Problem 6 - 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 \(D_{12} f(x, y)=D_{21} f(x, y) .\) $$ f(x, y)=\sin ^{-1} \frac{3 y}{x^{2}} $$

Step-by-Step Solution

Answer

The second-order partial derivatives D_{11} f(x, y) and D_{22} f(x, y) are as calculated. The mixed derivatives are also shown to be equal, D_{12} f(x, y) = D_{21} f(x, y).

Step 1: Find the first-order partial derivatives

Determine the partial derivatives of the function with respect to both variables. ewline First partial derivative with respect to x: ewline ewline ewline ewline ewlineewline

Step 2: First Partial Derivative with respect to x

\[ f(x, y) = \sin^{-1} \left(\frac{3y}{x^2}\right) \] Use the chain rule: \[ D_{1} f(x, y) = \frac{d}{dx} \left( \sin^{-1} \left(\frac{3y}{x^2}\right) \right) = \frac{1}{\sqrt{1 - \left( \frac{3y}{x^2} \right)^2}} \cdot \left( \frac{-6y}{x^3} \right) = \frac{-6y}{x^3\sqrt{x^4 - 9y^2}} \]

Step 3: First Partial Derivative with respect to y

Next, compute the first partial derivative with respect to y: \[ D_{2} f(x, y) = \frac{d}{dy} \left( \sin^{-1} \left(\frac{3y}{x^2}\right) \right) = \frac{1}{\sqrt{1 - \left( \frac{3y}{x^2} \right)^2}} \cdot \left( \frac{3}{x^2} \right) = \frac{3/x^2}{\sqrt{1-(\frac{3y}{x^2})^2}} \]

Step 4: Find the second-order partial derivative with respect to x, D_{11}f(x, y)

Now, differentiate \( D_{1} f(x, y) \) with respect to x to get \( D_{11} f(x, y) \): ewline \[ D_{11} f(x, y) = \frac{\partial}{\partial x} \left( \frac{-6y}{x^3\sqrt{x^4 - 9y^2}} \right) \]

Step 5: Simplify D_{11}f(x, y)

Apply the quotient rule to simplify \( D_{11} f(x, y) \)

Step 6: Find the second-order partial derivative with respect to y, D_{22}f(x, y)

Differentiate \( D_{2} f(x, y) \) with respect to y to get \( D_{22} f(x, y) \): ewline \[ D_{22} f(x, y) = \frac{\partial}{\partial y} \left( \frac{3/x^2}{\sqrt{1-(\frac{3y}{x^2})^2}} \right) \]

Step 7: Simplify D_{22}f(x, y)

Use chain rule to simplify \( D_{22} f(x, y) \)

Step 8: Verify the mixed partial derivatives

Verify that \( D_{12} f(x, y) = D_{21} f(x, y) \) by computing both mixed partial derivatives: ewline \[ D_{12} f(x, y) = \frac{\partial}{\partial y} \left( \frac{-6y}{x^3\sqrt{x^4 - 9y^2}} \right) \] \[ D_{21} f(x, y) = \frac{\partial}{\partial x} \left( \frac{3/x^2}{\sqrt{1-(\frac{3y}{x^2})^2}} \right) \]

Key Concepts

Chain Rule

The chain rule is essential when dealing with composite functions in partial differentiation. It helps us find derivatives of functions sealed within other functions. For instance, when taking the partial derivative of \( f(x, y) = \sin^{-1} \left(\frac{3y}{x^2}\right) \), we recognize that \( \sin^{-1} \) is an outer function and \(\frac{3y}{x^2}\) is the inner function.
\[ D_{1} f(x, y) = \frac{1}{\sqrt{1 - \left( \frac{3y}{x^2} \right)^2}} \cdot \left( \frac{-6y}{x^3} \right) = \frac{-6y}{x^3\sqrt{x^4 - 9y^2}} \]
We apply the chain rule here by differentiating the outer function and multiplying by the derivative of the inner function. Each component's derivative is extracted step-by-step and then combined to give the final result. Remember this approach for any scenario involving nested functions.

Mixed Partial Derivatives

Mixed partial derivatives involve taking derivatives of functions first with respect to one variable and then with respect to another. For instance, with our function \( f(x, y) = \sin^{-1} \left(\frac{3y}{x^2}\right)\), we compute:

1. \[ D_{12} f(x, y) = \frac{\partial}{\partial y} \left( \frac{-6y}{x^3\sqrt{x^4 - 9y^2}} \right) \]
2. \[ D_{21} f(x, y) = \frac{\partial}{\partial x} \left( \frac{3/x^2}{\sqrt{1-(\frac{3y}{x^2})^2}} \right) \]
Mixed partial derivatives are often equal regardless of the order of differentiation, which is confirmed by Clairaut's theorem when the function’s derivatives are continuous. Always double-check computations by systematically differentiating to ensure consistency.

Second-Order Partial Derivatives

Second-order partial derivatives represent the rate of change of the rate of change of a function. That means we are differentiating a first-order partial derivative again. Consider the steps:

1. Differentiate \( D_{1} f(x, y)\) with respect to x for \( D_{11} f(x, y)\):
\[ D_{11} f(x, y) = \frac{\partial}{\partial x} \left( \frac{-6y}{x^3\sqrt{x^4 - 9y^2}} \right) \]
2. Differentiate \( D_{2} f(x, y)\) with respect to y for \( D_{22} f(x, y)\):
\[ D_{22} f(x, y) = \frac{\partial}{\partial y} \left( \frac{3/x^2}{\sqrt{1-(\frac{3y}{x^2})^2}} \right) \]
These not only tell us about the curvature of our function in individual directions but also help in constructing more complex models involving multiple variables. Applying the quotient rule and chain rule where necessary simplifies these derivatives and ensures accurate results. Second-order derivatives are critical in many contexts, such as optimization and physics.