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设z=(sinx)^3+y^3-3xy^2,求二阶偏导数,急求,在线等.(请含过程)
人气:440 ℃ 时间:2020-06-27 10:09:36
解答
z=(sinx)^3+y^3-3xy^2
dz/dx=3(sinx)^2*cosx-3y^2
dz/dy=3y^2-6xy
d^2z/dx^2=d(dz/dx)/dx=d[3(sinx)^2*cosx-3y^2]/dx=6sinx*(cosx)^2-3(sinx)^3
d^2z/dx^2=d(dz/dy)/dy=d(3y^2-6xy)/dx=6y-6x
d^2z/dxdy=d(dz/dy)/dx=d(dz/dx)/dy=-6y
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