Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 12: Problem 7
$$ f(x, y)=x y+\frac{2}{x}+\frac{4}{y} $$
Short Answer
Expert verified
Find the partial derivatives: \( \frac{\partial f}{\partial x} = y - \frac{2}{x^2} \) and \( \frac{\partial f}{\partial y} = x - \frac{4}{y^2} \).
Step by step solution
01
Understand the Function
The function given is a two-variable function \( f(x, y) = xy + \frac{2}{x} + \frac{4}{y} \). This function consists of a term that involves the product of \( x \) and \( y \), and two rational terms, \( \frac{2}{x} \) and \( \frac{4}{y} \). The goal is typically to analyze or find specific values, behavior, or derivatives of the function.
02
Identify Objective
Since no specific task was provided with the function, determine a common task such as finding critical points, computing partial derivatives, or evaluating at a specific point. For this example, let's decide to find the partial derivatives to analyze how the function changes with respect to \( x \) and \( y \).
03
Compute Partial Derivative with Respect to x
To find \( \frac{\partial f}{\partial x} \), differentiate the function \( f(x, y) = xy + \frac{2}{x} + \frac{4}{y} \) with respect to \( x \). Using the power rule and the quotient rule, we get:\[\frac{\partial f}{\partial x} = y - \frac{2}{x^2}\]
04
Compute Partial Derivative with Respect to y
To find \( \frac{\partial f}{\partial y} \), differentiate the function \( f(x, y) = xy + \frac{2}{x} + \frac{4}{y} \) with respect to \( y \). This gives:\[\frac{\partial f}{\partial y} = x - \frac{4}{y^2}\]
05
Summary of Partial Derivatives
The partial derivatives of the function are:- With respect to \( x \): \( \frac{\partial f}{\partial x} = y - \frac{2}{x^2} \)- With respect to \( y \): \( \frac{\partial f}{\partial y} = x - \frac{4}{y^2} \)
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Two-variable function
A two-variable function is a function that depends on two independent variables, typically denoted as \(x\) and \(y\). These functions describe a relationship where the output or dependent variable depends on the values of both inputs. An example of such a function is \(f(x, y) = xy + \frac{2}{x} + \frac{4}{y}\). Here, each term in the function depends on either \(x\), \(y\), or both.
- The term \(xy\) shows direct multiplication of the two variables.
- The term \(\frac{2}{x}\) demonstrates a rational expression involving only \(x\).
- The term \(\frac{4}{y}\) features \(y\) in a similar context.
Critical points
Critical points in two-variable functions are similar to those in single-variable functions. They indicate where the function's rate of change is zero, which can be useful for finding maxima, minima, or saddle points.
To find critical points:
To find critical points:
- Calculate the partial derivatives of the function with respect to each variable.
- Set these partial derivatives equal to zero.
- Solve the system of equations to find points \((x, y)\) where the function's rate of change is zero.
Power rule
The power rule is a fundamental principle of differentiation used to find the derivative of functions with powers. It states that if \(f(x) = x^n\), then the derivative \(f'(x) = nx^{n-1}\). This rule is simple but incredibly powerful, allowing swift calculation of derivatives.
When applying it to two-variable functions for partial derivatives, it helps differentiate terms like \(x^n\) or \(y^n\) with respect to one variable:
When applying it to two-variable functions for partial derivatives, it helps differentiate terms like \(x^n\) or \(y^n\) with respect to one variable:
- For \(f(x, y) = xy\), treating \(y\) as a constant yields \(\frac{\partial }{\partial x}(xy) = y\).
- For \(\frac{2}{x} = 2x^{-1}\), using the power rule gives \(-2x^{-2}\).
Quotient rule
The quotient rule is used to differentiate functions where variables are divided by each other. Specifically, if \(f(x) = \frac{u(x)}{v(x)}\), then the derivative is:\[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}\]This rule is critical in handling terms like \(\frac{2}{x}\) when finding partial derivatives.
For two-variable functions, apply the quotient rule when dealing with terms that divide by a variable. For example:
For two-variable functions, apply the quotient rule when dealing with terms that divide by a variable. For example:
- The derivative with respect to \(x\) of \(\frac{2}{x}\) uses \(u = 2\) (which has derivative 0) and \(v = x\) to find \(-\frac{2}{x^2}\).
