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Chapter 12: Problem 40

Verify that \(f_{x y}=f_{y x}\) for the following functions. $$f(x, y)=x e^{y}$$

Short Answer

Expert verified
Question: Verify if the mixed partial derivatives of the function \(f(x, y) = xe^{y}\) are equal. Answer: Yes, the mixed partial derivatives of the function are equal, as \(f_{xy} = e^y = f_{yx}\).

Step by step solution

01

Calculate the partial derivative of f(x,y) with respect to x, \(f_x\)

To calculate the partial derivative of \(f(x, y) = x e^{y}\) with respect to \(x\), we keep \(y\) constant and differentiate \(f\) with respect to \(x\): $$f_{x}(x, y) = \frac{\partial}{\partial x}(x e^{y}) = e^{y}$$
02

Calculate the partial derivative of f(x,y) with respect to y, \(f_y\)

To calculate the partial derivative of \(f(x, y) = x e^{y}\) with respect to \(y\), we keep \(x\) constant and differentiate \(f\) with respect to \(y\): $$f_{y}(x, y) = \frac{\partial}{\partial y}(x e^{y}) = x e^{y}$$
03

Calculate the mixed partial derivative of f(x,y) with respect to x then y, \(f_{xy}\)

To calculate the mixed partial derivative of \(f(x, y) = x e^{y}\) with respect to \(x\) then \(y\), we take the partial derivative of \(f_x\) with respect to \(y\): $$f_{x y}(x, y) = \frac{\partial}{\partial y}(f_{x}(x, y)) = \frac{\partial}{\partial y}(e^{y}) = e^{y}$$
04

Calculate the mixed partial derivative of f(x,y) with respect to y then x, \(f_{yx}\)

To calculate the mixed partial derivative of \(f(x, y) = x e^{y}\) with respect to \(y\) then \(x\), we take the partial derivative of \(f_y\) with respect to \(x\): $$f_{y x}(x, y) = \frac{\partial}{\partial x}(f_{y}(x, y)) = \frac{\partial}{\partial x}(x e^{y}) = e^{y}$$
05

Compare \(f_{xy}\) and \(f_{yx}\)

Now, we compare the mixed partial derivatives \(f_{xy}\) and \(f_{yx}\) to verify if they are equal: $$f_{x y}(x, y) = e^{y} = f_{y x}(x, y)$$ Since \(f_{xy} = f_{yx}\), the mixed partial derivatives are equal.

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Key Concepts

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

Partial Derivative
In multivariable calculus, the partial derivative is a fundamental concept that deals with the rate of change of a function with respect to one of several variables while holding the others constant. Unlike ordinary derivatives for single-variable functions, a partial derivative focuses on changes along just one dimension of the input space.

For example, consider a function defined for two variables, say temperature and pressure, within a physical system. If you want to understand how temperature affects the system while keeping the pressure fixed, you would take the partial derivative of the function with respect to temperature. This is denoted as \( \frac{\partial}{\partial T} \) if 'T' represents temperature. In our exercise \( f(x, y) = x e^{y} \), the function describes how 'x' melds with an exponential function of 'y'. The partial derivatives \( f_x \) and \( f_y \) reveal how the function changes as 'x' varies with a fixed 'y' and vice versa, respectively.
Multivariable Calculus
Diving deeper into the realm of calculus with multiple variables, we encounter various types of derivatives that express the rates of change in multi-dimensional settings. Multivariable calculus is the extension of calculus in one variable to calculus with functions of several variables: the differentiation and integration of functions involving multiple variables, versus just one.

The beauty of multivariable calculus lies in its ability to analyze dynamic systems where inputs and outputs are not merely linear but can operate on planes and in spaces that have more than one dimension. In practice, this field is vital for various science and engineering disciplines that deal with phenomena having several degrees of freedom. In our exercise, differentiating \( f(x, y) = x e^{y} \) once returns first-order partial derivatives. Further differentiation leads to mixed partial derivatives like \( f_{xy} \) and \( f_{yx} \) which tell us how the function's rate of change itself changes, concerning one variable and then another.
Clairaut鈥檚 Theorem
Now to connect the dots between mixed partial derivatives, we summon Clairaut's theorem 鈥� also known as the symmetry of second derivatives. This theorem asserts that if we have a function like \( f(x, y) \) that is sufficiently smooth (technically, the function and its partial derivatives up to the second order are continuous), then the mixed partial derivatives are equal when calculated in either order; that is, \( f_{xy} = f_{yx} \).

In application, Clairaut鈥檚 theorem offers a significant simplification in tackling problems as it ensures that the order of differentiation does not matter for these mixed partials. The exercise provides a neat demonstration of Clairaut's theorem. As calculated, both \( f_{xy}(x, y) \) and \( f_{yx}(x, y) \) resulted in \( e^{y} \) proving their equality and confirming that the original function satisfies the conditions of Clairaut鈥檚 theorem.

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Most popular questions from this chapter

The pressure, temperature, and volume of an ideal gas are related by \(P V=k T,\) where \(k>0\) is a constant. Any two of the variables may be considered independent, which determines the third variable. a. Use implicit differentiation to compute the partial derivatives \(\frac{\partial P}{\partial V} \frac{\partial T}{\partial P},\) and \(\frac{\partial V}{\partial T}\) b. Show that \(\frac{\partial P}{\partial V} \frac{\partial T}{\partial P} \frac{\partial V}{\partial T}=-1 .\) (See Exercise 67 for a generalization.)

Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}-4 y^{2}+x y ; R=\left\\{(x, y): 4 x^{2}+9 y^{2} \leq 36\right\\}$$

An important derivative operation in many applications is called the Laplacian; in Cartesian coordinates, for \(z=f(x, y),\) the Laplacian is \(z_{x x}+z_{y y} .\) Determine the Laplacian in polar coordinates using the following steps. a. Begin with \(z=g(r, \theta)\) and write \(z_{x}\) and \(z_{y}\) in terms of polar coordinates (see Exercise 64). b. Use the Chain Rule to find \(z_{x x}=\frac{\partial}{\partial x}\left(z_{x}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. c. Use the Chain Rule to find \(z_{y y}=\frac{d}{\partial y}\left(z_{y}\right) .\) There should be two major terms, which, when expanded and simplified, result in five terms. d. Combine parts (b) and (c) to show that $$z_{x x}+z_{y y}=z_{r r}+\frac{1}{r} z_{r}+\frac{1}{r^{2}} z_{\theta \theta}$$

Use the definition of the gradient (in two or three dimensions), assume that \(f\) and \(g\) are differentiable functions on \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3}\), and let \(c\) be a constant. Prove the following gradient rules. a. Constants Rule: \(\nabla(c f)=c \nabla f\) b. Sum Rule: \(\nabla(f+g)=\nabla f+\nabla g\) c. Product Rule: \(\nabla(f g)=(\nabla f) g+f \nabla g\) d. Quotient Rule: \(\nabla\left(\frac{f}{g}\right)=\frac{g \nabla f-f \nabla g}{g^{2}}\) e. Chain Rule: \(\nabla(f \circ g)=f^{\prime}(g) \nabla g,\) where \(f\) is a function of one variable

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Generalize the procedure in Exercise 70 by assuming that \(n\) data points \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) are given. Write the function \(E(m, b)\) (summation notation allows for a more compact calculation). Show that the coefficients of the best-fit line are $$ \begin{aligned} m &=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}} \text { and } \\ b &=\frac{1}{n}\left(\sum y_{k}-m \Sigma x_{k}\right) \end{aligned}, $$ where all sums run from \(k=1\) to \(k=n\).
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