Explanation Explanation Given F(t) is a differentiable function of one variable. Proof The chain rule states that ∇ F ( f ( x , y , z ) ) = F ' ( f ( x , y , z ) ) ∇ f ( x , y , z ) Start from the expression on the right hand side. By the definition of gradient ∇ F ( f ( x , y , z ) ) = ( ∂ F ( f ( x , y , z ) ) ∂ x , ∂ F ( f ( x , y , z ) ) ∂ y , ∂ F ( f ( x , y , z ) ) ∂ z ) Consider the first term ∂ ( F ( f ( x , y , z ) ) ∂ x . By the differentiability of f ( x , y , z ) we can write f ( x + h , y , z ) = f ( x , y , z ) + ∂ f ∂ x h + h k 1 ( h ) (1) where k 1 (h) tends to zero as h tends to zero. Similarly using the differentiability of F we can write F ( u + h ) = F ( u ) + F ' ( u ) h + h k 2 ( h ) (2) where k 2 ( h ) tends to zero as h tends to zero. Apply the definition of partial derivative to ∂ ( F ( f ( x , y , z ) ) ) ∂ x . . That is ∂ F ( f ( x , y , z ) ) ∂ x = lim h → 0 F ( f ( x + h , y , z ) ) − F ( f ( x , y , z ) ) h = lim h → 0 F ( f ( x , y , z ) + ∂ f ∂ x h + h k 1 ( h ) ) − F ( f ( x , y , z ) ) h In the previous step we applied equation (1) to f ( x , y , z ) Now use equation (2) in the last expression of the previous limit...

3rd Edition

ISBN: 9781319030728

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 14.5, Problem 65E

To determine

To prove

The chain rule for multivariable functions given by $\nabla F (f (x, y, z)) = F' (f (x, y, z)) \nabla f (x, y, z)$

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Chapter 14.5, Problem 64E

Chapter 14.5, Problem 66E

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Chapter 14 Solutions

Calculus: Early Transcendentals

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E

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To prove The chain rule for multivariable functions given by ∇ F ( f ( x , y , z ) ) = F ' ( f ( x , y , z ) ) ∇ f ( x , y , z )

To prove The chain rule for multivariable functions given by ∇F(f(x,y,z))=F'(f(x,y,z))∇f(x,y,z)

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Chapter 14 Solutions

To prove

The chain rule for multivariable functions given by $\nabla F (f (x, y, z)) = F' (f (x, y, z)) \nabla f (x, y, z)$