高等数学上册第九章答案.doc

_高等数学上册第九章答案_高等数学上册第九章答案****题9-:(1)z=ln(y-x(2)u=:(1)lim;(x,y)→(0,0)(2)(x,ylim)→;(3)limtan(xy)(x,y)→(2,0)y.(4)lim(x2(x2+y2)x+y2)e-y→∞→∞令u=x2+y2,原式=limuu→∞eu=lim1u→∞eu=0(5)(x,ylim)→(0,01令t==limt-sint1-costx21t→0+t3=limt→0+3t2=limt→0+3t2=6****题9-:(1)z=sin(xy)+cos2(xy);(2)z=(1+xy)y;(3)u=arctan(x-y)z.(4)设z=y2∂z3x+ϕ(xy),其中ϕ(u)可导,证明x2∂x+y2=xy∂z∂y解∂z∂x=-y2∂z2y3x2+yϕ’(xy),∂y=3x+xϕ’(xy)左边=x2∂z∂x+y2=-y23+x2yϕ’(xy)+y2=xy⎡⎢2y⎣3x+xϕ’(xy)⎤⎥⎦=∂2z∂2z∂2z∂x2,∂y2和∂x∂y.(1)z=arctanyx;(2)z=-:y(1)z=ex;(2)u=xyz.(3)u=x+siny+∂u=1∂u=1cy+zeyz∂u=yeyz∂x∂y22∂z,所求的全微分为du=dx+⎛1y⎝2cos2+zeyz⎫⎪⎭dy+yeyzdz‘(4)u=tanx2+y2+z2)解∂u∂x=∂u,∂y=∂u=∂zdu=xdx+ydy+zdz)=y,当x=2,y=1,∆x=,∆y=-。(x,y,z)=df(1,1,1)1z-1解∂f1⎛x∂x=z⎫⎝y⎪⎭⋅1y,∴∂f∂x=1(1,1),11∂f1⎛xz-1∂y=z⎫⎝y⎪⎭⋅⎛-x⎫∂f⎝y2⎪⎭,∴∂y=-1(1,1,1)∂f∂f∂z=lnxy⋅⎛⎝-1⎫z2⎪⎭,∴∂z=0(1,1,1)故df(1,1,)1=d-xdy****题9-=u2lnv,而u=x