高等数学（第七版）同济大学习题9-5 个人解答

高等数学（第七版）同济大学习题9-5

1.设siny+ex−xy2=0，求dydx.\begin{aligned}&1. \ 设sin\ y+e^x-xy^2=0，求\frac{dy}{dx}.&\end{aligned}1. 设sin y+ex−xy2=0，求dxdy.

解：

设F(x,y)=siny+ex−xy2，则Fx=ex−y2，Fy=cosy−2xy，当Fy≠0时，有dydx=−FxFy=−ex−y2cosy−2xy=y2−excosy−2xy\begin{aligned} &\ \ 设F(x,\ y)=sin\ y+e^x-xy^2，则F_x=e^x-y^2，F_y=cos\ y-2xy，当F_y \neq 0时，\\\\ &\ \ 有\frac{dy}{dx}=-\frac{F_x}{F_y}=-\frac{e^x-y^2}{cos\ y-2xy}=\frac{y^2-e^x}{cos\ y-2xy} & \end{aligned} 设F(x, y)=sin y+ex−xy2，则Fx=ex−y2，Fy=cos y−2xy，当Fy=0时，有dxdy=−FyFx=−cos y−2xyex−y2=cos y−2xyy2−ex

2.设lnx2+y2=arctanyx，求dydx.\begin{aligned}&2. \ 设ln\sqrt{x^2+y^2}=arctan\frac{y}{x}，求\frac{dy}{dx}.&\end{aligned}2. 设lnx2+y2=arctanxy，求dxdy.

解：

设F(x,y)=lnx2+y2−arctanyx，则一阶偏导数分别为Fx=1x2+y2⋅2x2x2+y2−11+(yx)2⋅(−yx2)=x+yx2+y2，Fy=1x2+y2⋅2y2x2+y2−11+(yx)2⋅1x=y−xx2+y2，当Fy≠0时，有dydx=−FxFy=−x+yx2+y2y−xx2+y2=x+yx−y.\begin{aligned} &\ \ 设F(x, \ y)=ln\sqrt{x^2+y^2}-arctan\frac{y}{x}，则一阶偏导数分别为\\\\ &\ \ F_x=\frac{1}{\sqrt{x^2+y^2}}\cdot \frac{2x}{2\sqrt{x^2+y^2}}-\frac{1}{1+\left(\frac{y}{x}\right)^2}\cdot \left(-\frac{y}{x^2}\right)=\frac{x+y}{x^2+y^2}，\\\\ &\ \ F_y=\frac{1}{\sqrt{x^2+y^2}}\cdot \frac{2y}{2\sqrt{x^2+y^2}}-\frac{1}{1+\left(\frac{y}{x}\right)^2}\cdot \frac{1}{x}=\frac{y-x}{x^2+y^2}，\\\\ &\ \ 当F_y \neq 0时，有\frac{dy}{dx}=-\frac{F_x}{F_y}=-\frac{\frac{x+y}{x^2+y^2}}{\frac{y-x}{x^2+y^2}}=\frac{x+y}{x-y}. & \end{aligned} 设F(x, y)=lnx2+y2−arctanxy，则一阶偏导数分别为 Fx=x2+y21⋅2x2+y22x−1+(xy)21⋅(−x2y)=x2+y2x+y， Fy=x2+y21⋅2x2+y22y−1+(xy)21⋅x1=x2+y2y−x，当Fy=0时，有dxdy=−FyFx=−x2+y2y−xx2+y2x+y=x−yx+y.

3.设x+2y+z−2xyz=0，求∂z∂x及∂z∂y.\begin{aligned}&3. \ 设x+2y+z-2\sqrt{xyz}=0，求\frac{\partial z}{\partial x}及\frac{\partial z}{\partial y}.&\end{aligned}3. 设x+2y+z−2xyz=0，求∂x∂z及∂y∂z.

解：

设F(x,y,z)=x+2y+z−2xyz，则Fx=1−yzxyz，Fy=2−xzxyz，Fz=1−xyxyz，当Fz≠0时，有∂z∂x=−FxFz=yz−xyzxyz−xy，∂z∂y=−FyFz=xz−2xyzxyz−xy.\begin{aligned} &\ \ 设F(x, \ y, \ z)=x+2y+z-2\sqrt{xyz}，则F_x=1-\frac{yz}{\sqrt{xyz}}，F_y=2-\frac{xz}{\sqrt{xyz}}，F_z=1-\frac{xy}{\sqrt{xyz}}，\\\\ &\ \ 当F_z \neq 0时，有\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=\frac{yz-\sqrt{xyz}}{\sqrt{xyz}-xy}，\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=\frac{xz-2\sqrt{xyz}}{\sqrt{xyz}-xy}. & \end{aligned} 设F(x, y, z)=x+2y+z−2xyz，则Fx=1−xyzyz，Fy=2−xyzxz，Fz=1−xyzxy，当Fz=0时，有∂x∂z=−FzFx=xyz−xyyz−xyz，∂y∂z=−FzFy=xyz−xyxz−2xyz.

