Problem 23 - APEX Calculus

Question

Find \(f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}\) and \(f_{y x}\). $$ f(x, y)=3 x^{2}+1 $$

Step-by-Step Solution

Answer

\( f_x = 6x, f_y = 0, f_{xx} = 6, f_{yy} = 0, f_{xy} = 0, f_{yx} = 0 \).

Step 1: Find the Partial Derivative with Respect to x

To find the derivative of the function with respect to \( x \), treat \( y \) as a constant. The function is \( f(x, y) = 3x^2 + 1 \). The derivative of \( 3x^2 \) with respect to \( x \) is \( 6x \), and the derivative of 1 is 0. Thus, \( f_x = 6x \).

Step 2: Find the Partial Derivative with Respect to y

To find the derivative of the function with respect to \( y \), treat \( x \) as a constant. Since there is no \( y \) in the equation, the derivative is 0. Therefore, \( f_y = 0 \).

Step 3: Find the Second Partial Derivative with Respect to x

Take the derivative of \( f_x \) with respect to \( x \) again. We found \( f_x = 6x \). Differentiating \( 6x \) with respect to \( x \) gives \( 6 \). Thus, \( f_{xx} = 6 \).

Step 4: Find the Second Partial Derivative with Respect to y

Since \( f_y = 0 \), differentiate 0 with respect to \( y \). The result is still 0. Thus, \( f_{yy} = 0 \).

Step 5: Find the Cross Partial Derivative f(x, y) first with Respect to x and then y

Find \( f_{xy} \) by differentiating \( f_x = 6x \) with respect to \( y \). Since \( 6x \) does not depend on \( y \), \( f_{xy} = 0 \).

Step 6: Find the Cross Partial Derivative f(x, y) first with Respect to y and then x

Find \( f_{yx} \) by differentiating \( f_y = 0 \) with respect to \( x \). Since \( f_y = 0 \), \( f_{yx} = 0 \).

Key Concepts

Mixed Partial Derivatives

Mixed partial derivatives refer to the process of differentiating a function with respect to two or more different variables. In essence, you take the partial derivative with respect to one variable and then take another partial derivative of the result with respect to a different variable. For a function of two variables, such as \( f(x, y) \), the mixed partial derivatives are often denoted as \( f_{xy} \) and \( f_{yx} \). These represent differentiating first with respect to \( x \) and then \( y \), and vice versa.

A common feature of continuous functions is that the mixed partial derivatives \( f_{xy} \) and \( f_{yx} \) are equal, a result known as Clairaut's theorem, assuming the second partial derivatives are continuous. This is generally expressed as \( f_{xy} = f_{yx} \). However, it's essential to verify this condition before assuming equality. In contexts where the derivatives may not be continuous, it's possible for mixed partial derivatives to differ.

Given the function \( f(x, y) = 3x^2 + 1 \), the mixed partial derivatives \( f_{xy} \) and \( f_{yx} \) both equal 0 because each term in the function involving \( x \) or \( y \) is treated as a constant when differentiating with respect to the other variable. By applying the zero derivative rule, when a term does not depend on the variable, its derivative is zero.

Second Partial Derivatives

Second partial derivatives are derivatives of the first partial derivatives of a function. They provide information about the curvature of the function and how it changes as you move around its domain. For a function \( f(x, y) \), the second partial derivatives can be \( f_{xx} \), \( f_{yy} \), \( f_{xy} \), and \( f_{yx} \).

To compute these, differentiate the first partial derivative again. If you start with \( f_x \), differentiating it with respect to \( x \) again gives \( f_{xx} \). Similarly, \( f_y \) differentiated with respect to \( y \) gives \( f_{yy} \). The first gives insight into changes in the function in the direction of \( x \), while the second does so in the direction of \( y \).

In our exercise, for the function \( f(x, y) = 3x^2 + 1 \), we found \( f_x = 6x \). Differentiating this again with respect to \( x \) yields \( f_{xx} = 6 \). Meanwhile, \( f_y \) is 0, so its derivative with respect to \( y \) results in \( f_{yy} = 0 \). This indicates the change of \( f \) with respect to \( y \) remains constant since \( f \) doesn't depend on \( y \).

Calculus

Calculus is a foundational pillar of modern mathematics, dealing with continuous change. It consists of two main branches: differential calculus and integral calculus. Differential calculus focuses on the concept of a derivative, which is a measure of how a function changes as its input changes.

Partial derivatives are a key concept in differential calculus. They extend the idea of taking derivatives to functions of several variables, exposing how a function changes in different directions within its domain.

Understanding partial derivatives is crucial for multivariable calculus, which is widely used in various fields like physics, engineering, and economics. It allows the analysis of multidimensional systems by isolating how changes in each variable influence the system overall.

In practice, partial derivatives help determine tangent planes to surfaces, optimize functions to find maximum and minimum points, and solve systems of equations that arise in modeling real-world scenarios.