Problem 22 - The Calculus with Analytic Geometry

Question

In Exercises 21 through 23, find \(f_{12}(0,0)\) and \(f_{21}(0,0)\), if they exist. $$ f(x, y)= \begin{cases}\frac{x^{2} y^{2}}{x^{4}+y^{4}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0)\end{cases} $$

Step-by-Step Solution

Answer

Both \(f_{12}(0,0)\) and \(f_{21}(0,0)\) are 0.

Step 1: Identify the Partial Derivatives

First, determine the first partial derivatives of the function with respect to both variables, x and y. Let's denote these as \(f_x\) and \(f_y\), respectively.

Step 2: Compute \(f_x(x,y)\)

Use the definition of the partial derivative to compute \(f_x(x,y)\).\[f_x(x,y) = \frac{\partial}{\partial x}\left( \frac{x^{2} y^{2}}{x^{4}+y^{4}} \right)\]. Perform the differentiation considering the quotient rule.

Step 3: Compute \(f_y(x,y)\)

Similarly, compute \(f_y(x,y)\) using the definition of the partial derivative. \[f_y(x,y) = \frac{\partial}{\partial y}\left( \frac{x^{2} y^{2}}{x^{4}+y^{4}} \right)\]. Use the quotient rule for differentiation.

Step 4: Evaluate \(f_x(0,0)\) and \(f_y(0,0)\)

Now, compute \(f_x(0,0)\) and \(f_y(0,0)\). Plug in the values of \(x\) and \(y\) into the partial derivatives obtained. Notice that because \(x = 0\) and \(y = 0\) lead to an indeterminate form in the derivatives, \(f_x(0,0)\) and \(f_y(0,0)\) are computed directly from the definition of the function. Both \(f_x(0,0)\) and \(f_y(0,0)\) are 0.

Step 5: Compute Second Partial Derivative \(f_{12}(x, y)\)

Now, compute \(f_{12}(x,y)\) which is the partial derivative of \(f_x\) with respect to y, denoted as \(\frac{ \partial }{ \partial y } \left( f_x \right)\).

Step 6: Compute Second Partial Derivative \(f_{21}(x, y)\)

Similarly, compute \(f_{21}(x,y)\) which is the partial derivative of \(f_y\) with respect to x, denoted as \(\frac{ \partial }{ \partial x } \left( f_y \right)\).

Step 7: Evaluate \(f_{12}(0,0)\) and \(f_{21}(0,0)\)

Finally, evaluate \(f_{12}(0,0)\) and \(f_{21}(0,0)\) similarly to the evaluation of \(f_x(0,0)\) and \(f_y(0,0)\). Because the initial derivatives are 0, their second partial derivatives at \((0,0)\) are also 0.

Key Concepts

Quotient Rule

The quotient rule is essential when dealing with functions that are ratios of two differentiable functions. If you have a function defined as:
f(x) = \frac{u(x)}{v(x)}, then the derivative of f(x) is given by:
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}
In this exercise, our function is
f(x, y) = \frac{x^2 y^2}{x^4 + y^4}.
To find the first partial derivatives \(\frac{\partial f}{\partial x} \) and \(\frac{\partial f}{\partial y} \), we'll use the quotient rule for partial derivatives:

• For \( f_x(x, y): \) u(x, y) = x^2 * y^2 and v(x, y) = x^4 + y^4.
• For \( f_y(x, y): \) u(x, y) = x^2 * y^2 and v(x, y) = x^4 + y^4.

Differentiating these, we use the quotient rule to find the partial derivatives accurately.
The detailed computation involves applying the rule and then simplifying. This helps us determine the behavior of the function at specific points like (0,0).

Indeterminate Forms

In calculus, an indeterminate form is an expression involving two functions whose limit cannot be determined solely from the limits of the individual functions. The expression:
\( \frac{0}{0} \) is a typical example of such an indeterminate form. In our exercise, at the point (0,0), our function's partial derivatives lead to forms like \( \frac{0}{0} \):

• \( \frac{\partial}{\partial x} \left( \frac{x^2 y^2}{x^4 + y^4} \right) \) evaluated at (0,0).
• \( \frac{\partial}{\partial y} \left( \frac{x^2 y^2}{x^4 + y^4} \right) \) evaluated at (0,0).

These forms require special treatment. Typically, approaching them directly or using limits can resolve them. In our solution, we use the definition of the function to understand and evaluate these indeterminate partial derivatives at (0,0), resulting in 0 for both cases. This step ensures a proper evaluation of the derivatives precisely.

Second Partial Derivatives

Second partial derivatives involve taking the partial derivative of a already computed first partial derivative. For the given exercise, this means:

• First, compute \(f_x(x, y)\) and \(f_y(x, y)\).
• Next, find \(f_{12}(x, y)\) by differentiating \(f_x(x, y)\) with respect to y.
• Similarly, calculate \(f_{21}(x, y)\) by differentiating \(f_y(x, y)\) with respect to x.

In general, if the mixed second partial derivatives are continuous in the neighborhood of the point you are interested in (here, (0,0)), Clairaut's Theorem states:
\( \frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} \).
In this exercise, after computing both \( f_{12}(0, 0) \) and \( f_{21}(0, 0) \), we find they are equal, indicating that the calculations are consistent.

Partial Derivative Evaluation

Evaluation of partial derivatives at specific points can sometimes yield indeterminate forms. In this exercise:

• We need to evaluate \( f_x(0, 0) \) and \( f_y(0, 0) \).
• These point evaluations give us critical insights into the behavior of the function around that point.
• By definition of our function and proper limits, we determine these partials correctly.

For the point (0,0), the partial derivatives turn out to be zero due to the function's definition. Furthermore, since the second partial derivatives are derived from these first partial derivatives and also are about (0,0), they too simplify and stay zero. Studying partial derivatives and their evaluations sheds light on understanding the function's local behavior, continuity, and differentiability.