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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![### Partial Derivatives Using the Chain Rule
Consider the function \( z = (x + y) e^y \) where \( x = u^2 + v^2 \) and \( y = u^2 - v^2 \). We are required to determine the following partial derivatives using the chain rule and express the answers as functions of \( u \) and \( v \):
1. \( \frac{\partial z}{\partial u} \)
2. \( \frac{\partial z}{\partial v} \)
To find these derivatives, we will use the chain rule which relates the derivative of \( z \) with respect to \( u \) and \( v \) through \( x \) and \( y \). Here is the detailed breakdown of the steps involved:
#### Derivatives to Compute:
\[ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u} \]
\[ \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v} \]
#### Derivatives of Intermediate Functions:
1. Compute \( \frac{\partial x}{\partial u} \) and \( \frac{\partial x}{\partial v} \):
\[ x = u^2 + v^2 \Rightarrow \frac{\partial x}{\partial u} = 2u, \quad \frac{\partial x}{\partial v} = 2v \]
2. Compute \( \frac{\partial y}{\partial u} \) and \( \frac{\partial y}{\partial v} \):
\[ y = u^2 - v^2 \Rightarrow \frac{\partial y}{\partial u} = 2u, \quad \frac{\partial y}{\partial v} = -2v \]
### Final Expression:
Substitute the partial derivatives back into the chain rule equations and simplify to express the partial derivatives \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \) as functions of \( u \) and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa824ef54-0ee2-4591-9058-ff8a94975446%2F0b16498a-44a6-4cf5-ac96-4af769df7e9e%2Fqhkehuv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Partial Derivatives Using the Chain Rule
Consider the function \( z = (x + y) e^y \) where \( x = u^2 + v^2 \) and \( y = u^2 - v^2 \). We are required to determine the following partial derivatives using the chain rule and express the answers as functions of \( u \) and \( v \):
1. \( \frac{\partial z}{\partial u} \)
2. \( \frac{\partial z}{\partial v} \)
To find these derivatives, we will use the chain rule which relates the derivative of \( z \) with respect to \( u \) and \( v \) through \( x \) and \( y \). Here is the detailed breakdown of the steps involved:
#### Derivatives to Compute:
\[ \frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u} \]
\[ \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v} \]
#### Derivatives of Intermediate Functions:
1. Compute \( \frac{\partial x}{\partial u} \) and \( \frac{\partial x}{\partial v} \):
\[ x = u^2 + v^2 \Rightarrow \frac{\partial x}{\partial u} = 2u, \quad \frac{\partial x}{\partial v} = 2v \]
2. Compute \( \frac{\partial y}{\partial u} \) and \( \frac{\partial y}{\partial v} \):
\[ y = u^2 - v^2 \Rightarrow \frac{\partial y}{\partial u} = 2u, \quad \frac{\partial y}{\partial v} = -2v \]
### Final Expression:
Substitute the partial derivatives back into the chain rule equations and simplify to express the partial derivatives \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \) as functions of \( u \) and
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