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z sin xy

∂Z/∂x= y*cos(xy) -2cos(xy)*sin(xy)*y = y*cos(xy) - y*sin(2xy) ∂Z/∂y= x*cos(xy) -2cos(xy)*sin(xy)*x = x*cos(xy) - x*sin(2xy)

你是怎么把公式弄上去的,我用个公式编辑器编辑的公式,百度一直提示图片非法。郁闷。无奈,图片都弄不上去了。z对x的偏导为ycosxy,z对xy的二阶偏导为cosxy-xysinxy.

dz=cos(xy)xdy+cos(xy)ydx 由z/x=ycos(xy),z/y=xcos(xy)→dz=ycos(xy)dx + xcos(xy)dy. cos(x-y)dx-cos(x-y)dy

Zy=ycos(xy)+2cos(xy)*[-sin(xy)]*x =ycos(xy)-xsin(2xy)

由z=sin(xy)+cos2(xy),得? z? y=xcos(xy)-2xcos(xy)sin(xy)

这样,对x求偏导就把y视作常数,对y求偏导就把x视作常数

即z=e^ [ln(2+sinxy) *(x-2y)] 那么求偏导数得到 δz/δx=z *δ[ln(2+sinxy) *(x-2y)]/δx =z *[y* cosxy/(2+sinxy) *(x-2y) +ln(2+sinxy)] δz/δy=z *δ[ln(2+sinxy) *(x-2y)]/δy =z *[x *cosxy/(2+sinxy) *(x-2y) -2ln(2+sinxy)]

x=-8:.01:8; y=-8:.01:8; [X,Y]=meshgrid(x,y); Z=X.^2+Y.^2+sin(X*Y); mesh(X,Y,Z); meshc(X,Y,Z);

用D和Dt啊: (* 注意语法 *)z = y + Sin[x y](* 两个一阶导 *)D[z, {{x, y}}](* {y Cos[x y], 1 + x Cos[x y]} *)(* 四个二阶导,先y后x和先x后y在这里是一样的 *)D[z, {{x, y}, 2}](* {{-y^2 Sin[x y], Cos[x y] - x y Sin[x y]}, {Cos[x y] - ...

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