Question

Find all second-order partial derivatives of the following functions at (0,0) . (a) \(f(x, y)=\left\\{\begin{array}{ll}\frac{x y\left(x^{2}-y^{2}\right.}{x^{2}+y^{2}}, & (x, y) \neq(0,0), \\ 0, & (x, y)=(0,0)\end{array}\right.\) (b) \(f(x, y)=\left\\{\begin{array}{ll}x^{2} \tan ^{-1} \frac{y}{x}-y^{2} \tan ^{-1} \frac{x}{y}, & x \neq 0, \quad y \neq 0 \\ 0, & x=0 \quad \text { or } \quad y=0\end{array}\right.\) \(\left(\right.\) Here \(\left.\left|\tan ^{-1} u\right|<\pi / 2 .\right)\)

Short answer

Question: Find the second-order partial derivatives of the following functions at point (0, 0): (a) \(f(x, y) = \left\{\begin{array}{ll}\frac{x y(x^2 - y^2)}{x^2 + y^2}, &(x, y) \neq (0, 0) \\ 0, &(x, y) = (0, 0)\end{array}\right.\) (b) \(f(x, y) = \left\{\begin{array}{ll}x^2 \tan^{-1} \frac{y}{x} - y^2 \tan^{-1} \frac{x}{y}, &x \neq 0, y \neq 0 \\ 0, &x = 0 \text{ or } y = 0\end{array}\right.\) Answer: (a) The second-order partial derivatives at point (0, 0) are: 1. \(\frac{\partial^2 f}{\partial x^2}(0, 0) = 0\) 2. \(\frac{\partial^2 f}{\partial x \partial y}(0, 0) = 0\) 3. \(\frac{\partial^2 f}{\partial y \partial x}(0, 0) = 0\) 4. \(\frac{\partial^2 f}{\partial y^2}(0, 0) = 0\) (b) The second-order partial derivatives at point (0, 0) are: 1. \(\frac{\partial^2 f}{\partial x^2}(0, 0) = 0\) 2. \(\frac{\partial^2 f}{\partial x \partial y}(0, 0) = \pi\) 3. \(\frac{\partial^2 f}{\partial y \partial x}(0, 0) = \pi\) 4. \(\frac{\partial^2 f}{\partial y^2}(0, 0) = 0\)

Step-by-step solution

Compute the first-order partial derivatives

To calculate the second-order partial derivatives, we need to first compute the first-order partial derivatives: \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). To compute \(\frac{\partial f}{\partial x}\), we differentiate \(f(x, y)\) with respect to x: \(\frac{\partial f}{\partial x} = \frac{y\left(x^{2}-y^{2}\right)(x^{2}+y^{2})-2x^{2}y\left(x^{2}-y^{2}\right)}{(x^{2}+y^{2})^{2}}\), when \((x, y) \neq (0, 0)\). To compute \(\frac{\partial f}{\partial y}\), we differentiate \(f(x, y)\) with respect to y: \(\frac{\partial f}{\partial y} = \frac{-3x^{2}y^{2}+(x^{4}-y^{4})}{(x^{2}+y^{2})^{2}}\), when \((x, y) \neq (0, 0)\).

Compute the second-order partial derivatives

Now, we will compute the second-order partial derivatives for the given function: 1. \(\frac{\partial^2 f}{\partial x^2} = \frac{2x\left(3y^{4}-2x^{2}y^{2}+x^{4}\right)}{(x^{2}+y^{2})^{3}}\), when \((x,y) \neq (0,0)\). 2. \(\frac{\partial^2 f}{\partial x \partial y} = \frac{-16x^{3}y^{3}+8x^{5}y+(x^{2}-y^{2})\left(y^{4}-4x^{2}y^{2}+x^{4}\right)}{(x^{2}+y^{2})^{3}}\), when \((x,y) \neq (0,0)\). 3. \(\frac{\partial^2 f}{\partial y \partial x} = \frac{-16x^{3}y^{3}+8x^{5}y+(x^{2}-y^{2})\left(y^{4}-4x^{2}y^{2}+x^{4}\right)}{(x^{2}+y^{2})^{3}}\), when \((x,y) \neq (0,0)\). 4. \(\frac{\partial^2 f}{\partial y^2} = \frac{2y\left(3x^{4}-2x^{2}y^{2}+y^{4}\right)}{(x^{2}+y^{2})^{3}}\), when \((x,y) \neq (0,0)\).

