设函数y=y(x),由方程y=x*sin(y)所确定,求y'(0)y''(0)？
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两边对x求导
y'=xcosy*y'+siny
x=0得y=0
将x,y代入解得y'(0)=0
对y'=xcosy*y'+siny两边对x求导
y''=-sinyy'(xy')+cosy(y'+xy'')
将x=y=y'=0代入
得y''(0)=0,4,y'=sin(y)+xcos(y)y'
y'=sin(y)/(1-xcos(y)),y'(0)=sin(y)=0
y''=cos(y)y'+cos(y)y'+x(y''cos(y)-sin(y)y'y')
y''=(2cos(y)y'-xsin(y)(y')^2)/(1-xcos(y)),y''(0)=2sin(y)cos(y)=0,1,
y'=xcosy*y'+siny
x=0得y=0
将x,y代入解得y'(0)=0
对y'=xcosy*y'+siny两边对x求导
y''=-sinyy'(xy')+cosy(y'+xy'')
将x=y=y'=0代入
得y''(0)=0,4,y'=sin(y)+xcos(y)y'
y'=sin(y)/(1-xcos(y)),y'(0)=sin(y)=0
y''=cos(y)y'+cos(y)y'+x(y''cos(y)-sin(y)y'y')
y''=(2cos(y)y'-xsin(y)(y')^2)/(1-xcos(y)),y''(0)=2sin(y)cos(y)=0,1,
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