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Chapter 13: Problem 22

Evaluate \(f_{x}\) and \(f_{u}\) at the point. $$ f(x, y)=x^{2}-3 x y+y^{2} \quad(1,-1) $$

Short Answer

Expert verified
The partial derivative of the function \(f(x, y) = x^{2} - 3xy + y^{2}\) with respect to \(x\) evaluated at the point (1,-1) is 5, and with respect to \(y\) is -5.

Step by step solution

01

Finding Partial Derivative with respect to \(x\)

Firstly the partial derivative of \(f\) with respect to \(x\) is calculated termed as \(f_{x}\). This operation means taking the derivative of \(f\) with respect to \(x\), while treating \(y\) as a constant. Applying the power rule for differentiation (where the derivative of \(x^n\) with respect to \(x\) is \(nx^{n - 1}\)), calculate the result as: \(f_{x} = 2x - 3y\)
02

Evaluating \(f_{x}\) at the Point (1,-1)

After finding \(f_{x}\), the next step is to evaluate it at the point (1, -1). The \(x\) and \(y\) values from the point are substituted into \(f_{x}\) and yields \(f_{x}(1,-1) = 2*1 - 3*(-1) = 2 + 3 = 5\)
03

Finding Partial Derivative with respect to \(y\)

Next, calculate the partial derivative of \(f\) with respect to \(y\), denoted as \(f_{y}\). This operation means taking the derivative of \(f\) with respect to \(y\), while treating \(x\) as constant. So using the power rule for differentiation, get the result as: \(f_{y} = -3x + 2y\)
04

Evaluating \(f_{y}\) at the Point (1,-1)

After finding \(f_{y}\), the next step is to evaluate it at the point (1, -1). By filling in the values for \(x\) and \(y\) into \(f_{y}\), the resulting value will be \(f_{y}(1,-1) = -3*1 + 2*(-1) = -3 - 2 = -5\)

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that deals with the study of change and motion. It explores concepts such as limits, functions, derivatives, integrals, and the relationships between them. This mathematical framework is divided into two main parts: differential calculus and integral calculus. Differential calculus concerns itself with the rate at which quantities change, while integral calculus is about the accumulation of quantities. The exercise provided makes use of differential calculus, specifically the concept of partial derivatives, which is a fundamental tool for understanding how functions behave when they depend on more than one variable.
Power Rule for Differentiation
The power rule is a basic differentiation technique used in calculus. It states that if you have a function of the form f(x) = x^n, where n is any real number, then the derivative of this function with respect to x is f'(x) = nx^{n-1}. This rule is exceptionally handy because it simplifies the differentiation process, especially for polynomial functions. This rule is applied in our exercise to differentiate terms like x^2 and y^2, ending up with the derivatives 2x and 2y, respectively.
Multivariable Calculus
Multivariable calculus extends the principles of single-variable calculus to functions of several variables. It allows us to study phenomena where multiple factors influence the outcome. A core concept of multivariable calculus is the partial derivative, which is the derivative of a function with respect to one variable while holding the others constant. This concept is crucial in physics, engineering, and economics, where multiple variables often interact. In the exercise example, the function f(x, y) has two variables, x and y, and we find partial derivatives with respect to each variable individually.
Function Evaluation
Function evaluation involves finding the output of a function for specific values of its input variables. It is a fundamental concept not only in calculus but all of mathematics. Evaluating a function can tell us the value of a function at a particular point in its domain. In the context of the exercise, after determining the partial derivatives f_x and f_y, we perform function evaluation by substituting x = 1 and y = -1 into these derivatives to calculate their values at that particular point in the function's domain.

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Most popular questions from this chapter

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{3} \int_{0}^{1}(2 x+6 y) d y d x $$

Sketch the region of integration and evaluate the double integral. $$ \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} y d y d x $$

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