaves as we change \(x\) or \(y\) alone or simultaneously requires tools like partial derivatives.
We also talk about second-order partial derivatives, which provide insights into the concavity and curvature of the function. Calculating second-order partial derivatives involves taking the derivative of first-order derivatives, giving us a deeper understanding of how the function behaves locally.
differentiation
Differentiation is a core concept in calculus that deals with finding the rate at which a function changes at any point. When we have functions of multiple variables, differentiation isn't limited to one direction. We extend this process through partial differentiation, taking derivatives with respect to one variable at a time while keeping others constant.
For a multivariable function \( f(x, y) = x^4 y^3 - x^2 y^3 \), differentiation helps us break down the function's changes along the \(x\) and \(y\) axes. Through partial derivatives, we found:
  • First-order partial derivatives: \( f_x = 4x^3 y^3 - 2x y^3 \), \( f_y = 3x^4 y^2 - 3x^2 y^2 \)
  • Second-order partial derivatives: \( f_{xx} = 12x^2 y^3 - 2y^3 \), \( f_{xy} = 12x^3 y^2 - 6x y^2 \), \( f_{yy} = 6x^4 y - 6x^2 y \), \( f_{yx} = 12x^3 y^2 - 6x y^2 \)
Differentiation lays the groundwork for analyzing more complex functions, setting up optimizations, and solving practical problems in physics, engineering, and economics. Being comfortable with differentiation is crucial for delving into more advanced calculus topics and applications.

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