Explanation Given: The relation is, x y + x z + y z = 3 . The result of Exercise 48: “If F ( x , y , z ) = 0 is implicitly differentiable function then ∂ z ∂ x = − F x F z and ∂ z ∂ y = − F y F z ”. Calculation: Let the functionbe F ( x , y , z ) = x y + x z + y z − 3 (1) Obtain the equation ∂ z ∂ x by using the above result, ∂ z ∂ x = − F x F z (2) Take the partial derivative with respect to x in the equation (1), ∂ F ∂ x = ∂ ∂ x ( x y + x z + y z − 3 ) = ( 1 ) y + 1 ( z ) + 0 − 0 F x = y + z Thus, the partial derivate, F x = y + z . Take the partial derivative with respect to z in the equation (1), ∂ F ∂ z = ∂ ∂ y ( x y + x z + y z − 3 ) = 0 + x ( 1 ) + y ( 1 ) − 0 F z = x + y Thus, the partial derivate of z , F z = x + y

Calculus: Early Transcendentals, 2nd Edition

2nd Edition

ISBN: 9780321965165

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 12.5, Problem 49E

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To evaluate: The values of ∂z∂x and ∂z∂y using the result of Exercise 48, if xy+xz+yz=3 .

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Chapter 12.5, Problem 48E

Chapter 12.5, Problem 50E

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Chapter 12 Solutions

Calculus: Early Transcendentals, 2nd Edition

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The values of ∂ z ∂ x and ∂ z ∂ y using the result of Exercise 48, if x y + x z + y z = 3 .

Solution Summary: The author evaluates the value of partial x, if F(x,y,z)=0, using the result of Exercise 48.

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To evaluate: The values of ∂z∂x and ∂z∂y using the result of Exercise 48, if xy+xz+yz=3 .

Want to see the full answer?

Chapter 12 Solutions