Higher-order derivative?s? ? ?ind y?, ?? nd y?? or the following functions. y= x + 2(x-3)

Solution Step 1: Given y=x + 2(x-3) We have to determine y’,y’’,y’’’ For that we need following formulae d n n1 dx(x ) = nx d dv du dx(uv )=u dx + v dx v u dv d ( ) = dx 2dx dx v v Step 2 y’= dy dx d = dx ( x + 2(x 3)) d d =x + 2 dx (x 3) + (x 3) dx(x + 2) 1 d =x + 2 (1)+(x 3) 2 x+2 dx (x+2) (x3) = x + 2 + 2 x+2 2 x+2( x+2)+(x3) = 2 x+2 = 2(x+2)+(x3) 2 x+2 2x+4+x3 = 2 x+2 = 3x+1 2 x+2 Therefore y’ = 3x+1 2 x+2