### Partial Derivatives Using the Chain RuleConsider 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

Given, z=x+yeyx=u2+v2y=u2-v2 To find : ∂z∂u , ∂z∂v

Answered: If z=(X+Y) e and X-u * find the…

If z=(X+Y) e and X-u * find the tollawing partial derivatives using Chain ruleo Express answers as functions of aond v tv? and y=U?-v; Dz=

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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Question

### 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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