1)y=2(3x2+1)21y′=21(−2)(3x2+1)31⋅(6x)==(3x2+1)2−6x2)y=3+4x−x2y′=0+4−2x=4−2x3)y=2x3+1x2−2y′=(2x3+1)22x2−21⋅2x⋅(2x3+1)−x2−2⋅6x2==x2−2(2x3+1)2x⋅(2x3+1)−(x2−2)⋅6x2==x2−2(2x3+1)2−4x4+12x2+x4)y=3xx26x2y=3x6x2=36x2y′=362x=326x5)y=2x2x=2x25y′=2⋅25⋅x23=5x23
Comments
Leave a comment