Explanation Consider the given function. z = x 2 + 5 x − 2 y 8 Differentiate the above function on both sides with respect to x . ∂ z ∂ x = ∂ ∂ x x 2 + 5 x − 2 y 8 Use the chain rule ∂ ∂ x f x n = n f x n − 1 ⋅ f ′ x , here n is a real number and simplify the above expression. ∂ z ∂ x = 8 x 2 + 5 x − 2 y 8 − 1 ∂ ∂ x x 2 + 5 x − 2 y = 8 x 2 + 5 x − 2 y 7 ∂ ∂ x x 2 + 5 x − 2 y Use the distributive property of differentiation ∂ ∂ x A + B = ∂ ∂ x A + ∂ ∂ x B and differentiate the above expression. ∂ z ∂ x = 8 x 2 + 5 x − 2 y 7 ∂ ∂ x x 2 + ∂ ∂ x 5 x − ∂ ∂ x 2 y Use the property of differentiation ∂ x n ∂ x = n x n − 1 and simplify the above expression. ∂ z ∂ x = 8 x 2 + 5 x − 2 y 7 x 2 − 1 + 5 x 1 − 1 − 2 y 0 = 8 x 2 + 5 x − 2 y 7 2 x + 5 − 0 = 8 x 2 + 5 x − 2 y 7 2 x + 5 Again, consider the given function. z = x 2 + 5 x − 2 y 8 Differentiate the above function on both sides with respect to y . ∂ z ∂ y = ∂ ∂ y x 2 + 5 x − 2 y 8 Use the chain rule ∂ ∂ x f x n = n f x n − 1 ⋅ f ′ x , here n is a real number and simplify the above expression

Calculus: Early Transcendentals, Enhanced Etext

12th Edition

ISBN: 9781119777984

Author: Howard Anton; Irl C. Bivens; Stephen Davis

Publisher: Wiley Global Education US

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Chapter 13.3, Problem 5ES

To determine

The partial derivatives ∂z∂x and ∂z∂y of the function z=x2+5x−2y8 .

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Chapter 13.3, Problem 4ES

Chapter 13.3, Problem 6ES

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Chapter 13 Solutions

Calculus: Early Transcendentals, Enhanced Etext

Ch. 13.1 - Prob. 1QCE Ch. 13.1 - Let f(x,y)=xyx+y+1....Ch. 13.1 - Let f(x,y)=ex+y. (a) For what values of k will the...Ch. 13.1 - Let f(x,y,z)=1x2+y2+z2+1. (a) For what values of k...Ch. 13.1 - Prob. 1ES Ch. 13.1 - Prob. 2ES Ch. 13.1 - Prob. 3ES Ch. 13.1 - Prob. 4ES Ch. 13.1 - These exercises are concerned with functions of...Ch. 13.1 - Prob. 6ES

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The partial derivatives ∂ z ∂ x and ∂ z ∂ y of the function z = x 2 + 5 x − 2 y 8 .

Solution Summary: The author explains the partial derivatives of the function z=(x2+5x-2y)8.

The partial derivatives ∂z∂x and ∂z∂y of the function z=x2+5x−2y8 .

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Chapter 13 Solutions