Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The mixed partial derivatives are equal, confirming the equality.

Step 1: Compute First Partial Derivative with Respect to x

We begin by finding the partial derivative of \( f(x, y) = (x^3 + y^2)^5 \) with respect to \( x \). First, let \( u = x^3 + y^2 \), hence \( f(x, y) = u^5 \). The partial derivative of \( f \) with respect to \( x \) can be found using the chain rule: \( \frac{\partial f}{\partial x} = 5u^4 \cdot \frac{\partial u}{\partial x} \). Compute \( \frac{\partial u}{\partial x} = 3x^2 \). Thus, \( \frac{\partial f}{\partial x} = 5(x^3 + y^2)^4 \cdot 3x^2 = 15x^2(x^3 + y^2)^4 \).

Step 2: Compute First Partial Derivative with Respect to y

Next, we find the partial derivative of \( f \) with respect to \( y \). Using similar steps as before, we have \( \frac{\partial f}{\partial y} = 5u^4 \cdot \frac{\partial u}{\partial y} \). As \( \frac{\partial u}{\partial y} = 2y \), it follows that \( \frac{\partial f}{\partial y} = 5(x^3 + y^2)^4 \cdot 2y = 10y(x^3 + y^2)^4 \).

Step 3: Compute Second Mixed Partial Derivative \( \frac{\partial^2 f}{\partial y \partial x} \)

We now compute the mixed partial derivative \( \frac{\partial}{\partial y} \left( \frac{\partial f}{\partial x} \right) \). From Step 1, \( \frac{\partial f}{\partial x} = 15x^2(x^3 + y^2)^4 \). We differentiate this expression with respect to \( y \):\[\frac{\partial^2 f}{\partial y \partial x} = 15x^2 \cdot 4(x^3 + y^2)^3 \cdot 2y = 120x^2y(x^3 + y^2)^3.\]

Step 4: Compute Second Mixed Partial Derivative \( \frac{\partial^2 f}{\partial x \partial y} \)

Finally, compute \( \frac{\partial}{\partial x} \left( \frac{\partial f}{\partial y} \right) \) from Step 2, where \( \frac{\partial f}{\partial y} = 10y(x^3 + y^2)^4 \). Differentiating this with respect to \( x \), we have:\[\frac{\partial^2 f}{\partial x \partial y} = 10 \cdot 4(x^3 + y^2)^3 \cdot y \cdot 3x^2 = 120x^2y(x^3 + y^2)^3.\]

Step 5: Verify Equality of Mixed Partial Derivatives

From Steps 3 and 4, we obtained \( \frac{\partial^2 f}{\partial y \partial x} = 120x^2y(x^3 + y^2)^3 \) and \( \frac{\partial^2 f}{\partial x \partial y} = 120x^2y(x^3 + y^2)^3 \), respectively. Since these are equal, we have verified that \( \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with functions of multiple variables. It allows us to differentiate a composite function by linking the rates of change. When we have a function, say, \( f(x, y) = (x^3 + y^2)^5 \), it can be broken down into inner and outer functions for easier differentiation. Here, think of \( u = x^3 + y^2 \) as the inner function and \( f(u) = u^5 \) as the outer function.
By applying the chain rule, the derivative of \( f \) with respect to one of its variables, \( x \), involves the derivative of \( f \) with respect to \( u \) and the derivative of \( u \) with respect to \( x \). Thus, the chain rule gives us:

• First, find \( \frac{\partial u}{\partial x} = 3x^2 \).
• Then compute \( \frac{\partial f}{\partial x} = 5u^4 \cdot \frac{\partial u}{\partial x} \). Here, it becomes \( 15x^2(x^3 + y^2)^4 \) after substituting \( u \).

This method streamlines the process by breaking complex problems into manageable parts, maintaining focus on each step.

Partial Derivatives

Partial derivatives represent how a function changes as we vary one variable while keeping others constant. For functions of several variables, each variable can be treated individually, offering a glimpse of the function's behavior in one direction.
Consider the function \( f(x, y) = (x^3 + y^2)^5 \). We can take the partial derivative with respect to \( x \), treating \( y \) as a constant, and vice versa. This process helps us understand how \( f \) changes in each direction independently.
Things to note when calculating partial derivatives:

• Focus on the chosen variable and treat the rest as constants.
• Rely on derivative rules – helpfully, properties like linearity and product rules from basic calculus apply.

With practice, you will find that partial derivatives become a powerful tool for analyzing multivariable functions.

Order of Differentiation

In calculus, the order of differentiation refers to the sequence in which you apply partial derivatives to multiple variables. When you see notation like \( \frac{\partial^2 f}{\partial y \partial x} \), it indicates that you first take the partial derivative with respect to \( x \), and then \( y \).
The order should be followed accordingly:

• First, derive the expression in \( x \), then use that derivative with respect to \( y \) for \( \frac{\partial^2 f}{\partial y \partial x} \).
• The alternative is \( \frac{\partial^2 f}{\partial x \partial y} \), where you start with \( y \) and differentiate next in \( x \).

Understanding the order and interpreting it correctly is crucial in handling mixed partial derivatives, ensuring accuracy and consistency in analysis.

Equality of Mixed Partial Derivatives

The equality of mixed partial derivatives is an essential concept in multivariable calculus and often deals with the symmetry of second-order derivatives. The theorem states that under certain conditions, the order of differentiation does not affect the outcome, giving us \( \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} \).
This holds true if:

• The function and the necessary derivatives are continuous throughout a region.
• These conditions are fulfilled for well-behaved functions as used in physics and engineering.

In practice, verifying this equality ensures the correctness of analysis involving functions of several variables. In our example, \( 120x^2y(x^3 + y^2)^3 \) was obtained for both sequences of differentiation, confirming the equality validly expected for such problems. As such, the equality aids in simplifying multivariable calculus problems while providing deeper insights into the function's structure.