Question

设函数y=x^2+x,计算在x=2处当△x分别等于1,0.1,0.01时的△y及dy.

Answer 1

y=x^2+x,dy=2x+1

当X=2时,y=x^2+x=6

当△x=1,即x=3或者x=1,当x=1时y=x^2+x=2,△y=4,dy=3；当x=3,y=x^2+x=12,△y=12-6=6,dy=3*2+1=7

依次类推,可得到x=0.1,0.01时的结果