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Confirm that the mixed second-order partial derivatives of \(f\) are the same. $$ f(x, y)=\sqrt{x^{2}+y^{2}} $$

Short Answer

Expert verified
The mixed second-order partial derivatives are equal: \(\frac{-xy}{(x^2+y^2)^{3/2}}\).

Step by step solution

01

Identify the Function

Given the function \(f(x, y) = \sqrt{x^2 + y^2}\). We are tasked with confirming whether the mixed second-order partial derivatives \(\frac{\partial^2 f}{\partial x \partial y}\) and \(\frac{\partial^2 f}{\partial y \partial x}\) are the same.
02

Calculate the First Partial Derivatives

First, find the first partial derivatives of \(f\) with respect to \(x\) and \(y\):\[ \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} \]\[ \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2+y^2}} \]
03

Compute the Second Mixed Partial Derivative \(\frac{\partial^2 f}{\partial x \partial y}\)

Differentiate \( \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} \) with respect to \(y\):\[ \frac{\partial}{\partial y} \left( \frac{x}{\sqrt{x^2+y^2}} \right) = \frac{-xy}{(x^2+y^2)^{3/2}} \]
04

Compute the Second Mixed Partial Derivative \(\frac{\partial^2 f}{\partial y \partial x}\)

Differentiate \( \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2+y^2}} \) with respect to \(x\):\[ \frac{\partial}{\partial x} \left( \frac{y}{\sqrt{x^2+y^2}} \right) = \frac{-xy}{(x^2+y^2)^{3/2}} \]
05

Compare the Mixed Second Partial Derivatives

Both mixed partial derivatives computed in Steps 3 and 4 are equal: \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = \frac{-xy}{(x^2+y^2)^{3/2}} \] This confirms that the mixed second-order partial derivatives of \(f\) are the same.

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Key Concepts

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

Mixed Second-Order Partial Derivatives
Mixed second-order partial derivatives are derivatives which involve partial differentiation with respect to two different variables in succession. In other words, it is taking the derivative of a function in two different ways such as \( \frac{\partial^2 f}{\partial x \partial y} \) and \( \frac{\partial^2 f}{\partial y \partial x} \). These derivatives often appear when analyzing surfaces or multivariable functions.

An important property of mixed partial derivatives is Clairaut's theorem. This theorem states that if the function \( f(x, y) \) is smooth (specifically, if \( f \) has continuous second derivatives), then: \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \).

In essence, as long as the function is well-behaved, the order in which you take mixed derivatives does not affect the outcome. This property was confirmed in the exercise, which showed that the derivatives are indeed equal.
Function Differentiation
Differentiation is a critical concept in calculus, dealing with finding the rate at which a function changes at any point. When performing partial differentiation, you're focusing on one variable at a time while treating the others as constants.

For a function like \( f(x, y) = \sqrt{x^2 + y^2} \), you can calculate partial derivatives by treating \( x \) and \( y \) one at a time as you proceed through differentiation:
  • Finding the partial derivative with respect to \( x \), gives \( \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} \).
  • Similarly, deriving with respect to \( y \) yields \( \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2}} \).
The process of differentiation gets us information about the slope or rate of change of our function, crucial for understanding how changes in one variable affect the function.
Calculus Concepts
Calculus is the mathematical study of continuous change, with differentiation and integration being its core concepts. Understanding these methods offers valuable insights into how calculus can be applied to various problems.

Differentiation is primarily about finding slopes and rates of changes, offering a peek into how a function behaves as its variables move. For instance, the differentiation process uncovers highly detailed behavior of our given function \( f(x, y) = \sqrt{x^2 + y^2} \), such as how its value changes when either \( x \) or \( y \) changes.

  • Partial derivatives are about focusing on changes in one specific direction, holding the other variables constant.
  • Mixed second-order derivatives give insights into interactions between variables, showing how changes in one dimension relate to another.
Mastering these concepts empowers you not just to perform calculations, but to genuinely understand the underlying principles guiding the motion and change of complex systems.

