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第五题怎么做啊
1个回答
展开全部
dx=1/(1+t^2) dt
dy=3+3t^2=3(1+t^2) dt
y'=dy/dx=3(1+t^2)^2
d^2y/dx=[3(1+t^2)^2]'=6(1+t^2)*2t=12t(1+t^2) dt
y''=d^2y/dx^2=12t(1+t^2)/[1/(1+t^2)]=12t(1+t^2)^2
dy=3+3t^2=3(1+t^2) dt
y'=dy/dx=3(1+t^2)^2
d^2y/dx=[3(1+t^2)^2]'=6(1+t^2)*2t=12t(1+t^2) dt
y''=d^2y/dx^2=12t(1+t^2)/[1/(1+t^2)]=12t(1+t^2)^2
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