Explanation The partial derivatives by chain rule are given, f ( x , y ) = x 2 + y 2 , x = e u + v , y = u + v To find ∂ f ∂ u Given, f ( x , y ) = x 2 + y 2 ............. ( 1 ) ∂ f ∂ x = 2 x , ∂ f ∂ y = 2 y ∂ x ∂ u = e u + v , ∂ y ∂ u = 1 Different eqn. (1) partially w.r.t. u ∂ f ∂ u = ∂ f ∂ x . ∂ x ∂ u + ∂ f ∂ y . ∂ y ∂ u = ( 2 x ) e u + v + ( 2 y ) ( 1 ) = ( 2 e u + v ) e u + v + 2 ( u + v ) = 2 ( e u + v

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ISBN: 9781319308858

Author: Rogawski

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Chapter 14.6, Problem 8E

To determine

By using chain rule, we have to find the partial derivatives

$\begin{array}{l} \frac{\partial f}{\partial u}; \\ f (x, y) = x^{2} + y^{2} \\ x = e^{u + v} \\ y = u + v \end{array}$

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Chapter 14.6, Problem 7E

Chapter 14.6, Problem 9E

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Chapter 14 Solutions

CALCULUS:EARLY TRANS.(LL)-W/ACCESS

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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By using chain rule, we have to find the partial derivatives ∂ f ∂ u ; f ( x , y ) = x 2 + y 2 x = e u + v y = u + v

By using chain rule, we have to find the partial derivatives ∂f∂u;f(x,y)=x2+y2x=eu+vy=u+v

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Chapter 14 Solutions

By using chain rule, we have to find the partial derivatives

$\begin{array}{l} \frac{\partial f}{\partial u}; \\ f (x, y) = x^{2} + y^{2} \\ x = e^{u + v} \\ y = u + v \end{array}$