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等式左右两端对x求导：
d(ysin x-cos(x-y))/dx=d0/dx
dy/dx*sinx+y*cosx-dcos(x-y)/d(x-y)*d(x-y)/dx=0
y'sinx+ycosx+sin(x-y)(1-dy/dx)=0
y'sinx+ycosx+sin(x-y)-y'sin(x-y)=0
y'(sinx-sin(x-y))=-ycosx-sin(x-y)
y'=-(ycosx+sin(x-y))/(sinx-sin(x-y))
d(ysin x-cos(x-y))/dx=d0/dx
dy/dx*sinx+y*cosx-dcos(x-y)/d(x-y)*d(x-y)/dx=0
y'sinx+ycosx+sin(x-y)(1-dy/dx)=0
y'sinx+ycosx+sin(x-y)-y'sin(x-y)=0
y'(sinx-sin(x-y))=-ycosx-sin(x-y)
y'=-(ycosx+sin(x-y))/(sinx-sin(x-y))
1年前
6
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你能帮帮他们吗