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Partial derivatives Find the first partial derivatives of the following functions. $$f(x, y)=1-\cos (2(x+y))+\cos ^{2}(x+y)$$

Short Answer

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Question: Find the partial derivatives of the function $$f(x, y) = 1 - \cos (2(x+y)) + \cos ^2 (x+y)$$ with respect to x and y. Answer: The partial derivatives of the function f(x, y) are: $$\frac{\partial f}{\partial x} = -2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y)$$ $$\frac{\partial f}{\partial y} = -2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y)$$

Step by step solution

01

Write the given function and identify terms to differentiate separately

Given function is: $$f(x, y) = 1 - \cos (2(x+y)) + \cos ^2 (x+y)$$ We have three terms in this function: a constant term (1), a cosine term (-cos(2(x+y))) and a squared cosine term (cos^2(x+y)). We will find the derivatives of each term separately and then combine them to get the partial derivatives of the entire function.
02

Differentiation of constant term

The constant term here is 1. The derivative of a constant with respect to any variable is always 0.
03

Differentiation of cosine term

The second term is -cos(2(x+y)). We will find its partial derivatives with respect to x and y using the chain rule. Partial derivative with respect to x: $$\frac{\partial}{\partial x} (-\cos(2(x+y))) = -\sin(2(x+y)) \cdot \frac{\partial}{\partial x}(2(x+y))$$ and $$\frac{\partial}{\partial x}(2(x+y)) = 2$$ So, $$\frac{\partial}{\partial x} (-\cos(2(x+y))) = -2\sin(2(x+y))$$ Partial derivative with respect to y: $$\frac{\partial}{\partial y} (-\cos(2(x+y))) = -\sin(2(x+y)) \cdot \frac{\partial}{\partial y}(2(x+y))$$ and $$\frac{\partial}{\partial y}(2(x+y)) = 2$$ So, $$\frac{\partial}{\partial y} (-\cos(2(x+y))) = -2\sin(2(x+y))$$
04

Differentiation of squared cosine term

The third term is cos^2(x+y). We will find its partial derivatives with respect to x and y using the chain rule. Partial derivative with respect to x: $$\frac{\partial}{\partial x} (\cos^2 (x+y)) = 2\cos(x+y) \cdot \frac{\partial}{\partial x}(\cos(x+y))$$ and $$\frac{\partial}{\partial x}(\cos(x+y)) = -\sin(x+y)$$ So, $$\frac{\partial}{\partial x} (\cos^2 (x+y)) = -2\cos(x+y)\sin(x+y)$$ Partial derivative with respect to y: $$\frac{\partial}{\partial y} (\cos^2 (x+y)) = 2\cos(x+y) \cdot \frac{\partial}{\partial y}(\cos(x+y))$$ and $$\frac{\partial}{\partial y}(\cos(x+y)) = -\sin(x+y)$$ So, $$\frac{\partial}{\partial y} (\cos^2 (x+y)) = -2\cos(x+y)\sin(x+y)$$
05

Combine the partial derivatives for f(x, y)

Now we will combine the results from Steps 2-4 to get the complete partial derivatives for f(x, y). Partial derivative with respect to x: $$\frac{\partial f}{\partial x} = 0 - 2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y) = -2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y)$$ Partial derivative with respect to y: $$\frac{\partial f}{\partial y} = 0 - 2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y) = -2\sin(2(x+y)) - 2\cos(x+y)\sin(x+y)$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Chain Rule in Calculus
The chain rule is a fundamental concept in calculus, especially when dealing with complex functions. In essence, it’s a method used to calculate the derivative of a composite function. A composite function is simply a function made by composing two or more functions together. For example, if you have a function within another function (like a layered cake), the chain rule helps find the rate of change of this composition.
Let's break it down:
  • Consider two functions, say \( g(u) \) and \( u(x) \). The derivative of the composite function \( g(u(x)) \) with respect to \( x \) is given by \( \frac{dg}{du} \cdot \frac{du}{dx} \).
  • The chain rule is essential when differentiating expressions like \( \cos(2(x+y)) \) because they consist of a function inside another function.
In our example, we use the chain rule to differentiate terms like \( -\cos(2(x+y)) \), where the derivative involves both the outer and inner function derivatives.
Working with Trigonometric Functions
Trigonometric functions such as sine and cosine are frequently encountered in calculus. They describe periodic oscillations and are essential in many fields, such as physics and engineering. The basic trigonometric functions include \( \sin \), \( \cos \), and \( \tan \), each with specific properties and derivatives.
Here's a quick guide on their derivatives:
  • The derivative of \( \sin(u) \) is \( \cos(u) \cdot \frac{du}{dx} \).
  • The derivative of \( \cos(u) \) is \( -\sin(u) \cdot \frac{du}{dx} \).
In partial differentiation, these derivatives help us understand how a function changes with respect to one of its variables, while the others are held constant. For instance, in \( -\cos(2(x+y)) \), the derivative brings out the \( -\sin(2(x+y)) \) factor, demonstrating the chain rule in action again. It's essential to know these fundamental properties, as they form the building blocks for solving more complex calculus problems.
Basics of Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions of several variables. It aims to understand how these functions change in multiple dimensions. This area of calculus is essential in fields where systems depend on several factors simultaneously, like economics and fluid dynamics.
In multivariable calculus, we often deal with the following components:
  • **Partial Derivatives:** This represents the derivative of a function with respect to one variable while keeping others constant. It helps us understand the function's behavior in one specific direction.
  • **Gradient Vector:** It gives the direction of the steepest ascent in a multivariable function and points out where the function increases the fastest.
For the function \( f(x, y) = 1 - \cos(2(x+y)) + \cos^2(x+y) \), calculating partial derivatives involves looking at how changing one variable (either \( x \) or \( y \)) affects the function's output, while the other variable is held constant. This understanding is crucial for solving optimization problems and analyzing multidimensional data.

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Most popular questions from this chapter

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