1解当x≥0时,y=3√x^2=3x
则y'=(3x)'=3
当x＜0时,y=3√x^2=-3x
即y'=(-3x)'=-3
2
y'=(3√x)'
=3(√x)'
=3×1/2√x
=3/（2√x）1y=3√x^2=x^(2/3)则y'=[x^(2/3)]'=2/3×x^(2/3-1)=2/3×x^(-1/3)=2/[3x^(1/3)]y'=(3√x)'=(x^(1/3))'=1/3×x^(1/3-1)=1/3×x^(-2/3)=1/[3×x^(2/3)]=1/[3x^(2/3)]