Y 1 2 3 4x July 8, 2019 16 32 64 128 Y 1 2x 2 4x 3 Y 1 /( x 2 4x 3 Through 1 2 parallel to y 4x 3 Y x 3 4x 2 5x 1 X 2 1 y '+ 4xy 3 Y 0 2 1 3 2 4x dx Turunan pertama dari y 1 2 cos 3 4x π Y 3 4x 1 2 in standard form Y 1 3x 3 mx 2 4x m Y 3 4x 1 2 graph X y z 1 2 3 và 4x 3y 2z (- 2 3 y 1 4x 4 Y =( 4x 1 2 1 x 3 derivative
