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In Exercises \(55-60,\) verify that \(w_{x y}=w_{y x}\). $$\omega=\frac{x^{2}}{y^{3}}$$

Short Answer

Expert verified
Yes, \( w_{xy} = w_{yx} \) is verified.

Step by step solution

01

Find the partial derivative with respect to x

The given function is \( \omega = \frac{x^{2}}{y^{3}} \). To find the partial derivative of \( \omega \) with respect to \( x \), treat \( y \) as a constant and differentiate with respect to \( x \). This gives:\[ \omega_{x} = \frac{\partial}{\partial x} \left( \frac{x^{2}}{y^{3}} \right) = \frac{2x}{y^{3}}. \]
02

Differentiate \( \omega_{x} \) with respect to y

Now, take the partial derivative of \( \omega_{x} = \frac{2x}{y^{3}} \) with respect to \( y \). This involves treating \( x \) as a constant:\[ \omega_{xy} = \frac{\partial}{\partial y} \left( \frac{2x}{y^{3}} \right) = -\frac{6x}{y^{4}}. \]
03

Find the partial derivative with respect to y

Next, differentiate \( \omega = \frac{x^{2}}{y^{3}} \) with respect to \( y \), treating \( x \) as a constant:\[ \omega_{y} = \frac{\partial}{\partial y} \left( \frac{x^{2}}{y^{3}} \right) = -\frac{3x^{2}}{y^{4}}. \]
04

Differentiate \( \omega_{y} \) with respect to x

Finally, take the partial derivative of \( \omega_{y} = -\frac{3x^{2}}{y^{4}} \) with respect to \( x \), treating \( y \) as a constant:\[ \omega_{yx} = \frac{\partial}{\partial x} \left( -\frac{3x^{2}}{y^{4}} \right) = -\frac{6x}{y^{4}}. \]
05

Compare \( \omega_{xy} \) and \( \omega_{yx} \)

Compare the results from Step 2 and Step 4. We have:\[ \omega_{xy} = -\frac{6x}{y^{4}} \] and \[ \omega_{yx} = -\frac{6x}{y^{4}}. \]Since \( \omega_{xy} = \omega_{yx} \), the condition is verified.

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Key Concepts

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

Mixed Partial Derivatives
Mixed partial derivatives are a fascinating concept in calculus that involve taking derivatives of a function with more than one variable. Essentially, these are second-order derivatives where you differentiate with respect to one variable and then another.
For the function \( \omega = \frac{x^2}{y^3} \), it essentially requires determining \( \omega_{xy} \) and \( \omega_{yx} \). This is all about ensuring that your mixed partial derivatives are equal. When they are equal, as shown here with \( \omega_{xy} = \omega_{yx} \), it is referred to as being continuous on certain conditions, based on Clairaut's theorem.
This theorem asserts that if the mixed partial derivatives exist and are continuous on a neighborhood of a point, they are equal at that point. This is a fundamental similarity in several-variable functions, which makes it a critical area of study in understanding multivariable calculus.
Calculus
Calculus, often recognized as the mathematics of change, fundamentally builds upon derivatives and integrals. It's a vast field with two distinct branches: differential calculus and integral calculus.
Partial derivatives are an essential concept within differential calculus, particularly when dealing with functions of several variables. This branch allows us to find rates of change when function values depend on two or more variables. This ability to dissect functions into smaller parts is what makes calculus so powerful in science and engineering.
Partial differentiation, like we utilized in this exercise, involves differentiating a function of several variables with respect to one variable at a time, treating other variables as constants. This is a crucial step in deeply understanding the behavior of complex systems.
Functions of Several Variables
In the realm of mathematics, functions of several variables are an extension of traditional functions, involving two or more inputs instead of just one. This allows for much more dynamic and multidimensional representations, as seen in functions like \( \omega = \frac{x^2}{y^3} \).
These functions are applicable in diverse fields, enabling us to model real-world scenarios more accurately. For instance, they can describe how different physical quantities vary together, such as temperature changes within a room or pressure variations in a fluid.
Understanding these relationships requires different tools and approaches—one of which is partial differentiation. This process involves isolating and determining how a change in one input affects the output, while keeping other inputs constant. This comprehension allows us to mathematically explore more complex relationships in detail.
Differentiation Rules
Differentiation rules are the backbone for finding derivatives efficiently and accurately. In calculus, these rules help us apply partial differentiation effectively, as we did for \( \omega = \frac{x^2}{y^3} \).
Some crucial rules include:
  • Power Rule: which states \( \frac{d}{dx}(x^n) = nx^{n-1} \). This was useful in differentiating \( x^2 \) in the given function.
  • Constant Multiple Rule: where \( \frac{d}{dx}(c \cdot f(x)) = c \cdot f'(x) \) was used to handle constants like \( y^3 \).
These rules streamline the process, allowing us to tackle more complex equations with ease.
Additionally, the Chain Rule is often used with composite functions, aiding in the breakdown of nested functions into more manageable parts. These techniques and tricks are essential for handling all sorts of differentiations, including those beyond the single-variable realm.

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Most popular questions from this chapter

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