4.设xz=lnzy，求∂z∂x及∂z∂y.\begin{aligned}&4. \ 设\frac{x}{z}=ln\frac{z}{y}，求\frac{\partial z}{\partial x}及\frac{\partial z}{\partial y}.&\end{aligned}4. 设zx=lnyz，求∂x∂z及∂y∂z.

解：

设F(x,y,z)=xz−lnzy，则Fx=1z，Fy=−1zy⋅(−zy2)=1y，Fz=−xz2−1zy⋅1y=−x+zz2，当Fz≠0时，有∂z∂x=−FxFz=−1z−x+zz2=zx+z，∂z∂y=−FyFz=−1y−x+zz2=z2y(x+z).\begin{aligned} &\ \ 设F(x, \ y, \ z)=\frac{x}{z}-ln\frac{z}{y}，则F_x=\frac{1}{z}，F_y=-\frac{1}{\frac{z}{y}}\cdot \left(-\frac{z}{y^2}\right)=\frac{1}{y}，F_z=-\frac{x}{z^2}-\frac{1}{\frac{z}{y}}\cdot \frac{1}{y}=-\frac{x+z}{z^2}，\\\\ &\ \ 当F_z \neq 0时，有\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=\frac{-\frac{1}{z}}{-\frac{x+z}{z^2}}=\frac{z}{x+z}，\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=\frac{-\frac{1}{y}}{-\frac{x+z}{z^2}}=\frac{z^2}{y(x+z)}. & \end{aligned} 设F(x, y, z)=zx−lnyz，则Fx=z1，Fy=−yz1⋅(−y2z)=y1，Fz=−z2x−yz1⋅y1=−z2x+z，当Fz=0时，有∂x∂z=−FzFx=−z2x+z−z1=x+zz，∂y∂z=−FzFy=−z2x+z−y1=y(x+z)z2.

5.设2sin(x+2y−3z)=x+2y−3z，证明∂z∂x+∂z∂y=1.\begin{aligned}&5. \ 设2sin(x+2y-3z)=x+2y-3z，证明\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=1.&\end{aligned}5. 设2sin(x+2y−3z)=x+2y−3z，证明∂x∂z+∂y∂z=1.

解：

设F(x,y,z)=2sin(x+2y−3z)−x−2y+3z，则Fx=2cos(x+2y−3z)−1，Fy=2cos(x+2y−3z)⋅2−2=2Fx，Fz=2cos(x+2y−3z)⋅(−3)+3=−3Fx，当Fz≠0时，有∂z∂x+∂z∂y=−FxFz−FyFz=13+23=1.\begin{aligned} &\ \ 设F(x, \ y, \ z)=2sin(x+2y-3z)-x-2y+3z，则F_x=2cos(x+2y-3z)-1，\\\\ &\ \ F_y=2cos(x+2y-3z)\cdot 2-2=2F_x，F_z=2cos(x+2y-3z)\cdot (-3)+3=-3F_x，\\\\ &\ \ 当F_z \neq 0时，有\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=-\frac{F_x}{F_z}-\frac{F_y}{F_z}=\frac{1}{3}+\frac{2}{3}=1. & \end{aligned} 设F(x, y, z)=2sin(x+2y−3z)−x−2y+3z，则Fx=2cos(x+2y−3z)−1， Fy=2cos(x+2y−3z)⋅2−2=2Fx，Fz=2cos(x+2y−3z)⋅(−3)+3=−3Fx，当Fz=0时，有∂x∂z+∂y∂z=−FzFx−FzFy=31+32=1.

