Problem 6 - APEX Calculus

Question

Evaluate \(f_{x}(x, y)\) and \(f_{y}(x, y)\) at the indicated point. $$ f(x, y)=x^{3}-3 x+y^{2}-6 y \text { at }(-1,3) $$

Step-by-Step Solution

Answer

Both \( f_x(-1, 3) \) and \( f_y(-1, 3) \) are 0.

Step 1: Understand the Partial Derivatives

We need to calculate the partial derivatives of the function with respect to each variable. The function given is \( f(x, y) = x^3 - 3x + y^2 - 6y \). This means we need to find \( f_x(x, y) \) and \( f_y(x, y) \) by differentiating the function with respect to \( x \) and \( y \) respectively.

Step 2: Differentiate with Respect to x

To find \( f_x(x, y) \), differentiate \( f(x, y) = x^3 - 3x + y^2 - 6y \) with respect to \( x \). Based on basic differentiation rules, we have:\[ f_x(x, y) = \frac{d}{dx}(x^3) - \frac{d}{dx}(3x) + \frac{d}{dx}(y^2) - \frac{d}{dx}(6y) \]The result is:\[ f_x(x, y) = 3x^2 - 3 \] since terms involving \( y \) are treated as constants when differentiating with respect to \( x \).

Step 3: Differentiate with Respect to y

To find \( f_y(x, y) \), differentiate \( f(x, y) = x^3 - 3x + y^2 - 6y \) with respect to \( y \). Using basic differentiation rules, we have:\[ f_y(x, y) = \frac{d}{dy}(x^3) - \frac{d}{dy}(3x) + \frac{d}{dy}(y^2) - \frac{d}{dy}(6y) \]The result is:\[ f_y(x, y) = 2y - 6 \] as terms involving \( x \) are constants when differentiating with respect to \( y \).

Step 4: Evaluate \( f_x \) at \((-1, 3)\)

Plug \( x = -1 \) into the partial derivative \( f_x(x, y) = 3x^2 - 3 \):\[ f_x(-1, 3) = 3(-1)^2 - 3 = 3(1) - 3 = 3 - 3 = 0 \]

Step 5: Evaluate \( f_y \) at \((-1, 3)\)

Plug \( y = 3 \) into the partial derivative \( f_y(x, y) = 2y - 6 \):\[ f_y(-1, 3) = 2(3) - 6 = 6 - 6 = 0 \]

Key Concepts

Differentiation Rules

Differentiation forms the foundation of calculus, enabling us to understand how functions change. In the realm of partial derivatives, we apply these rules to functions with several variables. To determine the derivative of a function with respect to one of its variables, we treat all other variables as constants. This approach is crucial in multivariable calculus.When differentiating a function like \( f(x, y) = x^3 - 3x + y^2 - 6y \) with respect to \( x \), you only focus on terms that involve \( x \). The derivative of each term follows classical rules:

• Power rule: The derivative of \( x^n \) is \( nx^{n-1} \).
• Constant rule: The derivative of a constant is zero.
• Sum and Difference rule: Differentiate terms separately then combine results.

For \( f_x(x, y) \), using the power rule for \( x^3 \) gives \( 3x^2 \), and the derivative of \( -3x \) is \(-3\). Terms \( y^2 \) and \( -6y \) are constants regarding \( x \), hence they become zero. So, the differentiation rules aid in finding \( f_x(x, y) = 3x^2 - 3 \).

Function Evaluation

Function evaluation is the process of finding the output of a function for a specific input value. In this context, it involves substituting the values of \( x \) and \( y \) into the partial derivatives to determine their values at a specific point. For example, once the derivative \( f_x(x, y) = 3x^2 - 3 \) is found, evaluating it at \((-1, 3)\) requires substituting \( x = -1 \) into the expression. It simplifies to \( f_x(-1,3) = 3(-1)^2 - 3 = 0 \). Similarly, evaluate \( f_y(x, y) = 2y - 6 \) at \( y = 3 \) to get \( f_y(-1,3) = 2(3) - 6 = 0 \).Function evaluation is not just a mechanical process; it connects symbolic differentiation to practical applications. You gain insights into the behavior of the function at specific points, and it's these calculated values that inform subsequent decisions in problem-solving scenarios.

Multivariable Calculus

Multivariable calculus extends ordinary calculus into higher dimensions, dealing with functions dependent on more than one variable. This branch is vital for modeling and solving problems in physics, engineering, economics, and various other fields.In multivariable calculus, we often need to understand how functions change with respect to each variable independently, while considering the influence of all other variables. This leads to the concept of partial derivatives, like \( f_x(x, y) \) and \( f_y(x, y) \). Each partial derivative represents the rate of change of the function with respect to one variable, holding others constant.Consider the function \( f(x, y) = x^3 - 3x + y^2 - 6y \). This comprises terms that depend on \( x \) and terms that depend on \( y \), all interacting to determine the output. By taking partial derivatives, we unravel these interactions, providing insight into how changes in each variable affect the overall function.Understanding these nuances is crucial when analyzing surfaces or curves in three-dimensional space. Partial derivatives help in defining tangents, optimizing functions, and tackling complex integrals, making multivariable calculus a powerful tool in advanced mathematics.