(a) To determine Prove that “ ∂ f ∂ y ( x , 0 ) = x for all x and ∂ f ∂ x ( 0 , y ) = − y for all y ”. Explanation Given: The function is f ( x , y ) = { x y x 2 − y 2 x 2 + y 2 , if ( x , y ) ≠ 0 , 0 , if ( x , y ) = 0. . Calculation: At point x = 0 obtain f x ( 0 , y ) as follows. f x ( 0 , y ) = lim h → 0 f ( 0 + h , y ) − f ( 0 , y ) h = lim h → 0 f ( h , y ) − 0 h = lim h → 0 f ( h , y ) h The above expression does not exist. For the negative limits of h , lim h → 0 − f ( h , y ) h = lim h → 0 − h 2 h = 0 For the positve limits of h , (b) To determine Show that ∂ 2 f ∂ y ∂ x ( 0 , 0 ) ≠ ∂ 2 f ∂ x ∂ y ( 0 , 0 ) .

14th Edition

ISBN: 9780134438986

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Question

Chapter 14.3, Problem 82E

(a)

To determine

Prove that “∂f∂y(x,0)=x for all x and ∂f∂x(0,y)=−y for all y”.

(b)

To determine

Show that ∂2f∂y∂x(0,0)≠∂2f∂x∂y(0,0).

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Chapter 14.3, Problem 81E

Chapter 14.3, Problem 83E

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Thomas' Calculus (14th Edition)

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