Problem 50 - Calculus an Applied Approach

Question

Find the four second partial derivatives. Observe that the second mixed partials are equal. $$ z=\frac{x^{2}-y^{2}}{2 x y} $$

Step-by-Step Solution

Answer

The second partial derivatives of the function \(z=\frac{x^{2}-y^{2}}{2 x y}\) are \(z_{xx}=\frac{-y(x^2+4y^2)}{2x^3y^2}\), \(z_{xy}=\frac{-x(x^2+y^2)}{2x^3y^3}\), \(z_{yx}=\frac{-x(x^2+y^2)}{2x^3y^3}\), and \(z_{yy}=\frac{2x^4-5x^2y^2+y^4}{2x^3y^3}\). Furthermore, as expected from the Clairot’s theorem, it is confirmed that the mixed second partials \(z_{xy}\) and \(z_{yx}\) are equal.

Step 1: Find the first partial derivatives

Start by finding the first partial derivatives of the function. Use the quotient rule for differentiation. The first partial derivative of \(z\) with respect to \(x\) is noted as \(z_{x}\), and the first partial derivative of \(z\) with respect to \(y\) is noted as \(z_{y}\). After simplifying, you should get:\(z_{x}=\frac{y(x^2+y^2)}{2x^2y^2}\) and \(z_{y}=-\frac{x(x^2+3y^2)}{2x^2y^2}\).

Step 2: Find the four second partial derivatives

Next, differentiate each of the first partial derivatives with respect to both \(x\) and \(y\). This will give the four second partial derivatives: \(z_{xx}, z_{xy}, z_{yx}, z_{yy}\). Use the quotient rule and simplify the results. The final results are:\(z_{xx}=\frac{-y(x^2+4y^2)}{2x^3y^2}\), \(z_{xy}=\frac{-x(x^2+y^2)}{2x^3y^3}\), \(z_{yx}=\frac{-x(x^2+y^2)}{2x^3y^3}\), \(z_{yy}=\frac{2x^4-5x^2y^2+y^4}{2x^3y^3}\).

Step 3: Check that second mixed partials are equal

For a function of two variables, if all partial derivatives exist and are continuous in a neighborhood of a point, then the mixed partial derivatives are equal. By comparing the results from \(z_{xy}\) and \(z_{yx}\), it can be verified that the second mixed partials are equal.

Key Concepts

Quotient Rule

The quotient rule is a fundamental tool in calculus used specifically for differentiating functions that are divided by each other. In essence, when we have a function defined as the quotient of two functions, say \( f(x,y) = \frac{u(x,y)}{v(x,y)} \), the quotient rule helps us differentiate this type of function effectively.

According to the quotient rule, the derivative of \( z = \frac{u}{v} \) with respect to a variable (often \( x \) or \( y \)) is given by:

• \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)

In this formula, \( u' \) and \( v' \) represent the derivatives of \( u \) and \( v \), respectively, with respect to the variable you're differentiating. The approach involves three main steps:

• Differentiating the numerator \( u(x,y) \).
• Differentiating the denominator \( v(x,y) \).
• Applying these in the formula to obtain the derivative of the quotient.

Keep in mind that careful attention is required when simplifying the resulting expression to avoid errors and achieve the correct result. Differentiating each component correctly is crucial for applying this rule successfully in finding both first and second partial derivatives.

Mixed Partials

Mixed partial derivatives are derivatives where we differentiate a function with respect to more than one independent variable in succession. For instance, if you first differentiate \( f(x,y) \) with respect to \( x \) and then with respect to \( y \), the result is a mixed partial derivative, denoted as \( f_{xy} \).

When it comes to second partial derivatives, an important concept is the equality of mixed partial derivatives. According to Clairaut's theorem, under certain conditions, the mixed partials will be equal, meaning \( f_{xy} = f_{yx} \). These conditions generally include:

• The function must be continuous.
• All partial derivatives must exist and be continuous around a point.

This symmetry in mixed partial derivatives is a very useful property when analyzing functions of several variables. It means that we can often confirm whether calculations are correct by checking that our mixed partials match up. In the given exercise, this property was used to show that \( z_{xy} \) and \( z_{yx} \) are indeed equal, verifying the correctness of the differentiated terms.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents how the function changes as its inputs change. In the context of functions of several variables like \( z(x, y) \), differentiating involves finding partial derivatives with respect to each variable individually.

To differentiate the function given in the exercise, we start by finding first partial derivatives, \( z_x \) and \( z_y \), which measure the rate at which \( z \) changes as \( x \) and \( y \) change, respectively. The differentiation process involves simplifying the function and then applying derivative rules, such as the quotient rule.

After obtaining the first partial derivatives, we proceed to compute second partial derivatives. This continues the process of differentiation, applying it again to the results of the first differentiation. Second partial derivatives can reveal how the rate of change itself is changing, offering deeper insights into the behavior of the function. This multi-layered approach is fundamental in calculus, providing a comprehensive understanding of how variable changes impact a function. The calculated second derivatives, particularly the mixed ones, help confirm the broader properties and ensure the functionality adheres to the expected mathematical rules.