/*! 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 5 Find \(\partial f / \partial x\)... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 5

Find \(\partial f / \partial x\) and \(\partial f / \partial y\). \(f(x, y)=(x y-1)^{2}\)

Short Answer

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

Step by step solution

01

Understand the Problem

The function given is \( f(x, y) = (xy - 1)^2 \). To solve this problem, we need to find the partial derivatives of this function with respect to both \( x \) and \( y \).
02

Apply the Chain Rule for \( \frac{\partial f}{\partial x} \)

To find the partial derivative \( \frac{\partial f}{\partial x} \), apply the chain rule. Express \( f \) as: \( u = xy - 1 \) and \( f = u^2 \). First, find \( \frac{\partial u}{\partial x} = y \), then apply the chain rule: \[ \frac{\partial f}{\partial x} = 2u \cdot \frac{\partial u}{\partial x} = 2(xy - 1) \cdot y = 2y(xy - 1). \]
03

Compute the Partial Derivative \( \frac{\partial f}{\partial y} \)

Similarly, to find \( \frac{\partial f}{\partial y} \), use the same expression for \( u \): \( u = xy - 1 \). First, find \( \frac{\partial u}{\partial y} = x \). Then, apply the chain rule: \[ \frac{\partial f}{\partial y} = 2u \cdot \frac{\partial u}{\partial y} = 2(xy - 1) \cdot x = 2x(xy - 1). \]
04

Simplify the Partial Derivatives

The partial derivatives simplify down as follows: \( \frac{\partial f}{\partial x} = 2y(xy-1) = 2xy^2 - 2y \) and \( \frac{\partial f}{\partial y} = 2x(xy-1) = 2x^2y - 2x \).

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Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus used for differentiating composite functions. This rule is essential when dealing with functions that are formed by composing two or more functions. In simple words, it helps us to find the derivative of a function that depends on another variable which itself is a function of an independent variable.

Here's how the chain rule works:
  • If you have a composite function \( z = f(g(x)) \), the chain rule states that the derivative \( \frac{dz}{dx} = f'(g(x)) \cdot g'(x) \).
  • This rule also applies for partial derivatives when dealing with multivariable functions.

In the exercise given, the function \( f(x, y) = (xy - 1)^2 \) is a composite function of \( u = xy - 1 \).
By identifying \( u \) and expressing \( f \) as \( u^2 \), the chain rule helps in taking the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \), by treating \( u(x, y) \) appropriately.
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions involving multiple variables. It's crucial for understanding how changes in one variable influence others within a system where multiple variables interact. This branch of calculus allows us to solve problems in physics, engineering, and economic models.
  • Functions in multivariable calculus are often represented in terms of several variables, such as \( f(x, y, z) \).
  • Calculating derivatives in this context leads to partial derivatives, which measure the rate of change of the function with respect to one variable while keeping others constant.
This exercise involves a function \( f(x, y) = (xy - 1)^2 \), which depends on two variables \( x \) and \( y \).
The goal is to find out how changes in \( x \) or \( y \) affect the function, which is typically done using the concept of partial differentiation as learned in multivariable calculus.
Partial Differentiation
Partial differentiation is a technique used in multivariable calculus where we find the derivative of a function with respect to one variable, keeping the other variables constant. It's crucial for understanding how a particular variable influences the overall behavior of a function.
  • For a function \( f(x, y) \), the partial derivative with respect to \( x \) is denoted as \( \frac{\partial f}{\partial x} \).
  • Similarly, the partial derivative with respect to \( y \) is represented as \( \frac{\partial f}{\partial y} \).

When applying partial differentiation to the function \( f(x, y) = (xy - 1)^2 \), each variable is treated separately.
  • For \( \frac{\partial f}{\partial x} \), we use the chain rule, differentiating the inner function with respect to \( x \), resulting in \( \frac{\partial f}{\partial x} = 2xy^2 - 2y \).
  • For \( \frac{\partial f}{\partial y} \), the process is similar, yielding \( \frac{\partial f}{\partial y} = 2x^2y - 2x \).
This method explains how each variable impacts the function's output separately, thus providing a clearer picture of its behavior.

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Most popular questions from this chapter

To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Functions: a. \(f(x, y)=x^{2}+y^{2} \quad\) b. \(g(x, y)=1 /\left(x^{2}+y^{2}\right)\) Curves: i. The line \(x=t, \quad y=2-2 t\) ii. The line segment \(x=t, \quad y=2-2 t, \quad 0 \leq t \leq 1\)

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 $$ 1-\frac{x^{2} y^{2}}{3}<\frac{\tan ^{-1} x y}{x y}<1 $$ tell you anything about $$\lim _{(x, y) \rightarrow(0,0)} \frac{\tan ^{-1} x y}{x y} ?$$ Give reasons for your answer.

Designing a soda can A standard \(12-\mathrm{fl}\) oz can of soda is essentially a cylinder of radius \(r=1\) in. and height \(h=5\) in. a. At these dimensions, how sensitive is the can's volume to a small change in radius versus a small change in height? b. Could you design a soda can that appears to hold more soda but in fact holds the same 12 -fl oz? What might its dimensions be? (There is more than one correct answer.)

a. Maximum on a sphere Show that the maximum value of \(a^{2} b^{2} c^{2}\) on a sphere of radius \(r\) centered at the origin of a Cartesian \(a b c\) -coordinate system is \(\left(r^{2} / 3\right)^{3}\) b. Geometric and arithmetic means Using part (a), show that for nonnegative numbers \(a, b,\) and \(c,\) $$(a b c)^{1 / 3} \leq \frac{a+b+c}{3}$$ that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean.

In Exercises \(65-70,\) you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level critical plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. $$ f(x, y)=x^{3}-3 x y^{2}+y^{2}, \quad-2 \leq x \leq 2, \quad-2 \leq y \leq 2 $$
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