y=Insin(3x-1),求dx分之dy

y=lnsin（3x-1）
　　dy/dx=d【lnsin（3x-1）】/d【sin（3x-1）】×d【sin（3x-1）】/d（3x-1)×d（3x-1）/dx
　　=1/sin（3x-1）×cos（3x-1）×3
　　=3cot（3x-1）

y'=[Insin(3x-1)]'
=1/sin(3x-1)*[sin(3x-1)]'
=cos(3x-1)/sin(3x-1)*[(3x-1)]'
=3cot(3x-1)

y=lnsin(3x-1)
dy/dx=(lnsin(3x-1))'
=1/sin(3x-1) *(sin(3x-1))'
=1/sin(3x-1) *cos(3x-1)*(3x-1)'
=3cos(3x-1)/sin(3x-1)
=3ctan(3x-1)

设u=sin(3x-1)
则dy/dx=(dy/du)(du/dx)=(1/u)(du/dx)
设v=3x-1，则dv/dx=(3x-1)'=3
∴du/dx=(du/dv)(dv/dx)=3(cosv)=3cos(3x-1)
∴dy/dx=(1/u)(3cos(3x-1))=3cos(3x-1)/(sin(3x-1))=3cot(3x-1)
即：dy/dx=3cot(3x-1)
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