Problem 1 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 8 , do each of the following: (a) Find \(D_{11} f(x, y)\); (b) find \(D_{22} f(x, y) ;\) (c) show that \(D_{12} f(x, y)=D_{21} f(x, y) .\) $$ f(x, y)=\frac{x^{2}}{y}-\frac{y}{x^{2}} $$

Step-by-Step Solution

Answer

The second partial derivatives are \(D_{11} f(x, y) = \frac{2}{y} - \frac{6y}{x^4}\) and \(D_{22} f(x, y) = \frac{2x^2}{y^3}\). The mixed partial derivatives are equal: \(D_{12} f(x, y) = D_{21} f(x, y) = -\frac{2x}{y^2} + \frac{2}{x^3}\).

Step 1 - Find the first partial derivatives

To find the second partial derivatives, first find the first partial derivatives of the given function.\[ f(x, y) = \frac{x^2}{y} - \frac{y}{x^2} \]Compute the first partial derivative with respect to \(x\):\[ f_x = \frac{\partial}{\partial x} \left( \frac{x^2}{y} - \frac{y}{x^2} \right) \]Using the quotient rule and constant rule,\[ f_x = \frac{2x}{y} + \frac{2y}{x^3} \]Now, find the first partial derivative with respect to \(y\):\[ f_y = \frac{\partial}{\partial y} \left( \frac{x^2}{y} - \frac{y}{x^2} \right) \]Using the quotient rule and constant rule,\[ f_y = -\frac{x^2}{y^2} - \frac{1}{x^2} \]

Step 2 - Find the second partial derivative \(D_{11} f(x, y)\)

Next, take the partial derivative of \( f_x \) with respect to \(x\):\[ f_{xx} = \frac{\partial}{\partial x} \left( \frac{2x}{y} + \frac{2y}{x^3} \right) \]Better write it as,\[ f_{xx} = \frac{2}{y} - \frac{6y}{x^4} \]Thus,\[ D_{11} f(x, y) = \frac{2}{y} - \frac{6y}{x^4} \]

Step 3 - Find the second partial derivative \(D_{22} f(x, y)\)

Now, take the partial derivative of \( f_y \) with respect to \(y\):\[ f_{yy} = \frac{\partial}{\partial y} \left( -\frac{x^2}{y^2} - \frac{1}{x^2} \right) \]\[ f_{yy} = \frac{2x^2}{y^3} \]Therefore,\[ D_{22} f(x, y) = \frac{2x^2}{y^3} \]

Step 4 - Find the mixed partial derivatives \(D_{12} f(x, y)\) and \(D_{21} f(x, y)\)

First, find the mixed partial derivative \( f_{xy} \):\[ f_{xy} = \frac{\partial}{\partial y} \left( \frac{2x}{y} + \frac{2y}{x^3} \right) \]\[ f_{xy} = -\frac{2x}{y^2} + \frac{2}{x^3} \]Next, find the mixed partial derivative \( f_{yx} \):\[ f_{yx} = \frac{\partial}{\partial x} \left( -\frac{x^2}{y^2} - \frac{1}{x^2} \right) \]\[ f_{yx} = -\frac{2x}{y^2} + \frac{2}{x^3} \]Thus,\[ D_{12} f(x, y) = D_{21} f(x, y) = -\frac{2x}{y^2} + \frac{2}{x^3} \]

Key Concepts

Second Partial Derivatives

First, let’s dive into the concept of second partial derivatives. A second partial derivative is simply taking the partial derivative of a partial derivative.
For a given function, say, \(f(x, y)\), the notation \(D_{11} f(x, y)\) represents the second partial derivative with respect to \(x\). This means we first find the partial derivative of \(f\) with respect to \(x\), and then take the derivative of that result again with respect to \(x\).
For example, if we start with \(f(x, y) = \frac{x^2}{y} - \frac{y}{x^2}\), the first partial with respect to \(x\) is \(f_x = \frac{2x}{y} + \frac{2y}{x^3}\). To find the second partial derivative \(f_{xx}\), we take the partial derivative of \(f_x\) with respect to \(x\) again, resulting in \(f_{xx} = \frac{2}{y} - \frac{6y}{x^4}\).
It's important to practice this method as it builds a strong foundation for more complex topics.

Mixed Partial Derivatives

Mixed partial derivatives involve taking the partial derivative of a function with respect to different variables. For example, \(D_{12} f(x, y)\) means we first take the partial derivative of \(f\) with respect to \(x\), and then again with respect to \(y\).
In our function \(f(x, y) = \frac{x^2}{y} - \frac{y}{x^2}\), we already found \(f_{xy}\) by taking the partial derivative of \(f_x = \frac{2x}{y} + \frac{2y}{x^3}\) with respect to \(y\), resulting in \(f_{xy} = -\frac{2x}{y^2} + \frac{2}{x^3}\).
Similarly, \(D_{21} f(x, y)\) involves taking the partial derivative of \(f_y = -\frac{x^2}{y^2} - \frac{1}{x^2}\) with respect to \(x\), which gives \(f_{yx} = -\frac{2x}{y^2} + \frac{2}{x^3}\). As demonstrated, the mixed partial derivatives \(D_{12} f\) and \(D_{21} f\) are equal, confirming the symmetry property known as Clairaut's Theorem.
Ensuring mixed partial derivatives are correctly calculated is crucial since it validates the function's behavior under changing variables.

Quotient Rule

The Quotient Rule is vital in finding partial derivatives of functions expressed as ratios. This rule states that for a function \(h(x)\) defined as \( \frac{u(x)}{v(x)} \), where both \(u(x)\) and \(v(x)\) are differentiable functions, the derivative of \(h(x)\) with respect to \(x\) is:
i.e., \( \frac{d}{dx} \big( \frac{u}{v} \big) = \frac{u'v - uv'}{v^2} \).
When dealing with functions like \(f(x, y) = \frac{x^2}{y} - \frac{y}{x^2}\), we apply the quotient rule to each component separately.
For instance, to find \(f_x\) when \(f(x, y) = \frac{x^2}{y}\), we get:
Using the Quotient Rule: \( \frac{\frac{d}{dx}(x^2)y - x^2 \frac{d}{dx}y}{y^2} = \frac{2x y - 0}{y^2} \), further simplifying this gives \( \frac{2x}{y} \).
Similarly, apply the quotient rule to \( \frac{y}{x^2} \) in the same manner, confirming the derivation steps.
Understanding the Quotient Rule is essential for handling complex fractions in calculus and is indispensable for partial derivative calculations.