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Chapter 14: Problem 10

Find \(\partial f / \partial x\) and \(\partial f / \partial y\). \(f(x, y)=x /\left(x^{2}+y^{2}\right)\)

Short Answer

Expert verified
\( \partial f/\partial x = \frac{y^2 - x^2}{(x^2 + y^2)^2} \) and \( \partial f/\partial y = \frac{-2xy}{(x^2 + y^2)^2} \).

Step by step solution

01

Understanding Partial Derivatives

To find the partial derivatives \( \partial f/\partial x \) and \( \partial f/\partial y \), we need to differentiate the given function \( f(x, y) = \frac{x}{x^2 + y^2} \) with respect to \( x \) and \( y \) separately, keeping the other variable constant.
02

Differentiate with Respect to x

To find \( \partial f/\partial x \), consider \( y \) as a constant and differentiate. We use the quotient rule: if \( u(x) = x \) and \( v(x) = x^2 + y^2 \), then \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \). Here, \( u' = 1 \) and \( v' = 2x \). Thus,\[\frac{\partial f}{\partial x} = \frac{(1)(x^2 + y^2) - (x)(2x)}{(x^2 + y^2)^2} = \frac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2}.\]
03

Differentiate with Respect to y

For \( \partial f/\partial y \), consider \( x \) as a constant and differentiate. Again, apply the quotient rule. Here, \( u = x \) as before, so \( u' = 0 \), and the only change is \( v'(y) = 2y \). Thus,\[\frac{\partial f}{\partial y} = \frac{(0)(x^2 + y^2) - (x)(2y)}{(x^2 + y^2)^2} = \frac{-2xy}{(x^2 + y^2)^2}.\]

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Key Concepts

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

Quotient Rule
The quotient rule is a fundamental technique in calculus, particularly useful when differentiating quotients of two functions. When dealing with two functions, say \( u(x) \) and \( v(x) \), both dependent on \( x \), the rule helps determine the derivative of the division of these two functions. The quotient rule states that if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then its derivative \( f'(x) \) is computed as follows:
  • First, identify \( u \) and \( v \).
  • Then, calculate \( u' \) and \( v' \), the derivatives of \( u \) and \( v \) with respect to \( x \).
  • Finally, apply the formula: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \). This formula tells you how the quotient of two functions changes as \( x \) changes.
Using the quotient rule becomes second nature with practice. It's crucial where both components of a function are intricate or individually dependent on different variables, especially in multivariable functions.
Multivariable Calculus
Multivariable calculus expands the concepts of calculus from single-variable functions to functions involving several variables. This area of mathematics is incredibly useful in real-world applications, where problems involve more than one changing component.
  • In multivariable calculus, we deal with functions like \( f(x, y) \) or \( f(x, y, z) \).
  • The main idea is that changes in the outcome can be influenced by changes in more than one direction.
  • This branch includes techniques such as partial derivatives, which measure how a function changes as one of the variables changes, keeping others constant.
Understanding multivariable calculus is essential for fields like physics, engineering, and economics, where variables are often interdependent. It helps in modeling and solving complex problems with multiple factors in play.
Differentiation
Differentiation is the process of finding the derivative of a function, which is a way of measuring how the function's output changes as its input changes. In simpler terms, it tells you the slope of the function at any given point.
  • For single-variable functions, it's straightforward as we only have one direction of variability.
  • When we move to functions of several variables, partial differentiation is employed. This technique finds the rate of change with respect to one variable while keeping others constant.
  • Each function has its specific rules and techniques, such as the chain rule or product rule, that might be utilized in finding derivatives.
In the case of the exercise, we performed partial differentiation, which is essential for thoroughly analyzing functions that depend on more than one variable. Differentiation not only helps in identifying the behavior of functions but also in optimizing them, which is central to many scientific and engineering disciplines.

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Most popular questions from this chapter

Minimum distance to the origin Find the point closest to the origin on the line of intersection of the planes \(y+2 z=12\) and \(x+y=6 .\)

The Sandwich Theorem for functions of two variables states that if \(g(x, y) \leq f(x, y) \leq h(x, y)\) for all \((x, y) \neq\left(x_{0}, y_{0}\right)\) in a disk centered at \(\left(x_{0}, y_{0}\right)\) and if \(g\) and \(h\) have the same finite limit \(L\) as \((x, y) \rightarrow\left(x_{0}, y_{0}\right),\) then $$ \lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=L $$ Use this result to support your answers to the questions in Exercises \(45-48 .\) Does knowing that $$ 2|x y|-\frac{x^{2} y^{2}}{6}<4-4 \cos \sqrt{|x y|}<2|x y| $$ tell you anything about $$ \lim _{(x, y) \rightarrow(0,0)} \frac{4-4 \cos \sqrt{|x y|}}{|x y|} ? $$ Give reasons for your answer.

Extrema on a circle of intersection Find the extreme values of the function \(f(x, y, z)=x y+z^{2}\) on the circle in which the plane \(y-x=0\) intersects the sphere \(x^{2}+y^{2}+z^{2}=4\)

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Each of Exercises \(63-66\) gives a function \(f(x, y, z)\) and a positive number \(\epsilon .\) In each exercise, show that there exists a \(\delta>0\) such that for all \((x, y, z)\) , $$ \quad \sqrt{x^{2}+y^{2}+z^{2}}<\delta \Rightarrow|f(x, y, z)-f(0,0,0)|<\epsilon $$ $$ f(x, y, z)=x y z, \quad \epsilon=0.008 $$
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