Question

Verify that \(f_{x y}=f_{y x},\) for \(f(x, y)=2 x^{3}+3 y^{2}+1\).

Short answer

Answer: Yes, the second partial derivatives are equal. Both \(f_{xy}\) and \(f_{yx}\) are equal to \(0\).

Step-by-step solution

Find the first-order partial derivatives

We will now compute the first-order partial derivatives of the given function. $$ f(x, y) = 2x^3 + 3y^2 + 1 $$ The partial derivative with respect to \(x\) is: $$ f_x = \frac{\partial}{\partial x}(2x^3 + 3y^2 + 1) = 6x^2 $$ The partial derivative with respect to \(y\) is: $$ f_y = \frac{\partial}{\partial y}(2x^3 + 3y^2 + 1) = 6y $$

Find the second-order partial derivatives

Now we will compute the second-order partial derivatives of the function. The second-order partial derivative with respect to \(x\) and then \(y\) is: $$ f_{xy} = \frac{\partial}{\partial y}(f_x) = \frac{\partial}{\partial y}(6x^2) = 0 $$ The second-order partial derivative with respect to \(y\) and then \(x\) is: $$ f_{yx} = \frac{\partial}{\partial x}(f_y) = \frac{\partial}{\partial x}(6y) = 0 $$

Compare \(f_{xy}\) and \(f_{yx}\)

As both \(f_{xy}\) and \(f_{yx}\) are equal to \(0\), we can conclude that \(f_{xy} = f_{yx}\). Therefore, the second partial derivatives of the given function are equal, as expected.

Key concepts

Partial Derivative

In the world of multivariable calculus, the partial derivative plays a crucial role. It measures how a function changes as one of its variables changes, while keeping all other variables constant. Imagine you are in a garden with varying terrain, analyzing how steep the ground is if you walk north. That would be similar to taking the partial derivative with respect to the north direction, regardless of the east-west slope.

Let's apply this to our problem. For the function f(x, y) = 2x^3 + 3y^2 + 1, the partial derivative with respect to x, denoted as f_x, is found by treating y as a constant and differentiating only with respect to x. This results in f_x = 6x². Similarly, the partial derivative with respect to y, noted as f_y, is obtained by treating x as a constant, yielding f_y = 6y. These operations provide insight into the function's behavior regarding individual variables.

Mixed Partial Derivative

The concept of a mixed partial derivative occurs when you're interested in how a function's rate of change with respect to one variable is itself changing with respect to another variable. In simpler terms, it tells you how the slope of the hill is changing as you walk north while you're facing east.

To find the mixed partial derivative of f(x, y) from our exercise, we first took the derivative with respect to x to get f_x = 6x², and then we took the derivative of this result with respect to y. Since 6x² does not contain y, its derivative with respect to y is zero, which gives us f_xy = 0. The reverse process involves computing f_y = 6y, and then differentiating this result with respect to x, which also gives us zero. Hence, f_yx = 0. The symmetry observed in these mixed partial derivatives isn't just coincidental, as explained by Clairaut's theorem.

Clairaut's Theorem in Calculus

The symmetry we observed in mixed partial derivatives isn't a happy accident but instead is described by a fundamental result known as Clairaut's theorem. This theorem states that if a function's mixed partial derivatives are continuous around a point, then the order of differentiation does not matter—the mixed partial derivatives are equal.

In the context of our function f(x, y) = 2x^3 + 3y^2 + 1, we've calculated both f_xy and f_yx and found them both to be zero. Clairaut's theorem guarantees us this equality because the involved functions are smooth and continuous, meaning the partial derivatives can be interchanged without affecting the outcome. This property assures us of the consistency and reliability of mixed partial derivatives in analyzing multivariable functions.

Multivariable Calculus

The broader subject that encapsulates these concepts is multivariable calculus. It's the extension of calculus to functions of more than one variable. Just as standard calculus involves the study of rates and cumulative changes in single-variable functions, multivariable calculus generalizes these concepts to multiple dimensions.

For example, in analyzing our function f(x, y), multivariable calculus allows us to explore how changes in x and y independently and collectively affect the value of f. It involves not just partial derivatives but also multiple integrals, vector calculus, and the analysis of vector fields. Multivariable calculus plays a vital role in many disciplines including physics, engineering, economics, and more, providing powerful tools for modeling and solving real-world problems in various spaces and dimensions.