### Calculating Partial Derivatives Using the Chain RuleTo 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

We have to calculate partial derivative by using chain rule. And we have to express derivative in…

Answered: Use the chcin rule to calcuuate the…

Use the chcin rule to calcuuate the partial derivetives. 4v-2u g(x.y)= Sin (x?sy? ), X= u-V Express in terms of the independent variables u,Ve

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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