已知z=(2x2+1y2)e-arctanyx 求一阶及二阶偏导数.docx

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y2y已知z=(22+1y2)eFta,求W,噂解：对函数直接求偏导数有:1yz-arctan-=4xeX4y-y-(221y2)e-arctan;1+(x)2X=4xe-arctan-,22.x+(2x+1y)e-arctan-xx2+y2-arctan-=exy(2x21y2)2z2-arctan-=ex1*(-1)*NX1+(2x*4xy(2x2+1y2)2.2x+y-arctan-ex*44xy(x2+y2)-y(2x2+1y2)*2x(x2y2)2,-arctan-.fy,r.y(22+1y2)2xy3飞X*西*4x+Fy+N+西丁，-arctan-一1y4+6xy3+2x2y2+4x3y=e*4+语斤,4x+2+y2O由fy再求偏导数，得:OXz-arctan-X*4xy(22+)x2+y22zy1y一=earctan*(7)*q*4x-+(x)2Xy(221y2)2,2X+y-arctan-eX*(2x2+1y2)+y*2y(x2y2)-y(2x2+1y2)*2y(2y2)2y-arctan-1X1=eX*272*4x+X+yy(221y2)2.2X+y(2+y2)3-arctan-a2x4+1x2y2+1y4X(x2y2)2-arctan-.=eX*6x4+2x3y+5x2y2+1xy3+1y4