解ï¼x=3t²=Ï(t)ï¼y=3t-t³=Ï(t)ï¼
dy/dx=(dy/dt)/(dx/dt)=(3-3t²)/6t=(1/2)[(1/t)-t]=y'ï¼
d²y/dx²=dy'/dx=(dy'/dt)/(dx/dt)=(1/2)[(-1/t²)-1]/(6t)=-(1+t²)/(12t³).
Ïâ(t)=3-3t²ï¼Ïââ(t)=-6tï¼ Ï'(t)=6tï¼Ïââ(t)=6ï¼å¥å
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d²y/dx²=[Ïââ(t) Ï'(t)-Ïâ(t)Ïââ(t)]/[Ï'(t)]³=[(-6t)(6t)-(3-3t²)(6)]/(6t)³=(-36t²-18+18t²)/(216t³)
=(-18t²-18)/(216t³)=-(t²+1)/(12t³)
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