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Second partial derivatives Find the four second partial derivatives of the following functions. $$f(x, y)=2 x^{5} y^{2}+x^{2} y$$

Short Answer

Expert verified
Question: Find the four second partial derivatives of the function $$f(x, y)=2 x^{5} y^{2}+x^{2} y$$. Answer: The four second partial derivatives are: 1. $$f_{xx} = 40x^{3}y^{2} + 2y$$ 2. $$f_{yy} = 4x^{5}$$ 3. $$f_{xy} = 20x^{4}y + 2x$$ 4. $$f_{yx} = 20x^{4}y + 2x$$

Step by step solution

01

Find the first partial derivatives with respect to x and y

To find the first partial derivatives, we differentiate the given function with respect to each variable, treating the other variable as a constant: First Partial derivative with respect to x (denoted as \(f_x\)): $$f_x = \frac{\partial}{\partial x} (2x^{5}y^{2} + x^{2}y) = 10x^{4}y^{2} + 2xy$$ First Partial derivative with respect to y (denoted as \(f_y\)): $$f_y = \frac{\partial}{\partial y} (2x^{5}y^{2} + x^{2}y) = 4x^{5}y + x^{2}$$
02

Find second partial derivatives with respect to x (denoted as \(f_{xx}\)) and y (denoted as \(f_{yy}\))

Now, we differentiate the first partial derivatives with respect to their respective variables once more: Second Partial derivative with respect to x (\(f_{xx}\)): $$f_{xx} = \frac{\partial^2}{\partial x^2} (2x^{5}y^{2} + x^{2}y) = \frac{\partial}{\partial x}(10x^{4}y^{2} + 2xy) = 40x^{3}y^{2} + 2y$$ Second Partial derivative with respect to y (\(f_{yy}\)): $$f_{yy} = \frac{\partial^2}{\partial y^2} (2x^{5}y^{2} + x^{2}y) = \frac{\partial}{\partial y}(4x^{5}y + x^{2}) = 4x^{5}$$
03

Find mixed second partial derivatives with respect to x and y (denoted as \(f_{xy}\) and \(f_{yx}\))

Finally, we need to find the mixed second partial derivatives. We do this by differentiating the first partial derivatives with respect to the other variable: Mixed second partial derivative with respect to x and y (\(f_{xy}\)): $$f_{xy} = \frac{\partial^2}{\partial x \partial y} (2x^{5}y^{2} + x^{2}y) = \frac{\partial}{\partial x}(4x^{5}y + x^{2}) = 20x^{4}y + 2x$$ Mixed second partial derivative with respect to y and x (\(f_{yx}\)): $$f_{yx} = \frac{\partial^2}{\partial y \partial x} (2x^{5}y^{2} + x^{2}y) = \frac{\partial}{\partial y}(10x^{4}y^{2} + 2xy) = 20x^{4}y + 2x$$ The four second partial derivatives are: $$f_{xx} = 40x^{3}y^{2} + 2y,$$ $$f_{yy} = 4x^{5},$$ $$f_{xy} = 20x^{4}y + 2x,$$ $$f_{yx} = 20x^{4}y + 2x.$$

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Key Concepts

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

Partial Derivative
A partial derivative represents how a function changes as one variable is altered while keeping all other variables constant. In the realm of multivariable calculus, functions depend on more than one variable, such as f(x, y), which depends on both x and y. When finding a partial derivative, say with respect to x, you treat y as a constant and differentiate f(x, y) only with respect to x. This is designated as f_x or \( \frac{\partial f}{\partial x} \). In the given exercise, the first partial derivative of the function with respect to x is calculated, while y is considered a constant.

Understanding partial derivatives is crucial as it allows us to analyze functions with multiple variables individually and understand the behavior of the function with respect to each variable. It forms the foundation of more complex concepts such as gradient, divergence, curl, and optimization in higher dimensions.
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions with more than one variable. It deals with functions like f(x, y) and how to perform operations such as differentiation and integration in this context. The exercise involves second partial derivatives, which is an important concept within this field. For example, while the first derivative can tell you the rate of change at a point, the second partial derivatives give information about the curvature and concavity of the function in different directions.

Applications of multivariable calculus are abundant in fields ranging from physics and engineering to economics and biology. It allows for the analysis of systems where multiple factors are simultaneously in flux, necessitating a more complex and nuanced approach than single-variable calculus.
Mixed Partial Derivatives
In the world of multivariable functions, mixed partial derivatives are a particular type of second partial derivatives. They measure the rate at which the partial derivative of a function with respect to one variable changes as another variable is changed. Essentially, it is the partial derivative of a partial derivative, where the variables of differentiation are distinct. For instance, \( f_{xy} \) represents the derivative of \( f_x \) concerning y.

The equality of mixed partial derivatives, known as Clairaut's Theorem, states that for a function with continuous second partial derivatives, \( f_{xy} = f_{yx} \). This is illustrated in the provided exercise where \( f_{xy} \) and \( f_{yx} \) are computed to be the same. Understanding mixed partial derivatives is important for higher-level applications, such as when you're dealing with higher-order differential equations or optimizing functions of multiple variables.
Differentiation
Differentiation is a key operation in calculus that measures how a function changes as its input changes. It's all about finding that instant rate of change - formally, the derivative. In single-variable calculus, this is straightforward, but differentiation takes on greater complexity in multivariable calculus. Here, apart from calculating first-order derivatives, we also compute second-order derivatives, such as in the case of this exercise.

The process involves applying rules of differentiation, such as the power rule, product rule, and chain rule, sometimes repeatedly, to find both regular and mixed second partial derivatives. Differentiation not only helps in finding the slope of the tangent line at a given point on the graph but also in many application areas such as finding minimums and maximums, solving optimization problems, and modeling various physical phenomena.

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Most popular questions from this chapter

Batting averages Batting averages in bascball are defined by \(A=x / y,\) where \(x \geq 0\) is the total number of hits and \(y>0\) is the total number of at- bats. Treat \(x\) and \(y\) as positive real numbers and note that \(0 \leq A \leq 1\) a. Use differentials to estimate the change in the batting average if the number of hits increases from 60 to 62 and the number of at-bats increases from 175 to 180 . b. If a batter currently has a batting average of \(A=0.350,\) does the average decrease more if the batter fails to get a hit than it increases if the batter gets a hit? c. Does the answer to part (b) depend on the current batting average? Explain.

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