Chapter 4: Problem 4
Find all four second-order partial derivatives of the given function \(f\). $$ f(x, y)=\frac{y}{x+y} $$
Short Answer
Expert verified
The second-order partial derivatives are: \( f_{xx} = \frac{2y}{(x+y)^3} \), \( f_{xy} = \frac{x}{(x+y)^3} \), \( f_{yx} = \frac{x}{(x+y)^3} \), \( f_{yy} = \frac{-2x}{(x+y)^3} \).
Step by step solution
01
First-Order Partial Derivatives
To find the second-order partial derivatives, we must first compute the first-order derivatives. The partial derivative of \(f\) with respect to \(x\), denoted \(f_x\), is obtained by differentiating \(f(x, y) = \frac{y}{x+y}\) with respect to \(x\). Using the quotient rule, we get:\[ f_x = \frac{0 \cdot (x+y) - y \cdot 1}{(x+y)^2} = -\frac{y}{(x+y)^2} \]Similarly, the partial derivative with respect to \(y\), denoted \(f_y\), is:\[ f_y = \frac{(x+y) \cdot 1 - y \cdot 1}{(x+y)^2} = \frac{x}{(x+y)^2} \]
02
Second-Order Partial Derivative with Respect to x
Differentiate the first-order partial derivative \(f_x = -\frac{y}{(x+y)^2}\) with respect to \(x\) to find \(f_{xx}\). Using the chain and quotient rules, we obtain:\[ f_{xx} = \frac{2y}{(x+y)^3} \]
03
Mixed Partial Derivative f_xy
Next, find the mixed partial derivative \(f_{xy}\) by differentiating \(f_x = -\frac{y}{(x+y)^2}\) with respect to \(y\):\[ f_{xy} = -\frac{1 \cdot (x+y)^2 - y \cdot 2(x+y)}{(x+y)^4} = \frac{x}{(x+y)^3} \]
04
Mixed Partial Derivative f_yx
Now, find the mixed partial derivative \(f_{yx}\) by differentiating \(f_y = \frac{x}{(x+y)^2}\) with respect to \(x\):\[ f_{yx} = \frac{(x+y)^2 - x \cdot 2(x+y)}{(x+y)^4} = \frac{x}{(x+y)^3} \]By Clairaut's Theorem of mixed partial derivatives, \(f_{xy} = f_{yx}\).
05
Second-Order Partial Derivative with Respect to y
Finally, find the second-order partial derivative \(f_{yy}\) by differentiating \(f_y = \frac{x}{(x+y)^2}\) with respect to \(y\):\[ f_{yy} = \frac{-2x}{(x+y)^3} \]
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
Partial derivatives are a key concept in calculus when dealing with functions of multiple variables. They represent the rate of change of a function with respect to one variable, while keeping the other variables constant.
For example, given a function like our exercise function, \( f(x, y)=\frac{y}{x+y} \), we have two partial derivatives: \( f_x \) and \( f_y \), indicating derivatives with respect to \( x \) and \( y \) respectively.
For example, given a function like our exercise function, \( f(x, y)=\frac{y}{x+y} \), we have two partial derivatives: \( f_x \) and \( f_y \), indicating derivatives with respect to \( x \) and \( y \) respectively.
- \( f_x \) tells us how \( f \) changes as \( x \) changes, with \( y \) being constant.
- \( f_y \) tells us how \( f \) changes as \( y \) changes, with \( x \) being constant.
Quotient Rule
The quotient rule is an essential tool in calculus for differentiating functions that are expressed as a division of two other functions. If you have a function like \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule states:\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]This rule is vital because it allows us to compute the derivative of a function efficiently when it is structured as a fraction.
In our exercise, we used the quotient rule to find the partial derivatives of the function \( f(x, y) = \frac{y}{x+y} \). Differentiating with respect to \( x \) and \( y \), we set up and solved for \( f_x \) and \( f_y \).
This calculation is crucial as it simplifies the process of finding both first-order and second-order partial derivatives, allowing us to handle complex functions with ease.
In our exercise, we used the quotient rule to find the partial derivatives of the function \( f(x, y) = \frac{y}{x+y} \). Differentiating with respect to \( x \) and \( y \), we set up and solved for \( f_x \) and \( f_y \).
This calculation is crucial as it simplifies the process of finding both first-order and second-order partial derivatives, allowing us to handle complex functions with ease.
Clairaut's Theorem
Clairaut's Theorem is a fascinating result in the study of partial derivatives, especially when exploring mixed partial derivatives. The theorem states that if the mixed partial derivatives of a function are continuous, then they are equal.
For a function \( f(x, y) \), this means:
For a function \( f(x, y) \), this means:
- \( f_{xy} \) and \( f_{yx} \) are equal.
Mixed Partial Derivatives
Mixed partial derivatives are derivatives taken with respect to more than one variable. For instance, given a function \( f(x, y) \), the mixed derivatives \( f_{xy} \) and \( f_{yx} \) involve differentiating with respect to \( x \) and then \( y \), or vice versa.
In our exercise, finding \( f_{xy} \) and \( f_{yx} \) showed how the interplay between \( x \) and \( y \) affects the function \( f(x, y) = \frac{y}{x+y} \). Calculating mixed derivatives is useful for understanding critical points and behavior of multivariable functions in a variety of scientific and engineering applications.
- \( f_{xy} = \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) \)
- \( f_{yx} = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) \)
In our exercise, finding \( f_{xy} \) and \( f_{yx} \) showed how the interplay between \( x \) and \( y \) affects the function \( f(x, y) = \frac{y}{x+y} \). Calculating mixed derivatives is useful for understanding critical points and behavior of multivariable functions in a variety of scientific and engineering applications.