6.设x=x(y,z)，y=y(x,z)，z=z(x,y)都是由方程F(x,y,z)=0所确定的具有连续偏导数的函数，证明∂x∂y⋅∂y∂z⋅∂z∂x=−1.\begin{aligned}&6. \ 设x=x(y, \ z)，y=y(x, \ z)，z=z(x, \ y)都是由方程F(x, \ y, \ z)=0所确定的具有连续偏导数的函数，\\\\&\ \ \ \ 证明\frac{\partial x}{\partial y}\cdot \frac{\partial y}{\partial z}\cdot \frac{\partial z}{\partial x}=-1.&\end{aligned}6. 设x=x(y, z)，y=y(x, z)，z=z(x, y)都是由方程F(x, y, z)=0所确定的具有连续偏导数的函数，证明∂y∂x⋅∂z∂y⋅∂x∂z=−1.

解：

因为∂x∂y=−FyFx，∂y∂z=−FzFy，∂z∂x=−FxFz，所以∂x∂y⋅∂y∂z⋅∂z∂x=(−FyFx)⋅(−FzFy)⋅(−FxFz)=−1\begin{aligned} &\ \ 因为\frac{\partial x}{\partial y}=-\frac{F_y}{F_x}，\frac{\partial y}{\partial z}=-\frac{F_z}{F_y}，\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}，所以\frac{\partial x}{\partial y}\cdot \frac{\partial y}{\partial z}\cdot \frac{\partial z}{\partial x}=\left(-\frac{F_y}{F_x}\right)\cdot \left(-\frac{F_z}{F_y}\right)\cdot \left(-\frac{F_x}{F_z}\right)=-1 & \end{aligned} 因为∂y∂x=−FxFy，∂z∂y=−FyFz，∂x∂z=−FzFx，所以∂y∂x⋅∂z∂y⋅∂x∂z=(−FxFy)⋅(−FyFz)⋅(−FzFx)=−1

7.设Φ(u,v)具有连续偏导数，证明由方程Φ(cx−az,cy−bz)=0所确定的函数z=f(x,y)满足a∂z∂x+b∂z∂y=c.\begin{aligned}&7. \ 设\varPhi(u, \ v)具有连续偏导数，证明由方程\varPhi(cx-az, \ cy-bz)=0所确定的函数z=f(x, \ y)满足\\\\&\ \ \ \ a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}=c.&\end{aligned}7. 设Φ(u, v)具有连续偏导数，证明由方程Φ(cx−az, cy−bz)=0所确定的函数z=f(x, y)满足 a∂x∂z+b∂y∂z=c.

解：

令u=cx−az，v=cy−bz，则Φx=Φu⋅∂u∂x=cΦu，Φy=Φv⋅∂v∂y=cΦv，Φz=Φu⋅∂u∂z+Φv⋅∂v∂z=−aΦu−bΦv，当Φz≠0时，有∂z∂x=−ΦxΦz=cΦuaΦu+bΦv，∂z∂y=−ΦyΦz=cΦvaΦu+bΦv，得a∂z∂x+b∂z∂y=a⋅cΦuaΦu+bΦv+b⋅cΦvaΦu+bΦv=c.\begin{aligned} &\ \ 令u=cx-az，v=cy-bz，则\varPhi_x=\varPhi_u\cdot \frac{\partial u}{\partial x}=c\varPhi_u，\varPhi_y=\varPhi_v\cdot \frac{\partial v}{\partial y}=c\varPhi_v，\\\\ &\ \ \varPhi_z=\varPhi_u\cdot \frac{\partial u}{\partial z}+\varPhi_v\cdot \frac{\partial v}{\partial z}=-a\varPhi_u-b\varPhi_v，\\\\ &\ \ 当\varPhi_z \neq 0时，有\frac{\partial z}{\partial x}=-\frac{\varPhi_x}{\varPhi_z}=\frac{c\varPhi_u}{a\varPhi_u+b\varPhi_v}，\frac{\partial z}{\partial y}=-\frac{\varPhi_y}{\varPhi_z}=\frac{c\varPhi_v}{a\varPhi_u+b\varPhi_v}，\\\\ &\ \ 得a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}=a\cdot \frac{c\varPhi_u}{a\varPhi_u+b\varPhi_v}+b\cdot \frac{c\varPhi_v}{a\varPhi_u+b\varPhi_v}=c. & \end{aligned} 令u=cx−az，v=cy−bz，则Φx=Φu⋅∂x∂u=cΦu，Φy=Φv⋅∂y∂v=cΦv， Φz=Φu⋅∂z∂u+Φv⋅∂z∂v=−aΦu−bΦv，当Φz=0时，有∂x∂z=−ΦzΦx=aΦu+bΦvcΦu，∂y∂z=−ΦzΦy=aΦu+bΦvcΦv，得a∂x∂z+b∂y∂z=a⋅aΦu+bΦvcΦu+b⋅aΦu+bΦvcΦv=c.

