y=(x+1)(x+2)2(x+3)3
y=(x+1)(x2+4x+4)(x+3)3
y=(x3+5x2+8x+4)(x3+9x2+27x+27)
Ta có:
y'=(3x2+2.5x+8+0)(x3+9x2+27x+27)+(x3+5x2+8x+4)(3x2+2.9x+27+0)=(3x2+10x+8)(x3+9x2+27x+27)+(x3+5x2+8x+4)(3x2+18x+27)=(3x+4)(x+2)(x+3)3+3(x+1)(x+2)(x+3)2=(x+2)(x+3)2[(3x+4)(x+3)+3(x+1)]=(x+2)(x+3)2(3x2+13x+12+3x+3)=(x+2)(x+3)2(3x2+16x+15)