Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Verify that $$\frac{\partial^{2} f}{\partial y \partial x}=\frac{\partial^{2} f}{\partial x \partial y}$$ \(f(x, y)=2 x^{2} y^{3}-x^{3} y^{5}\)

Step-by-Step Solution

Answer

The mixed partial derivatives are equal: \( \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} \).

Step 1: Find the First Partial Derivative with Respect to x

Consider the function \( f(x, y) = 2x^2 y^3 - x^3 y^5 \). We need to first find the partial derivative of \( f \) with respect to \( x \). When differentiating with respect to \( x \), treat \( y \) as a constant. Use the power rule to differentiate each term:\[\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (2x^2 y^3) - \frac{\partial}{\partial x} (x^3 y^5)\]Calculating each term separately:- For \( 2x^2 y^3 \), the derivative is \( 4xy^3 \) because it is \( y^3 \times 2 \times 2x \).- For \( x^3 y^5 \), the derivative is \( 3x^2 y^5 \) because it is \( y^5 \times 3 \times x^2 \).Thus, we have:\[\frac{\partial f}{\partial x} = 4xy^3 - 3x^2y^5\]

Step 2: Find the Second Partial Derivative with Respect to y

Now, take the derivative of \( \frac{\partial f}{\partial x} = 4xy^3 - 3x^2y^5 \) with respect to \( y \). Treat \( x \) as a constant:\[\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} (4xy^3) - \frac{\partial}{\partial y} (3x^2 y^5)\]Calculating each term separately:- For \( 4xy^3 \), the derivative is \( 12xy^2 \) because it is \( 4x \times 3y^2 \).- For \( 3x^2 y^5 \), the derivative is \( 15x^2 y^4 \) because it is \( 3x^2 \times 5y^4 \).Thus, we have:\[\frac{\partial^2 f}{\partial y \partial x} = 12xy^2 - 15x^2y^4\]

Step 3: Find the First Partial Derivative with Respect to y

Next, find the partial derivative of \( f(x, y) = 2x^2 y^3 - x^3 y^5 \) with respect to \( y \). Treat \( x \) as a constant. Use the power rule to differentiate each term:\[\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} (2x^2 y^3) - \frac{\partial}{\partial y} (x^3 y^5)\]Calculating each term separately:- For \( 2x^2 y^3 \), the derivative is \( 6x^2 y^2 \) because it is \( 2x^2 \times 3y^2 \).- For \( x^3 y^5 \), the derivative is \( 5x^3 y^4 \) because it is \( x^3 \times 5y^4 \).Thus, we have:\[\frac{\partial f}{\partial y} = 6x^2 y^2 - 5x^3 y^4\]

Step 4: Find the Second Partial Derivative with Respect to x

Now, take the derivative of \( \frac{\partial f}{\partial y} = 6x^2 y^2 - 5x^3 y^4 \) with respect to \( x \). Treat \( y \) as a constant:\[\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} (6x^2 y^2) - \frac{\partial}{\partial x} (5x^3 y^4)\]Calculating each term separately:- For \( 6x^2 y^2 \), the derivative is \( 12xy^2 \) because it is \( 6 \times 2x \times y^2 \).- For \( 5x^3 y^4 \), the derivative is \( 15x^2 y^4 \) because it is \( 5 \times 3x^2 \times y^4 \).Thus, we have:\[\frac{\partial^2 f}{\partial x \partial y} = 12xy^2 - 15x^2y^4\]

Step 5: Verify Equality of Mixed Partial Derivatives

Compare the second mixed partial derivatives you calculated:- \( \frac{\partial^2 f}{\partial y \partial x} = 12xy^2 - 15x^2y^4 \)- \( \frac{\partial^2 f}{\partial x \partial y} = 12xy^2 - 15x^2y^4 \)Since both expressions are equal, the mixed partial derivatives are the same.

Key Concepts

Mixed Partial Derivatives

Mixed partial derivatives involve differentiating a multivariable function with respect to different variables in sequence. For example, if we have a function \( f(x, y) \), a mixed partial derivative could be \( \frac{\partial^2 f}{\partial y \partial x} \). This notation indicates that you first differentiate with respect to \( x \) and then \( y \).

A fascinating property of mixed partial derivatives is when they are equal irrespective of the order of differentiation, provided certain conditions like continuity are met.

• For a function \( f(x, y) \), under regular conditions, \( \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} \).
• This equality holds due to Clairaut's theorem, which states that if the mixed partial derivatives are continuous at a point, they can be interchanged.

Understanding this property helps in solving problems involving multi-variable functions and can simplify complex expressions significantly.

Power Rule

The power rule is a fundamental principle in calculus regarding how to differentiate expressions involving powers of variables. It states that if you have a function in the form \( x^n \), the derivative with respect to \( x \) is \( nx^{n-1} \). In partial derivatives, this rule applies while differentiating with respect to a specific variable.

When applying the power rule:

• You multiply the power by the coefficient.
• Subtract one from the original power to get the new power.

For instance, in the function \( 2x^2 y^3 \), applying the power rule with respect to \( x \), you would focus on \( x^2 \) so it becomes \( 2 \times 2x^{2-1}y^3 = 4xy^3 \).

This rule simplifies the differentiation process and is crucial for finding partial derivatives efficiently.

Function Differentiation

Function differentiation involves finding the derivative of a function, which represents how the function's value changes as its input varies. For functions dependent on multiple variables, such as \( f(x, y) \), we focus on partial derivatives.

Partial derivatives indicate the rate of change of the function with respect to one variable while keeping other variables constant.

• For \( f(x, y) = 2x^2 y^3 - x^3 y^5 \), differentiating with respect to \( x \) and \( y \) separately helps understand how \( f(x, y) \) changes when each variable changes independently.
• This differentiation is carried out by treating one variable as constant while finding the derivative with respect to the other.

This approach gives us insight into the function's behavior and is essential for optimizing problems or understanding underlying patterns.

Order of Differentiation

When differentiating a function of multiple variables, the order in which you take partial derivatives can sometimes affect the result. However, under the right conditions, notably continuity, the order does not matter.

For instance, if we take the partial derivatives of \( f(x, y) \) with respect to \( x \) first and \( y \) second (\( \frac{\partial^2 f}{\partial y \partial x} \)) or \( y \) first and \( x \) second (\( \frac{\partial^2 f}{\partial x \partial y} \)), the results are often the same. This is due to:

• Clairaut's theorem, which states that mixed derivatives are equal if the second partial derivatives are continuous around that point.
• This means if you find either \( \frac{\partial^2 f}{\partial y \partial x} \) or \( \frac{\partial^2 f}{\partial x \partial y} \), you can be confident in the result's validity when the conditions are met.

The order of differentiation is a powerful concept since it assures the reliability and symmetry of calculations involving multivariable functions.