8.设ez−xyz=0，求∂2z∂x2.\begin{aligned}&8. \ 设e^z-xyz=0，求\frac{\partial^2 z}{\partial x^2}.&\end{aligned}8. 设ez−xyz=0，求∂x2∂2z.

解：

设F(x,y,z)=ez−xyz，则Fx=−yz，Fz=ez−xy，当Fz≠0时，有∂z∂x=−FxFz=yzez−xy，∂2z∂x2=∂∂x(∂z∂x)=y∂z∂x(ez−xy)−yz(ez∂z∂x−y)(ez−xy)2=y2z−yz(ez⋅yzez−xy−y)(ez−xy)2=2y2zez−2xy3z−y2z2ez(ez−xy)3.\begin{aligned} &\ \ 设F(x, \ y, \ z)=e^z-xyz，则F_x=-yz，F_z=e^z-xy，当F_z \neq 0时，有\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=\frac{yz}{e^z-xy}，\\\\ &\ \ \frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{y\frac{\partial z}{\partial x}(e^z-xy)-yz\left(e^z\frac{\partial z}{\partial x}-y\right)}{(e^z-xy)^2}=\frac{y^2z-yz\left(e^z\cdot \frac{yz}{e^z-xy}-y\right)}{(e^z-xy)^2}=\frac{2y^2ze^z-2xy^3z-y^2z^2e^z}{(e^z-xy)^3}. & \end{aligned} 设F(x, y, z)=ez−xyz，则Fx=−yz，Fz=ez−xy，当Fz=0时，有∂x∂z=−FzFx=ez−xyyz， ∂x2∂2z=∂x∂(∂x∂z)=(ez−xy)2y∂x∂z(ez−xy)−yz(ez∂x∂z−y)=(ez−xy)2y2z−yz(ez⋅ez−xyyz−y)=(ez−xy)32y2zez−2xy3z−y2z2ez.

9.设z3−3xyz=a3，求∂2z∂x∂y.\begin{aligned}&9. \ 设z^3-3xyz=a^3，求\frac{\partial^2 z}{\partial x\partial y}.&\end{aligned}9. 设z3−3xyz=a3，求∂x∂y∂2z.

解：

设F(x,y,z)=z3−3xyz−a3，则Fx=−3yz，Fy=−3xz，Fz=3z2−3xy，当Fz≠0时，有∂z∂x=−FxFz=yzz2−xy，∂z∂y=−FyFz=xzz2−xy，∂2z∂x∂y=∂∂y(∂z∂x)=∂∂y(yzz2−xy)=(z+y∂z∂y)(z2−xy)−yz(2z∂z∂y−x)(z2−xy)2=(z+xyzz2−xy)⋅(z2−xy)−yz(2xz2z2−xy−x)(z2−xy)2=z(z4−2xyz2−x2y2)(z2−xy)3\begin{aligned} &\ \ 设F(x, \ y, \ z)=z^3-3xyz-a^3，则F_x=-3yz，F_y=-3xz，F_z=3z^2-3xy，当F_z \neq 0时，\\\\ &\ \ 有\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=\frac{yz}{z^2-xy}，\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}=\frac{xz}{z^2-xy}，\\\\ &\ \ \frac{\partial^2 z}{\partial x\partial y}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial y}\left(\frac{yz}{z^2-xy}\right)=\frac{\left(z+y\frac{\partial z}{\partial y}\right)(z^2-xy)-yz\left(2z\frac{\partial z}{\partial y}-x\right)}{(z^2-xy)^2}=\\\\ &\ \ \frac{\left(z+\frac{xyz}{z^2-xy}\right)\cdot(z^2-xy)-yz\left(\frac{2xz^2}{z^2-xy}-x\right)}{(z^2-xy)^2}=\frac{z(z^4-2xyz^2-x^2y^2)}{(z^2-xy)^3} & \end{aligned} 设F(x, y, z)=z3−3xyz−a3，则Fx=−3yz，Fy=−3xz，Fz=3z2−3xy，当Fz=0时，有∂x∂z=−FzFx=z2−xyyz，∂y∂z=−FzFy=z2−xyxz， ∂x∂y∂2z=∂y∂(∂x∂z)=∂y∂(z2−xyyz)=(z2−xy)2(z+y∂y∂z)(z2−xy)−yz(2z∂y∂z−x)= (z2−xy)2(z+z2−xyxyz)⋅(z2−xy)−yz(z2−xy2xz2−x)=(z2−xy)3z(z4−2xyz2−x2y2)

