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Find the indicated derivative for the following functions. $$\begin{aligned}&\partial w / \partial x, \text { where } w=\cos z-\cos x \cos y+\sin x \sin y, \text { and } z=x+y \end{aligned}$$

Short Answer

Expert verified
Answer: The partial derivative of the function \(w\) with respect to \(x\) is \(\frac{\partial w}{\partial x} = -\sin (x + y)\).

Step by step solution

01

Identify the given functions

We are given the following functions: $$ w = \cos z - \cos x \cos y + \sin x \sin y $$ and $$ z = x + y $$ The task is to find the partial derivative of \(w\) with respect to \(x\), which can be represented as \(\partial w / \partial x\).
02

Apply the chain rule to the function \(w\) and find its partial derivative with respect to \(z\)

Using the chain rule we get, $$ \frac{\partial w}{\partial x} = \frac{\partial w}{\partial z} \frac{\partial z}{\partial x} $$ Now, we need to find the first part of this expression, which is the partial derivative of \(w\) with respect to \(z\). This can be computed as follows: $$ \frac{\partial w}{\partial z} = \frac{\partial (\cos z - \cos x \cos y + \sin x \sin y)}{\partial z} $$ Differentiating each term one by one, we get: $$ \frac{\partial w}{\partial z} = -\sin z - 0 + 0 = -\sin z $$
03

Find the partial derivative of \(z\) with respect to \(x\)

Next, we need to find the partial derivative of \(z\) with respect to \(x\): $$ \frac{\partial z}{\partial x} = \frac{\partial (x + y)}{\partial x} $$ Differentiating with respect to \(x\), we get: $$ \frac{\partial z}{\partial x} = 1 $$
04

Combine the two partial derivatives to find the final answer

Now that we have both partial derivatives, we can multiply them together to find the partial derivative of \(w\) with respect to \(x\): $$ \frac{\partial w}{\partial x} = \frac{\partial w}{\partial z} \frac{\partial z}{\partial x} = (-\sin z) (1) = -\sin z $$
05

Replace \(z\) with the expression \(x + y\) to find the final answer

Finally, we know that \(z = x + y\), so we can replace \(z\) in the expression we just found with this relationship: $$ \frac{\partial w}{\partial x} = -\sin (x + y) $$ Therefore, the partial derivative of the function \(w\) with respect to \(x\) is: $$ \frac{\partial w}{\partial x} = -\sin (x + y) $$

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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 Partial Derivatives
The chain rule is a fundamental formula in calculus that is crucial when dealing with functions of multiple variables. It helps us find the derivative of composite functions, where one function is nested inside another.
When applying the chain rule to partial derivatives, it allows us to determine how a change in one variable affects another through an intermediary variable. This is particularly important in multivariable calculus.
Consider a function like the one in our exercise: a function, \(w\), that depends indirectly on \(x\) through another variable, \(z\). The chain rule helps us calculate how \(w\) changes with respect to \(x\), even though the function is not directly expressing \(w\) in terms of \(x\).
  • The formula used in our example is \( \frac{\partial w}{\partial x} = \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial x} \).
  • We first compute \( \frac{\partial w}{\partial z} \), which gives how \(w\) changes with \(z\).
  • Then, \( \frac{\partial z}{\partial x} \), which shows how \(z\) changes with \(x\).
  • Multiplying these provides the rate at which \(w\) changes with \(x\), capturing the indirect relationship through \(z\).
Through these steps, the chain rule becomes a powerful tool to handle complex dependencies in differentiating multivariable functions.
Key Properties of Trigonometric Functions
Trigonometric functions like sine and cosine are essential in mathematics, especially in calculus and physics, because they model periodic phenomena and have distinctive derivative properties.
In our specific problem, the function \(w\) contains trigonometric expressions such as \( \cos z \) and \( \sin y \). Understanding the derivatives of these functions is key to solving the exercise.
For trigonometric functions:
  • The derivative of \( \cos x \) with respect to \(x\) is \(-\sin x\).
  • The derivative of \( \sin x \) with respect to \(x\) is \(\cos x\).

In this exercise, when solving \( \frac{\partial w}{\partial z} \), we differentiate \( \cos z \) as \(-\sin z\). Knowing these basic derivative rules is fundamental for tackling more complex calculus problems.
Moreover, trigonometric functions often occur in composite forms, as seen with \(w\). Practicing their differentiation helps develop a strong foundational skill set in calculus.
Understanding Functions of Multiple Variables
A function of multiple variables involves more than one input, such as \(w(x, y, z)\), where \(w\) is defined in terms of \(x\), \(y\), and \(z\). These types of functions model situations where outcomes depend on several factors.
In the current exercise, \(w\) is expressed as a combination of sine and cosine functions of \(x\), \(y\), and \(z\).
When evaluating partial derivatives for such functions:
  • We look at how the function changes with respect to one variable, keeping the others constant.
  • This approach allows us to see the effect of each separate input.
  • Understanding the partial derivative \( \frac{\partial w}{\partial x} \) involves calculating how \(w\) depends on \(x\) while \(y\) and \(z\) remain fixed.

This concept is vital in fields like physics and engineering, where systems depend on various factors. Mastering the handling of these derivatives paves the way for solving equations and modeling complex real-world relationships.

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