Problem 2 - Active Calculus Multivariable

Question

Find the first partial derivatives of \(f(x, y)=\sin (x-y)\) at the point \((-4,\) -4) A. \(f_{x}(-4,-4)=\) _________. B. \(f_{y}(-4,-4)=\) _________.

Step-by-Step Solution

Answer

A. \(f_{x}(-4,-4)=1\) B. \(f_{y}(-4,-4)=-1\)

Step 1: Compute the partial derivative with respect to x

To compute the first partial derivative of the function with respect to \(x\), we will treat all other variables (in this case, just \(y\)) as constants. We differentiate the function with respect to \(x\): \(f_x(x, y) = \frac{\partial}{\partial x} \sin(x - y)\) Now, apply the chain rule to find the derivative: \(f_x(x, y) = \cos(x - y) \cdot \frac{\partial}{\partial x}(x - y)\) Next, compute the derivative of the expression inside the cosine function: \(\frac{\partial}{\partial x}(x - y) = 1\) Substitute this result back into the expression for the derivative: \(f_x(x, y) = \cos(x - y) \cdot 1 = \cos(x - y)\)

Step 2: Compute the partial derivative with respect to y

Now, we will compute the first partial derivative of the function with respect to \(y\), treating all other variables (in this case, just \(x\)) as constants. We differentiate the function with respect to \(y\): \(f_y(x, y) = \frac{\partial}{\partial y} \sin(x - y)\) Again, apply the chain rule to find the derivative: \(f_y(x, y) = \cos(x - y) \cdot \frac{\partial}{\partial y}(x - y)\) Compute the derivative of the expression inside the cosine function: \(\frac{\partial}{\partial y}(x - y) = -1\) Substitute this result back into the expression for the derivative: \(f_y(x, y) = \cos(x - y) \cdot (-1) = -\cos(x - y)\)

Step 3: Evaluate the partial derivatives at the point (-4, -4)

Now that we have computed the partial derivatives, we can evaluate them at the point \((-4, -4)\): A. For the partial derivative with respect to \(x\): \(f_x(-4, -4) = \cos((-4) - (-4)) = \cos(0) = 1\) B. For the partial derivative with respect to \(y\): \(f_y(-4, -4) = -\cos((-4) - (-4)) = -\cos(0) = -1\) Therefore, the first partial derivatives of the function at the point \((-4, -4)\) are: A. \(f_{x}(-4,-4)=1\) B. \(f_{y}(-4,-4)=-1\)

Key Concepts

Chain Rule in Calculus

The chain rule is an essential tool in calculus, particularly when dealing with composite functions. If you have a function nested within another, the chain rule helps you differentiate it. Imagine you want to differentiate a function like \(\sin(x-y)\). Since this function has a composition of sine and subtraction, you apply the chain rule.
The chain rule states: If \(u(x, y) = x - y\), and \(f(u) = \sin(u)\), then the derivative \(f'(u)\) is given as:

• \(f_x = f'(u) \cdot u_x\)
• \(f_y = f'(u) \cdot u_y\)

Here, \(f'(x) = \cos(x-y)\), so you apply the derivative of \(u\) with respect to each variable. Thus, if you differentiate with respect to \(x\), you treat \(y\) as a constant, and vice-versa when differentiating with respect to \(y\).
In essence, the chain rule allows us to navigate through functions in different layers, making it possible to work with complex compositions systematically.

Introduction to Multivariable Calculus

Multivariable calculus extends the principles of calculus from functions of a single variable to functions of two or more variables. This field explores how functions behave in higher dimensions. Unlike single-variable functions, multivariable functions like \(f(x, y) = \sin(x - y)\) need partial derivatives.
Partial derivatives are taken by differentiating each variable independently while keeping all other variables constant. For example:

• \(f_x\) denotes the partial derivative with respect to \(x\), treating \(y\) as a constant.
• \(f_y\) denotes the partial derivative with respect to \(y\), treating \(x\) as a constant.

By finding partial derivatives, we can understand how the function changes along each variable independently. These variations are crucial for analyzing surfaces, gradients, and optimizing multivariable functions.
Multivariable calculus is abundant in applications like economics, engineering, and physics where systems are governed by multiple inputs. Understanding how to approach these functions is a cornerstone of advanced calculus.

Trigonometric Functions and Their Derivatives

Trigonometric functions, such as sine and cosine, are fundamental in calculus and appear frequently in both theoretical and applied mathematics. One useful property is their periodicity and smooth, wave-like nature, which is vital in modeling cyclic behaviors.
Differentiating trigonometric functions involves some typical results:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).

When these functions appear in expressions like \(\sin(x - y)\), you use derivatives by following differentiation rules and properties such as the chain rule. Evaluating these derivatives at specific points, like \((-4, -4)\), can further simplify problems and obtain values essential for assessments and interpretations.
These trigonometric concepts are foundational in fields ranging from electrical engineering to signal processing, often helping to analyze waves, oscillations, and even quantum phenomena. Understanding their derivatives allows us to explore complex calculus problems involving periodic components effectively.