Explanation There is given below the partial derivatives by chain rule. F ( u , v ) = e u + v u = x 2 , v = x y To find ∂ F ∂ y Given F ( u , v ) = e u + v ............. ( 1 ) ∂ f ∂ u = e u + v , ∂ F ∂ v = e u + v ∂ u ∂ x = 2 x , ∂ v ∂ x = y ∂ u ∂ y = 0 , ∂ v ∂ y = x Diff. eqn. (1) partially w.r.t. y ∴ ∂ F ∂ y = ∂ F ∂ u . ∂ u ∂ y + ∂ F ∂ v . ∂ v ∂ y = e u + v ( 0 ) + e u + v ( x ) = x . e u + v = x ...

ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

4th Edition

ISBN: 9781319411657

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 14.6, Problem 7E

To determine

By using chain rule to calculate partial derivatives,

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

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Chapter 14.6, Problem 6E

Chapter 14.6, Problem 8E

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Chapter 14 Solutions

ACHIEVE FOR CALCULUS 1 TERM >CSI CUSTOM<

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 to calculate partial derivatives, ∂ F ∂ y . F ( u , v ) = e u + v u = x 2 v = x y

By using chain rule to calculate partial derivatives, ∂F∂y.F(u,v)=eu+vu=x2v=xy

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Chapter 14 Solutions

By using chain rule to calculate partial derivatives,

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