a. Explanation Given: z = ln x y Where, x = e u v 2 , y = e u v Calculation: To find the partial derivatives ∂ z ∂ u , ∂ z ∂ v , where z = ln x y Where, x = e u v 2 , y = e u v , first lets express z in terms of u and v , as z = ln x y = ln ( e u v 2 e u v ) = ln ( e u v 2 + u v ) = u v 2 + u v Now, to find ∂ z ∂ u differentiate above with respect to u treating v as constant, as ∂ z ∂ u = ∂ b. To determine To find the partial derivatives ∂ z ∂ u , ∂ z ∂ v , where z = ln x y Where, x = e u v 2 , y = e u v

Student's Solution and Survival Manual for Calculus

7th Edition

ISBN: 9781524934040

Author: STRAUSS MONTY J, TODA MAGDALENA DANIELE, SMITH KARL J

Publisher: Kendall Hunt Publishing

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Chapter 11.5, Problem 11PS

To determine

To find the partial derivatives ∂z∂u,∂z∂v , where

z=lnxy

Where, x=euv2,y=euv

To determine

To find the partial derivatives ∂z∂u,∂z∂v , where

z=lnxy

Where, x=euv2,y=euv

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Chapter 11.5, Problem 10PS

Chapter 11.5, Problem 12PS

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Chapter 11 Solutions

Student's Solution and Survival Manual for Calculus

Ch. 11.1 - Prob. 1PS Ch. 11.1 - Prob. 2PS Ch. 11.1 - Prob. 3PS Ch. 11.1 - Prob. 4PS Ch. 11.1 - Prob. 5PS Ch. 11.1 - Prob. 6PS Ch. 11.1 - Prob. 7PS Ch. 11.1 - Prob. 8PS Ch. 11.1 - Prob. 9PS Ch. 11.1 - Prob. 10PS

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To find the partial derivatives ∂ z ∂ u , ∂ z ∂ v , where z = ln x y Where, x = e u v 2 , y = e u v

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To find the partial derivatives ∂z∂u,∂z∂v , where z=lnxy Where, x=euv2,y=euv

To find the partial derivatives ∂z∂u,∂z∂v , where z=lnxy Where, x=euv2,y=euv

Want to see the full answer?

Chapter 11 Solutions

To find the partial derivatives ∂z∂u,∂z∂v , where

z=lnxy

Where, x=euv2,y=euv

To find the partial derivatives ∂z∂u,∂z∂v , where

z=lnxy

Where, x=euv2,y=euv