# X y 0 x y if f has first order partial derivatives at

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( x 0 , y 0 ) x y If f has first-order partial derivatives at each point of some open circular region centered at ( x 0 , y 0 ), and if these partial derivatives are continuous at ( x 0 , y 0 ), then f is differentiable and therefore continuous at ( x 0 , y 0 ). [2] The partial derivatives can exist at ( x 0 , y 0 ), even if they are not continuous at ( x 0 , y 0 ). Note [1] The continuity of f x ( x , y) and f y ( x , y ) are essential to guarantee the differentiability, and hence, continuity of z = f ( x , y ) at ( x 0 , y 0 ). [See handout]
1 MA211 Week 4: LECTURE 1 Chapter 13.5: The Chain Rule 2 dt dy y z dt dx x z dt dz y x f z t y y t x x ) , ( ) ( ) ( If then The t -Chain Rule ) , ( y x f z ) ( t x t ) ( t y t x z y z dt dx dt dy Tree Diagram
3 The uv -Chain Rule v y y z v x x z v z u y y z u x x z u z y x f z v u x y v u x x ) , ( ) , ( ) , ( If then ) , ( y x f z ) , ( v u x ) , ( v u y u v u v x z y z u x u y v x v y Tree Diagram 4 Example (3,2) Metal plate 4 , 1 4 j i v particle (3,2). at Find . in , 1 ln , is plate the of re temperatu The . 2 , 3 point at the (cm/s) 4 city with velo plane - in the plate metal a along moves particle A 0 2 dt dT C x, x y y x T xy j i v Solution The particle must move in the direction of v , which gives a straight path. Hence, at any time t between (3,2) and (4,-2), its position is (4,-2) 1 0 , 4 2 , 3 4 1 2 3 t t y t x t y x t v x x 0
5 Hence, given x y y x T t t y y t t x x ln , 4 2 ) ( 3 ) ( 2 we use the t -Chain Rule to get x y x y x y x y dt dy y T dt dx x T dt dT ln 8 4 ln 2 1 2 2 so that     s C dt dT y x 0 2 2 , 3 , 3 ln 16 3 4 3 ln 2 8 3 2 6 Example . 0 that Show . , Let y z x z x y y x f z Solution Let ) , ( , , v u f z x y y x v v y x y x u u Then using the xy -Chain Rule, we have   v z u z v z u z x v v z x u u z x z 1 1 and   v z u z v z u z y v v z y u u z y z 1 1 Thus, 0 y z x z ) , ( v u f z ) , ( y x u ) , ( y x v x y x y u z v z x u x v y u y v
7 Equality of Mixed Partials . is, that , , all for then plane, - over the everywhere continuous are , , , If y f x x f y y x f f xy f f f f yx xy yx xy y x Example that Show . , and ) ( that Suppose y x g u u f z 2 2 2 2 2 2 2 z dz u d z u x du x du x 8 Solution . for expression an need First we x z So we need a tree diagram. ) ( u f z u ( x , y ) du dz x y x u y u We can then see that x u du dz x z Now, using the product rule, we have du dz x x u x u du dz x u du dz x x z x 2 2 partials, mixed of equality the and rule product the again via yields, term second the , Since u z du dz 2 2 2 2 2 0 du z d x u u z x u du dz x u x u du dz u x u x z u x u u z x x u du dz x x u