Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 17-20, verify that $$ \frac{\partial^{2} f}{\partial y \partial x}=\frac{\partial^{2} f}{\partial x \partial y} $$ \(f(x, y)=\tan ^{-1} x y\)

Step-by-Step Solution

Answer

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

Step 1: Find \( \frac{\partial f}{\partial x} \)

First, differentiate the function \( f(x, y) = \tan^{-1}(xy) \) with respect to \( x \). Using the chain rule, we get: \[ \frac{\partial f}{\partial x} = \frac{1}{1 + (xy)^2} \cdot y. \]

Step 2: Differentiate \( \frac{\partial f}{\partial x} \) with respect to \( y \)

Now, we need to find \( \frac{\partial^2 f}{\partial y \partial x} \) by differentiating the expression from Step 1 with respect to \( y \): \[ \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial}{\partial y} \left( \frac{y}{1 + (xy)^2} \right). \] Using the quotient rule, we get: \[ \frac{\partial^2 f}{\partial y \partial x} = \frac{1 \cdot (1+(xy)^2) - 2x^2y^2}{(1+(xy)^2)^2}. \] Simplifying gives: \[ \frac{1 - x^2y^2}{(1+(xy)^2)^2}. \]

Step 3: Find \( \frac{\partial f}{\partial y} \)

Next, differentiate \( f(x, y) \) with respect to \( y \). Again, by using the chain rule, we get: \[ \frac{\partial f}{\partial y} = \frac{1}{1 + (xy)^2} \cdot x. \]

Step 4: Differentiate \( \frac{\partial f}{\partial y} \) with respect to \( x \)

Now, find \( \frac{\partial^2 f}{\partial x \partial y} \) by differentiating the expression from Step 3 with respect to \( x \): \[ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x} \left( \frac{x}{1 + (xy)^2} \right). \] Using the quotient rule, we get: \[ \frac{(1+(xy)^2) - 2x^2y^2}{(1+(xy)^2)^2}. \] Simplifying gives: \[ \frac{1 - x^2y^2}{(1+(xy)^2)^2}. \]

Step 5: Verify the Equality

Compare the results from Step 2 and Step 4: \( \frac{1 - x^2y^2}{(1+(xy)^2)^2} \). Since both are equal, we verify \( \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} \).

Key Concepts

Chain Rule in Calculus

The chain rule in calculus is a fundamental principle for differentiating composite functions. When you have a function composed of other functions, the chain rule allows you to find the derivative of the outer function while considering the inner functions.
For instance, in the case of the function \( f(x, y) = \tan^{-1}(xy) \), \( xy \) is the inner function, while \( \tan^{-1} \) represents the outer function.
When differentiating \( f(x, y) \) with respect to either variable, we first differentiate the outer function then multiply it by the derivative of the inner function.

**Key Points:**

• The chain rule is essential when dealing with compositions of functions.
• It involves differentiating the outer function and then multiplying by the derivative of the inner function.

This rule is not only crucial for single-variable calculus but also forms the basis for finding partial derivatives in multivariable calculus.

Quotient Rule in Calculus

The quotient rule is particularly useful when differentiating functions that are divided by each other. It helps find the derivative of a function of the form \( \frac{u}{v} \).
In our exercise, after obtaining the first partial derivative, we arrived at expressions such as \( \frac{y}{1+(xy)^2} \).
To find the change with respect to another variable, we apply the quotient rule.

**Quotient Rule Formula:**

• If \( u \) and \( v \) are functions of \( x \), then the derivative of \( \frac{u}{v} \) is \( \frac{v \cdot u' - u \cdot v'}{v^2} \).

This rule allows us to systematically find the derivative of complex fractions, as seen in the simplification process of our given exercise.

Partial Differentiation

Partial differentiation is a process used to find the derivative of a multivariable function with respect to one variable, treating all other variables as constants.
In the function \( f(x, y) = \tan^{-1}(xy) \), we independently take the derivative with respect to \( x \) while treating \( y \) as a constant and vice versa.
This method allows us to explore how a change in one specific variable can affect the function as a whole.

**Important Aspects of Partial Derivatives:**

• They reveal the rate of change of a function with respect to one of its variables.
• The notation \( \frac{\partial}{\partial x} \) indicates the partial derivative with respect to \( x \).

Partial differentiation is essential in many fields such as physics, engineering, and economics where functions of multiple variables are common.

Anticommutativity of Partial Derivatives

In calculus, the anticommutativity of partial derivatives states that for certain well-behaved functions, the order of taking partial derivatives does not impact the result.
This property is usually referred to as being commutative for partial derivatives, despite the misleading name.

**Concept Explanation:**

• The property holds for functions that are continuous and have continuous mixed partial derivatives.
• This means \( \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} \), which was the purpose of our exercise.

This property is invaluable as it saves computation time by confirming that the order of differentiation doesn't affect the end result, as was seen when both orders of differentiation yielded the same simplified expression in the given exercise.