/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Evaluate \(f_{x}\) and \(f_{y}\)... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate \(f_{x}\) and \(f_{y}\) at the given point. $$ f(x, y)=\frac{x y}{x-y}, \quad(2,-2) $$

Short Answer

Expert verified
The values of the partial derivatives \(f_x\) and \(f_y\) at the point (2, -2) are 1 and -1 respectively.

Step by step solution

01

Compute Partial Derivative with Respect to x

The partial derivative of \(f(x, y)\) with respect to \(x\) is given by the formula: \[f_x = y(x(x-y)' - (x-y)x') / (x-y)^2 \] Simplifying this by keeping \(y\) as a constant and simplifying afterwards, we get: \(f_x = \frac{-2xy}{(x-y)^{2}}\)
02

Compute Partial Derivative with Respect to y

The partial derivative of \(f(x, y)\) with respect to \(y\) is given by the formula: \[f_y = x(x(x-y)' - (x-y)y') / (x-y)^2 \] Simplifying this by keeping \(x\) as a constant and simplifying afterwards, we get: \(f_y=\frac{2xy}{(x-y)^{2}}\)
03

Evaluate the Partial Derivatives at the Given Point

Evaluate \(f_x\) and \(f_y\) at the point (2, -2): \[ f_x(2,-2) = \frac{-2*2*-2}{(2--2)^{2}} = 1 \] \[f_y(2,-2) = \frac{2*2*-2}{(2--2)^{2}} = -1 \]

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Key Concepts

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

Multivariable Calculus
In the world of mathematics, multivariable calculus extends the concepts of one-variable calculus to functions that have multiple inputs, like the functions of x and y in our exercise. When dealing with functions like f(x, y) that depend on more than one variable, partial derivatives become an essential tool.

A partial derivative of a function is a measure of how the function changes as one of its input variables changes, while all other input variables are held constant. In the problem given, we are examining how the function f(x, y) changes with respect to x and y independently. To find these derivatives, we apply the same principles of differentiation used in single-variable calculus but only focus on the variable of interest, treating all other variables as constants.
Chain Rule
The chain rule is a pivotal theorem in calculus that facilitates the differentiation of composite functions. When applying the chain rule in multivariable contexts, it's essential to consider how each variable interacts with the others. In the case of the function f(x, y) = x y / (x - y), we treat y as a constant when differentiating with respect to x and vice versa.

This approach allows us to break down complex expressions into simpler parts, which we can then differentiate independently before reassembling them to find the overall derivative. Understanding and applying the chain rule effectively ensures that we can handle the intricacies of partial derivative computation in functions that might initially seem challenging to differentiate.
Derivative Evaluation
Once we've computed the partial derivatives of a function, the next step is to evaluate them at a specific point. This evaluation is crucial as it gives us insight into the function's behavior at that precise location. In our current exercise, we're looking to evaluate f_x and f_y at the point (2, -2).

Executing this step correctly involves substituting the x and y values from the point of interest into the formulas we derived for the partial derivatives. The results tell us the slope of the tangent line to the function's surface in the direction of the x and y-axis, respectively, at the given point. The accurate evaluation of derivatives at specific points is a fundamental skill in calculus, which underpins many practical applications, like predicting the behavior of physical systems or optimizing engineering designs.

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