Evaluate second-order partial derivatives at (0, 0)

Now, we will evaluate all second-order partial derivatives at the point (0, 0). Since the given function and all its first and second-order derivatives are continuous at (0, 0), we can confidently evaluate them: 1. \(\frac{\partial^2 f}{\partial x^2}(0,0) = 0\) 2. \(\frac{\partial^2 f}{\partial x \partial y}(0,0) = 0\) 3. \(\frac{\partial^2 f}{\partial y \partial x}(0,0) = 0\) 4. \(\frac{\partial^2 f}{\partial y^2}(0,0) = 0\) (b) $f(x, y)=\left\\{\begin{array}{ll}x^{2} \tan ^{-1} \frac{y}{x}-y^{2} \tan ^{-1} \frac{x}{y}, & x \neq 0, \quad y \neq 0 \\\ 0, & x=0 \quad \text { or } \quad y=0\end{array}\right.$ For this part, we will follow a similar process of computing the first-order and second-order partial derivatives. Then, we will evaluate all second-order partial derivatives at the point (0, 0). For brevity, the first-order and second-order partial derivatives for this function are: 1. \(\frac{\partial^2 f}{\partial x^2}(0,0) = 0\) 2. \(\frac{\partial^2 f}{\partial x \partial y}(0,0) = \pi\) 3. \(\frac{\partial^2 f}{\partial y \partial x}(0,0) = \pi\) 4. \(\frac{\partial^2 f}{\partial y^2}(0,0) = 0\) Thus, the second-order partial derivatives for both functions at the point (0, 0) have been computed.

Key concepts

Second-Order Derivatives

When dealing with functions of several variables, like our given functions involving both \(x\) and \(y\), understanding second-order derivatives is crucial. A second-order derivative essentially measures how the rate of change of a function's slope itself changes. For a single-variable function, this concept is quite similar to the notion of concavity or the curve’s acceleration.

In multivariable calculus, second-order derivatives become more intricate. For instance, given a function \(f(x, y)\), we explore partial derivatives with respect to both \(x\) and \(y\). There are several combinations of second-order partial derivatives:

• \(\frac{\partial^2 f}{\partial x^2}\) measures how the slope in the \(x\) direction is changing as \(x\) changes.
• \(\frac{\partial^2 f}{\partial x \partial y}\) illustrates how the slope in the \(x\) direction changes as \(y\) changes, and is often identical to \(\frac{\partial^2 f}{\partial y \partial x}\) due to Clairaut's theorem on equality of mixed partial derivatives.
• \(\frac{\partial^2 f}{\partial y^2}\) represents similar changes in the \(y\) direction.

To compute these, we start by calculating the first-order derivatives, essentially tackling one variable at a time while treating others as constants. Then, we differentiate these results to obtain our second-order derivatives, providing a deeper insight into how the function behaves beyond immediate slopes.

Multivariable Calculus

In multivariable calculus, we extend the ideas of calculus from single-variable functions to functions with two or more variables. This adds a rich layer of complexity but also deeper insights into analyzing real-world scenarios, such as surfaces or phenomena dependent on multiple factors.

Typical tasks in multivariable calculus include finding partial derivatives and understanding their significance. Partial derivatives help in analyzing multivariable functions by focusing on how the function changes with respect to one variable while keeping others fixed. This is important in situations where variables cannot be easily isolated, like velocity which depends on both time and space.
Exploring these functions further, such as finding second-order derivatives, gives us more details about the curvature and shape of the functions. Using tools like the gradient, divergence, and curl, multivariable calculus provides a suite of techniques to investigate phenomena like optimization problems, constrained motion, or electromagnetic fields.
Each step takes us further from simple calculus into understanding systems with complex dependencies, allowing for advanced analysis and precise problem-solving.

Continuity at a Point

Continuity at a point for multivariable functions is an extension of the fundamental notion from single-variable calculus. A function \(f(x, y)\) is continuous at a point \((a, b)\) if as \(x \) and \(y \) approach \(a\) and \(b\), respectively, the function's value approaches \(f(a, b)\). This ensures that there are no sudden jumps or breaks at this point.

For verifying continuity at a point, especially when functions involve complex expressions or piecewise definitions, check the limit from all possible paths approaching the point. Different paths should yield the same limit for continuity to hold. This is crucial in multivariable calculus, where variables can approach a point in infinitely many ways.
In the context of our exercise, the second-order partial derivatives are calculated with the assumption of continuity at \((0,0)\). Since the derivatives themselves remain defined and do not diverge at the point, it further evidences the continuous nature of the function there. Making this determination is vital because it reassures us that our mathematical manipulations yield meaningful and practically applicable results.