dy=dx+dsiny=dx+cosydy
即y'=dy/dx=1/(1-cosy)
对x求导
y''=-1/(1-cosy)²*(1-cosy)'
=-siny*y'/(1-cosy)²
=-siny/(1-cosy)³
所以d²y/dx²=y''=-siny/(1-cosy)³
设y=x+siny,则[(d^2)y]/[dx^2]=?
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