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Sprendimas.1. (y = const),∂z∂x = (x3 + 2xy 2 + 3x 2 y + y 3 ) ′ x = 3x 2 + 2y 2 + 6yx;∂z(x 0 ; y 0 )=∂x(x = const);∂z(1; −1)∂x= 3 · 1 2 + 2 · (−1) 2 + 6 · (−1) · 1 = −1;∂z∂y = (x3 + 2xy 2 + 3x 2 y + y 3 ) ′ y = 4xy + 3x 2 + 3y 2 ;∂z(x 0 ; y 0 )∂y2. (y = const)=(x = const),∂z(1; −1)∂y= 4 · 1 · (−1) + 3 · 1 2 + 3 · (−1) 2 = 2.( )∂z∂x = 1 1( ) x 2 ·y1 +y∂z(−1; 1)∂x∂z∂y = 11 +∂z(−1; 1)∂y==yy 2 + x 2 ;11 2 + 1 2 = 1 2 .( ) x 2 ·(− x )y 2y= − xy 2 + x 2 ;= − 11 2 + 1 2 = −1 2 .3. Kai y = const, z = x y yra laipsninė funkcija, todėl∂z∂x = (xy ) ′ x = yx y−1 .76