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已知x=f(t)-π;y=f(e^3t-1),求dy/dx︱t=0;d²y/dx²︱t=0.
解:dy/dx=(dy/dt)/(dx/dt)=3e^(3t)f′(e^3t-1)/f′(t)
d²y/dx²={f′(t)[9e^(3t)f′(e^3t-1)]-[3e^(3t)f′(e^3t-1)f″(t)]}/[f′(t)]².
故dy/dx︱(t=0)=3f′(0)/f′(0)=3
d²y/dx²︱(t=0)={f′(0)(9f′(0)-3f′(0)f″(0)}/[f′(0)]²=[9f′(0)-3f″(0)]/f′(0)=9-[3f″(0)]/f′(0)]
解:dy/dx=(dy/dt)/(dx/dt)=3e^(3t)f′(e^3t-1)/f′(t)
d²y/dx²={f′(t)[9e^(3t)f′(e^3t-1)]-[3e^(3t)f′(e^3t-1)f″(t)]}/[f′(t)]².
故dy/dx︱(t=0)=3f′(0)/f′(0)=3
d²y/dx²︱(t=0)={f′(0)(9f′(0)-3f′(0)f″(0)}/[f′(0)]²=[9f′(0)-3f″(0)]/f′(0)=9-[3f″(0)]/f′(0)]
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