Problem 58 - Calculus Early Transcendentals

Question

Verify that the conclusion of Clairaut's Theorem holds, that is, \(u_{x y}=u_{y x}\) $$u=x^{4} y^{2}-2 x y^{5}$$

Step-by-Step Solution

Answer

Clairaut's Theorem holds; \(u_{xy} = u_{yx}\).

Step 1: Compute the first partial derivatives

First, find the partial derivative of the function \(u\) with respect to \(x\) and \(y\).\[ u_x = \frac{\partial}{\partial x}(x^4 y^2 - 2xy^5) = 4x^3 y^2 - 2y^5 \]\[ u_y = \frac{\partial}{\partial y}(x^4 y^2 - 2xy^5) = 2x^4 y - 10xy^4 \]

Step 2: Compute the mixed partial derivative \(u_{xy}\)

Using the partial derivative \(u_x\), now compute \(u_{xy}\) by differentiating \(u_x\) with respect to \(y\).\[ u_{xy} = \frac{\partial}{\partial y}(4x^3 y^2 - 2y^5) = 8x^3 y - 10y^4 \]

Step 3: Compute the mixed partial derivative \(u_{yx}\)

Using the partial derivative \(u_y\), compute \(u_{yx}\) by differentiating \(u_y\) with respect to \(x\).\[ u_{yx} = \frac{\partial}{\partial x}(2x^4 y - 10xy^4) = 8x^3 y - 10y^4 \]

Step 4: Verify the condition of Clairaut's Theorem

Compare the results from steps 2 and 3. Both \(u_{xy}\) and \(u_{yx}\) should be equal for Clairaut's Theorem to hold.The computed values are:\[ u_{xy} = 8x^3 y - 10y^4 \]\[ u_{yx} = 8x^3 y - 10y^4 \]Since \(u_{xy} = u_{yx}\), the conclusion of Clairaut's Theorem holds.

Key Concepts

Understanding Mixed Partial Derivatives

Mixed partial derivatives involve computing the derivative of a function with respect to two distinct variables. Imagine you have a function of two variables, such as in the given exercise where the function is defined as \( u = x^4 y^2 - 2xy^5 \).
When we talk about mixed partial derivatives, we are referring to the second-order derivatives where you first differentiate the function with respect to one variable and then with respect to another.

• The symbol \( u_{xy} \) refers to first differentiating \( u \) with respect to \( x \), and then differentiating the resulting expression with respect to \( y \).
• Similarly, \( u_{yx} \) means first differentiating \( u \) with respect to \( y \), and then differentiating that result with respect to \( x \).

Clairaut's Theorem on mixed partial derivatives states that, under certain conditions, the mixed partial derivatives are equal, i.e., \( u_{xy} = u_{yx} \). This is precisely what the exercise aims to demonstrate.

The Basics of Partial Derivatives

Partial derivatives are a way of finding out how a multi-variable function changes when we alter one of its variables, holding others constant.
In simpler terms, if you have a function \( u(x, y) \), a partial derivative with respect to \( x \) measures how \( u \) changes as \( x \) changes, keeping \( y \) constant.

• The notation \( u_x \) denotes the partial derivative of \( u \) with respect to \( x \).
• Similarly, \( u_y \) indicates the partial derivative of \( u \) with respect to \( y \).

In the original exercise, you first computed these derivatives: \[ u_x = 4x^3y^2 - 2y^5 \] \[ u_y = 2x^4y - 10xy^4 \] These calculations help step up to computing the mixed partial derivatives, essential for verifying Clairaut's Theorem.

Insight into Function Differentiation

Function differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes as its inputs change.
For functions of a single variable, this means finding the slope of the tangent line to the curve at any given point. But when dealing with functions of multiple variables, like in bivariate functions, differentiation becomes more complex with partial derivatives.
The process involves:

• Identifying how a change in one variable affects the overall change in the function while holding the other variables constant.
• Using differential calculus techniques to handle each variable separately.

This concept becomes particularly important when examining conditions like those in Clairaut's Theorem, where we ensure that changes through different variable paths yield the same outcome.
Differentiating functions stepwise helps manage this complexity, especially for verifying relationships between mixed derivatives as done in the original solution.