2022年用Mathematica求偏导数与多元函数的极值练习参考解答.docx

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1、精选学习资料 - - - - - - - - - 10 用 Mathematica 求偏导数与多元函数的极值 练习参考解答1 求以下函数的偏导数；1 zyx21y22 zexyz3 uzx4 uxyxxz2 求以下函数的偏导数或导数；1 设zarctgxy,yex,求dz ；dxz ；v2 设zxln xy,求3z,3zx2yxy23 设z2 xlny ,xu,y3 u2 v ,求z ,uv4 设ufx,y,求u ,zu ,yu ；xyz5 设zfxy,xy ,x,求zx,zxx,zxy；y3 求以下方程所确定的隐函数的导数；1 x2y3x2y340,求dy ；dx2 exy2zez0,求z

2、,xz ；y3 zfxyz ,xyz ,求z ,xx ,yz ；y4 x2y2z2a2,x2y2ax,求dy ,dxdz ；dx4 求函数fx ,yx25y26x10y6的极值；5 求函数zx2y2,在x,y |x2y24范畴内的最大最小值；练习参考解答1 求以下函数的偏导数；名师归纳总结 1 zx21y22 zexy3 uyzx4 uxy z第 1 页,共 7 页xxz- - - - - - -精选学习资料 - - - - - - - - - 解 1 In1:=D1/Sqrtx2+y2,x In2:= D1/Sqrtx2+y2,yOut1= 2 x2 3 / 2 x y yOut2= 2 2

3、 3 / 2 x y 2 In3:= DEx*y,x In4:= DEx*y,x Out3= e xyyxyOut4= e x3 In5:= Dy/x+z/x-x/z,x In6:= Dy/x+z/x-x/z,y In7:= Dy/x+z/x-x/z,z Out5= y2 1 z2x z zOut6= 1xOut7= 1 x2x z4 In8:= Dx*yz,x In9:= Dx*yz,xIn10:= Dx*yz,z 1 zOut8= y xy z1 zOut9= x xy zzOut10= xy Log xy 2 求以下函数的偏导数或导数；1 设zarctgxy,yx e,求dz ；dx解I

4、n1:= yx_:Ex; zx_,y_:=ArcTanx*y; Dzx,y,x名师归纳总结 - - - - - - -第 2 页,共 7 页精选学习资料 - - - - - - - - - Out1= 1yy2x22 设zxln xy,求3z,3zx2yxy2解In1:= zx_,y_:=x*Logx*yDzx,y,x,2,y; Simplify% Dzx,y,x,y,3; Simplify%Out1= 2xy 23x2y2u3 v ,求z ,uz ；v 1x2y233 设zx6 x26x46 xy4 1x2y243siny,x1v,yu解In1:= xu_,v_:=1-v/u; yu_,v_

5、:=u+3v; zx_,y_:=xu,v2*Sinyu,v; Dzx,y,u; Simplify% Dzx,y,v; Simplify% Out1= u2v uu2v Cos u3v2u2vSin u3v u33vu2v3 u2vCosu3 v2Sin uu24 设ufx,y,求u ,zu ,yu ；xyz解In1:= ux_,y_,z_:=fx/y,y/z; Dux,y,z,x 名师归纳总结 - - - - - - -第 3 页,共 7 页精选学习资料 - - - - - - - - - Dux,y,z,y Dux,y,z,z f,10x,yzxy；yzOut1= yf0,1 x,yxf 1

6、,0x,yyzyzyfzy2,01x,yyz5 设zfz2xy,xy ,x,求zx,zxx,y解In1:= zx_,y_:=fx+y,x*y,x/y; Dzx,y,x Dzx,y,x,x Dzx,y,x,y Out1= 1yy2211y2x2 12 xy32x2y2 xy3 1x2y22x2三 求以下方程所确定的隐函数的导数；1 x2y3x2y340,求dy ；dx解In1:= Dx2*yx+3x2yx3-4= =0,x; Solve%,y x 名师归纳总结 Out1= yx 2xy x 26 xy x3第 4 页,共 7 页x29 xy x 2- - - - - - -精选学习资料 - -

7、- - - - - - - 2 exy2zez0,求z ,xz ；y解In1:= DE-x*y-2*zx+Ezx= =0,x; SimplifySolve%,z x DE-x*y-2*zy+Ezy= =0,y; SimplifySolve%,z y Out1= zx exyy x 2ezzyexyx2ez y3 zfxyz ,xyz ,求z ,xx ,yz ；y解In1:= Dzx-fx+y+zx,x*y*zx= =0,x; SimplifySolve%,z x Dzy-fx+y+zy,x*y*zy= =0,y; SimplifySolve%,z y Out1= 4 zxyz x f,01xy

8、z x,xyz xzf,10xyz x ,xyz x/1xyf01,xyz x ,xyz x f,10 xyz x,xyz x yxz y f01,xyz y ,xyz yx2y2f,10xyz y ,xyz y/1xyf01,xyz y ,xyz y f ,10xyz y ,xyz y z2a2,x2y2ax,求dy ,dxdz ；dx解In1:= Dx2+yx2+zx2-a= =0,x2+yx2-a*x= =0,x; SimplifySolve%,yx SimplifySolve%,zx 名师归纳总结 - - - - - - -第 5 页,共 7 页精选学习资料 - - - - - - -

9、 - - Out1= yxxz xzxy xzxxyxy x6的极值；z x 4 求函数fx ,yx25y26x10y解In1:= Clearx,y,z,a,b,c,d,t; fx_,y_:=x2+5y2-6x+10y+6; a=Dfx,y,x,2; b=Dfx,y,x,y; c=Dfx,y,y,2; d=a*c-b2; t=SloveDfx,y= =0,x,Dfx,y= =0,y,x,y; l=Lengthtl Fori=1,i0&a10&a10,Print“fmin= ”,z, d1= =0,Print “ No Sure ”,z, d1= =0,PrintNo Out1= x-3,y-1

10、 fmin=-8 5 求函数zx2y2,在x,y |x2y24范畴内的最大最小值；名师归纳总结 解先求zx2y2在圆域内x2y24的最大最小值 : 第 6 页,共 7 页In1:= fx_,y_:=x2-y2; - - - - - - -精选学习资料 - - - - - - - - - t=SolveDfx,y= =0,x,Dfx,y= =0,y,x,y Out1= In2:= x-0,y-0 * 驻点*x2+y2-4/.t1 Out2= -4 该驻点在圆外 ,圆内无驻点 ,故不取极值 .下面考虑圆x2y24上的最值 .这是在约束条x2y24下的条件极值 ,用 Lagrange乘数法求解 . In3:= Clearx,y,F,t; Fx_,y_,t_:=fx,y+tx2+y2-4; s=SolveDfx,y,t=0,Dfx,y,t=0,y,DFx,y,t= =0,t,x,y,t Out3= t-1,x-2,y-0,t-1,x-2,y-0, t-1,x-2,y-0 , t-1,x-2,y-0 In4:= Fx,y/.s1 Out4= 4 In5:= Fx,y/.s2 Out5= 4 In6:= Fx,y/.s3 Out6= -4 In7:= Fx,y/.s4 Out7= -4名师归纳总结 - - - - - - -第 7 页,共 7 页