Chapter 13: Problem 29
Find \(f_{x y}\) and \(f_{y x}\) $$ f(x, y)=3 x^{2}-\sqrt{2} x y^{2}+y^{5}-2 $$
Short Answer
Expert verified
Both \(f_{xy}\) and \(f_{yx}\) are \(-2\sqrt{2}y\).
Step by step solution
01
Find the First Partial Derivative with respect to x
To find \(f_x\), we must differentiate \(f(x, y) = 3x^2 - \sqrt{2}xy^2 + y^5 - 2\) with respect to \(x\). Treat \(y\) as a constant.\[ f_x = \frac{\partial}{\partial x}(3x^2 - \sqrt{2}xy^2 + y^5 - 2) \]Differentiating term by term gives:\[ f_x = 6x - \sqrt{2}y^2 \]
02
Find the First Partial Derivative with respect to y
To find \(f_y\), differentiate \(f(x, y) = 3x^2 - \sqrt{2}xy^2 + y^5 - 2\) with respect to \(y\). Treat \(x\) as a constant.\[ f_y = \frac{\partial}{\partial y}(3x^2 - \sqrt{2}xy^2 + y^5 - 2) \]Differentiating term by term gives:\[ f_y = -2\sqrt{2}xy + 5y^4 \]
03
Find the Second Mixed Partial Derivative \(f_{xy}\)
Now differentiate the expression for \(f_x = 6x - \sqrt{2}y^2\) with respect to \(y\).\[ f_{xy} = \frac{\partial}{\partial y}(6x - \sqrt{2}y^2) \]Calculate the derivative:\[ f_{xy} = -2\sqrt{2}y \]
04
Find the Second Mixed Partial Derivative \(f_{yx}\)
Differentiate the expression for \(f_y = -2\sqrt{2}xy + 5y^4\) with respect to \(x\).\[ f_{yx} = \frac{\partial}{\partial x}(-2\sqrt{2}xy + 5y^4) \]Calculate the derivative:\[ f_{yx} = -2\sqrt{2}y \]
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 fundamental concept in multivariable calculus. When you're dealing with functions of multiple variables, like \( f(x, y) \), each variable can change independently of the others. The partial derivative of a function with respect to one variable gives us the rate at which the function changes as that variable changes, while the other variables are held constant. To compute a partial derivative, you treat the variable of interest as the usual variable in differentiation, while treating all other variables as constants.
- For example, in the function \( f(x, y) = 3x^2 - \sqrt{2}xy^2 + y^5 - 2 \), computing \( f_x \) means differentiating with respect to \( x \), and considering \( y \) as a constant. This results in \( f_x = 6x - \sqrt{2}y^2 \).
- Similarly, to find \( f_y \), we differentiate with respect to \( y \), treating \( x \) as a constant. This gives \( f_y = -2\sqrt{2}xy + 5y^4 \).
Mixed Partial Derivatives
Mixed partial derivatives involve taking partial derivatives with respect to multiple variables, one after the other. For a function \( f(x, y) \), mixed partial derivatives like \( f_{xy} \) and \( f_{yx} \) represent how the function changes first with respect to \( x \) and then \( y \), or vice versa.The order in which the derivatives are taken can initially seem to matter, but a fascinating result called Clairaut's Theorem tells us that, under certain conditions of continuity, the mixed partial derivatives are equal: \( f_{xy} = f_{yx} \).
- In the original exercise, we have \( f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = -2\sqrt{2}y \).
- Calculating \( f_{yx} \), or \( \frac{\partial^2 f}{\partial x \partial y} \), also yields \( -2\sqrt{2}y \), showcasing this equality.
Differentiation
Differentiation in calculus refers to the process of finding a derivative, which measures how a function changes as its input changes. In the context of multivariable calculus, differentiation extends beyond simple one-variable functions to include partial derivatives for multi-variable functions. This allows us to explore how a function evolves as each variable changes.Differentiation techniques remain similar to single-variable calculus but adapted for multiple variables. Here are some key elements to understand:
- Identify the variable with respect to which you're differentiating. Treat other variables as constants during this process.
- Apply standard differentiation rules, such as the power rule or product rule, to each term independently.