Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Calculating Partial Derivatives Using the Chain Rule
To illustrate the process of calculating partial derivatives using the chain rule, consider the following functions:
\[ g(x, y) = \sin(x^2 + y^2) \]
\[ x = u - v \]
\[ y = 4v - 2u \]
We aim to express the partial derivatives in terms of the independent variables \( u \) and \( v \).
#### Steps:
1. **Define the Function and Substitutions:**
- The main function is \( g(x, y) = \sin(x^2 + y^2) \).
- The variable \( x \) is defined as \( x = u - v \).
- The variable \( y \) is defined as \( y = 4v - 2u \).
2. **Apply the Chain Rule:**
- For the partial derivative of \( g \) with respect to \( u \):
\[
\frac{\partial g}{\partial u} = \frac{\partial g}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial g}{\partial y} \frac{\partial y}{\partial u}
\]
- For the partial derivative of \( g \) with respect to \( v \):
\[
\frac{\partial g}{\partial v} = \frac{\partial g}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial g}{\partial y} \frac{\partial y}{\partial v}
\]
3. **Calculate Required Derivatives:**
- Compute \(\frac{\partial x}{\partial u}\) and \(\frac{\partial x}{\partial v}\):
\[
\frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = -1
\]
- Compute \(\frac{\partial y}{\partial u}\) and \(\frac{\partial y}{\partial v}\):
\[
\frac{\partial y}{\partial u} = -2, \quad \frac{\partial y}{\partial v} = 4
\]
- Compute \(\frac{\partial g}{\partial x}\) and \(\frac{\partial g}{\partial y](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa824ef54-0ee2-4591-9058-ff8a94975446%2F7e094488-cda5-48cb-ae2a-98eafc75f93f%2Fhyf9c2t_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculating Partial Derivatives Using the Chain Rule
To illustrate the process of calculating partial derivatives using the chain rule, consider the following functions:
\[ g(x, y) = \sin(x^2 + y^2) \]
\[ x = u - v \]
\[ y = 4v - 2u \]
We aim to express the partial derivatives in terms of the independent variables \( u \) and \( v \).
#### Steps:
1. **Define the Function and Substitutions:**
- The main function is \( g(x, y) = \sin(x^2 + y^2) \).
- The variable \( x \) is defined as \( x = u - v \).
- The variable \( y \) is defined as \( y = 4v - 2u \).
2. **Apply the Chain Rule:**
- For the partial derivative of \( g \) with respect to \( u \):
\[
\frac{\partial g}{\partial u} = \frac{\partial g}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial g}{\partial y} \frac{\partial y}{\partial u}
\]
- For the partial derivative of \( g \) with respect to \( v \):
\[
\frac{\partial g}{\partial v} = \frac{\partial g}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial g}{\partial y} \frac{\partial y}{\partial v}
\]
3. **Calculate Required Derivatives:**
- Compute \(\frac{\partial x}{\partial u}\) and \(\frac{\partial x}{\partial v}\):
\[
\frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = -1
\]
- Compute \(\frac{\partial y}{\partial u}\) and \(\frac{\partial y}{\partial v}\):
\[
\frac{\partial y}{\partial u} = -2, \quad \frac{\partial y}{\partial v} = 4
\]
- Compute \(\frac{\partial g}{\partial x}\) and \(\frac{\partial g}{\partial y
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