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Most popular questions from this chapter

Find \(f_{x}, f_{y},\) and \(f_{z}\) $$ f(x, y, z)=\tan ^{-1}\left(\frac{1}{x y^{2} z^{3}}\right) $$

Express the derivatives in "subscript" notation. $$\begin{array}{llll}{\text { (a) } \frac{\partial^{3} f}{\partial y^{2} \partial x}} & {\text { (b) } \frac{\partial^{4} f}{\partial x^{4}}} & {\text { (c) } \frac{\partial^{4} f}{\partial y^{2} \partial x^{2}}} & {\text { (d) } \frac{\partial^{5} f}{\partial x^{2} \partial y^{3}}}\end{array}$$

A common problem in experimental work is to obtain a mathematical relationship \(y=f(x)\) between two variables \(x\) and \(y\) by "fitting" a curve to points in the plane that correspond to experimentally determined values of \(x\) and \(y,\) say $$ \left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right) $$ The curve \(y=f(x)\) is called a mathematical model of the data. The general form of the function \(f\) is commonly determined by some underlying physical principle, but sometimes it is just determined by the pattern of the data. We are concerned with fitting a straight line \(y=m x+b\) to data. Usually, the data will not lie on a line (possibly due to experimental error or variations in experimental conditions), so the problem is to find a line that fits the data "best" according to some criterion. One criterion for selecting the line of best fit is to choose \(m\) and \(b\) to minimize the function $$ g(m, b)=\sum_{i=1}^{n}\left(m x_{i}+b-y_{i}\right)^{2} $$ This is called the method of least squares, and the resulting line is called the regression line or the least squares line of best fit. Geometrically, \(\left|m x_{i}+b-y_{i}\right|\) is the vertical distance between the data point \(\left(x_{i}, y_{i}\right)\) and the line \(y=m x+b\) These vertical distances are called the residuals of the data points, so the effect of minimizing \(g(m, b)\) is to minimize the sum of the squares of the residuals. In these exercises, we will derive a formula for the regression line. The purpose of this exercise is to find the values of \(m\) and \(b\) that produce the regression line. (a) To minimize \(g(m, b),\) we start by finding values of \(m\) and \(b\) such that \(\partial g / \partial m=0\) and \(\partial g / \partial b=0 .\) Show that these equations are satisfied if \(m\) and \(b\) satisfy the conditions $$ \left(\sum_{i=1}^{n} x_{i}^{2}\right) m+\left(\sum_{i=1}^{n} x_{i}\right) b=\sum_{i=1}^{n} x_{i} y_{i} $$ \(\left(\sum_{i=1}^{n} x_{i}\right) m+n b=\sum_{i=1}^{n} y_{i}\) (b) Let \(\bar{x}=\left(x_{1}+x_{2}+\cdots+x_{n}\right) / n\) denote the arithmetic average of \(x_{1}, x_{2}, \ldots, x_{n} .\) Use the fact that $$ \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} \geq 0 $$ to show that $$ n\left(\sum_{i=1}^{n} x_{i}^{2}\right)-\left(\sum_{i=1}^{n} x_{i}\right)^{2} \geq 0 $$ with equality if and only if all the \(x_{i}\) 's are the same. (c) Assuming that not all the \(x_{i}\) 's are the same, prove that the equations in part (a) have the unique solution $$ \begin{aligned} m=& \frac{n \sum_{i=1}^{n} x_{i} y_{i}-\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} y_{i}}{n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2}} \\ b=& \frac{1}{n}\left(\sum_{i=1}^{n} y_{i}-m \sum_{i=1}^{n} x_{i}\right) \end{aligned} $$ [Note: We have shown that \(g\) has a critical point at these values of \(m\) and \(b\). In the next exercise we will show that \(g\) has an absolute minimum at this critical point. Accepting this to be so, we have shown that the line \(y=m x+b\) is the regression line for these values of \(m \text { and } b .]\)

Solve using Lagrange multipliers. Find the point on the plane \(x+2 y+z=1\) that is closest to the origin.

Confirm that the mixed second-order partial derivatives of \(f\) are the same. $$ f(x, y)=(x-y) /(x+y) $$

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