Explanation Formula used: If h = f ( x , y , z , t ) , then the partial derivative functions are, f x ( x , y , z , t ) = f x = ∂ ∂ x f ( x , y , z , t ) f y ( x , y , z , t ) = f y = ∂ ∂ y f ( x , y , z , t ) f z ( x , y , z , t ) = f z = ∂ ∂ z f ( x , y , z , t ) f t ( x , y , z , t ) = f t = ∂ ∂ t f ( x , y , z , t ) Calculation: The given function is, h ( x , y , z , t ) = x 2 y cos ( z t ) (1) Take the partial derivative with respect to x for the equation (1), ∂ h ∂ x = ∂ ∂ x ( h ( x , y , z , t ) ) = ∂ ∂ x ( x 2 y cos ( z t ) ) = y cos ( z t ) ( 2 x ) = 2 x y cos ( z t ) Take the partial derivative with respect to y for the equation (1), ∂ h ∂ y = ∂ ∂ y ( h ( x , y , z , t ) ) = ∂ ∂ y ( x 2 y cos ( z t ) ) = x 2 cos ( <

8th Edition

ISBN: 9781337771474

Author: Stewart

Publisher: Cengage

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Question

Chapter 14.3, Problem 37E

To determine

To find: The first order partial derivatives of the function h(x,y,z,t)=x2ycos(zt) .

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Chapter 14.3, Problem 36E

Chapter 14.3, Problem 38E

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Chapter 14 Solutions

Calculus:Early...-Webassign Access

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The first order partial derivatives of the function h ( x , y , z , t ) = x 2 y cos ( z t ) .

Solution Summary: The author explains the first order partial derivatives of the function h(x,y,z,t).

To find: The first order partial derivatives of the function h(x,y,z,t)=x2ycos(zt) .

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Chapter 14 Solutions