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Chapter 6: Problem 7

$$\text { Find } f_{x}, f_{y}, f_{x}(-2,4), \text { and } f_{y}(4,-3)$$ $$f(x, y)=\sqrt{x^{2}+y^{2}}$$

Short Answer

Expert verified
f_x = \frac{x}{\sqrt{x^2 + y^2}}, f_y = \frac{y}{\sqrt{x^2 + y^2}}, f_x(-2, 4) = -\frac{1}{\sqrt{5}}, f_y(4, -3) = -\frac{3}{5}

Step by step solution

01

- Find the partial derivative with respect to x, denoted as f_x

To find the partial derivative of the function with respect to x, apply the chain rule. The function is given by \( f(x, y) = \sqrt{x^{2}+y^{2}} \). The partial derivative of this function with respect to x is: \[ f_x = \frac{\partial}{\partial x} \sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}} \]
02

- Find the partial derivative with respect to y, denoted as f_y

Similarly, apply the chain rule to find the partial derivative with respect to y. The partial derivative of the function with respect to y is: \[ f_y = \frac{\partial}{\partial y} \sqrt{x^2 + y^2} = \frac{y}{\sqrt{x^2 + y^2}} \]
03

- Evaluate f_x at (-2, 4)

Substitute \ x = -2 \ and \ y = 4 \ into the partial derivative \ f_x \ to evaluate at the given point: \[ f_x(-2, 4) = \frac{-2}{\sqrt{(-2)^2 + 4^2}} = \frac{-2}{\sqrt{4 + 16}} = \frac{-2}{\sqrt{20}} = \frac{-2}{2\sqrt{5}} = -\frac{1}{\sqrt{5}} \]
04

- Evaluate f_y at (4, -3)

Substitute \ x = 4 \ and \ y = -3 \ into the partial derivative \ f_y \ to evaluate at the given point: \[ f_y(4, -3) = \frac{-3}{\sqrt{4^2 + (-3)^2}} = \frac{-3}{\sqrt{16 + 9}} = \frac{-3}{\sqrt{25}} = \frac{-3}{5} \]

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Key Concepts

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

Chain Rule in Calculus
The chain rule is a fundamental concept in calculus that links the derivative of a composite function to the derivatives of its component functions. When we are dealing with functions of several variables like in our given function, the chain rule helps us take the derivative with respect to one of those variables while keeping the others constant. For example, in our exercise, we use the chain rule to find the partial derivatives of \( f(x, y) = \sqrt{x^{2}+y^{2}} \) with respect to x and y. By recognizing the function as a composition of inner and outer functions, the chain rule allows us to differentiate efficiently.

Benefits of using the chain rule:
  • Simplifies complex differentiation problems.
  • Applicable to functions of multiple variables.
  • Ensures that composite functions are handled correctly.
Multivariable Calculus
Multivariable calculus is an extension of single-variable calculus to functions of several variables. It deals with functions that depend on more than one variable and includes the study of partial derivatives, multiple integrals, and more. In our exercise, \( f(x, y) \) is a function of two variables, x and y. In multivariable calculus, we study how changes in these variables affect the output of the function.

Key topics within multivariable calculus:
  • Partial Derivatives: The rate of change of a function with respect to one variable, holding others constant.
  • Gradient: A vector consisting of partial derivatives indicating the direction of the steepest ascent.
  • Multiple Integrals: Extending the concept of integration to functions of more than one variable.
In our problem, we specifically look at partial derivatives, finding the rate of change of \( f(x, y) \) with respect to x and y.
Partial Derivative Evaluation
Evaluating partial derivatives is a crucial skill in multivariable calculus. A partial derivative of a function with respect to one of its variables shows how the function changes as that particular variable changes, keeping the others fixed. In our exercise, we start by finding the partial derivatives \( f_x \) and \( f_y \), and then evaluate them at specific points.

Steps to evaluate partial derivatives:
  • Calculate the partial derivative: Use rules such as the chain rule to find expressions for \( f_x \) and \( f_y \).
  • Substitute given points: Plug in the given values for x and y into the partial derivatives to get numerical results.
To find \( f_x \) at (-2, 4), we substitute x = -2 and y = 4 into \( \frac{x}{\sqrt{x^2 + y^2}} \), resulting in \ -\frac{1}{\sqrt{5}} \. Similarly, to find \( f_y \) at (4, -3), we substitute x = 4 and y = -3 into \ \frac{y}{\sqrt{x^2 + y^2}} \, yielding \ -\frac{3}{5} \.

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Most popular questions from this chapter

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