10.求由下列方程组所确定的函数的导数或偏导数：\begin{aligned}&10. \ 求由下列方程组所确定的函数的导数或偏导数：&\end{aligned}10. 求由下列方程组所确定的函数的导数或偏导数：

(1)设{z=x2+y2，x2+2y2+3z2=20，求dydx，dzdx；(2)设{x+y+z=0，x2+y2+z2=1，求dxdz，dydz；(3)设{u=f(ux,v+y)，v=g(u−x,v2y)，其中f，g具有一阶连续偏导数，求∂u∂x，∂v∂x；(4)设{x=eu+usinv，y=eu−ucosv，求∂u∂x，∂u∂y，∂v∂x，∂v∂y.\begin{aligned} &\ \ (1)\ \ 设\begin{cases}z=x^2+y^2，\\\\x^2+2y^2+3z^2=20，\end{cases}求\frac{dy}{dx}，\frac{dz}{dx}；\\\\ &\ \ (2)\ \ 设\begin{cases}x+y+z=0，\\\\x^2+y^2+z^2=1，\end{cases}求\frac{dx}{dz}，\frac{dy}{dz}；\\\\ &\ \ (3)\ \ 设\begin{cases}u=f(ux, \ v+y)，\\\\v=g(u-x, \ v^2y)，\end{cases}其中f，g具有一阶连续偏导数，求\frac{\partial u}{\partial x}，\frac{\partial v}{\partial x}；\\\\ &\ \ (4)\ \ 设\begin{cases}x=e^u+usin\ v，\\\\y=e^u-ucos\ v，\end{cases}求\frac{\partial u}{\partial x}，\frac{\partial u}{\partial y}，\frac{\partial v}{\partial x}，\frac{\partial v}{\partial y}. & \end{aligned} (1) 设⎩⎨⎧z=x2+y2，x2+2y2+3z2=20，求dxdy，dxdz； (2) 设⎩⎨⎧x+y+z=0，x2+y2+z2=1，求dzdx，dzdy； (3) 设⎩⎨⎧u=f(ux, v+y)，v=g(u−x, v2y)，其中f，g具有一阶连续偏导数，求∂x∂u，∂x∂v； (4) 设⎩⎨⎧x=eu+usin v，y=eu−ucos v，求∂x∂u，∂y∂u，∂x∂v，∂y∂v.

解：

(1)对两方程两端对x求导，得{dzdx=2x+2ydydx，2x+4ydydx+6zdzdx=0.，整理得{2ydydx−dzdx=−2x，2ydydx+3zdzdx=−x.，当D=∣2y−12y3z∣=6yz+2y≠0时，解方程组得dydx=∣−2x−1−x3z∣D=−6xz−x6yz+2y=−x(6z+1)2y(3z+1)，dzdx=∣2y−2x2y−x∣D=2xy6yz+2y=x3z+1.(2)方程组确定两个一元隐函数：x=x(z)和y=y(z)，对方程两端对z求导，整理得{dxdz+dydz=−1，2xdxdz+2ydydz=−2z.，当D=∣112x2y∣=2(y−x)≠0时，解方程组得dxdz=∣−11−2z2y∣D=−2y+2z2(y−x)=y−zx−y，dydz=∣1−12x−2z∣D=−2z+2x2(y−x)=z−xx−y.(3)方程组确定两个二元隐函数：u=u(x,y)，v=v(x,y)，分别对方程两端对x求偏导数，得{∂u∂x=f1′⋅(u+x∂u∂x)+f2′⋅∂v∂x，∂v∂x=g1′⋅(∂u∂x−1)+2g2′yv⋅∂v∂x.，整理得{(xf1′−1)∂u∂x+f2′∂v∂x=−uf1′，g1′∂u∂x+(2yvg2′−1)∂v∂x=g1′.，当D=∣xf1′−1f2′g1′2yvg2′−1∣=(xf1′−1)(2yvg2′−1)−f2′g1′≠0时，解方程组得∂u∂x=∣−uf1′f2′g1′2yvg2′−1∣D=−uf1′(2yvg2′−1)−f2′g1′(xf1′−1)(2yvg2′−1)−f2′g1′，∂v∂x=∣xf1′−1−uf1′g1′g1′∣D=g1′(xf1′+uf1′−1)(xf1′−1)(2yvg2′−1)−f2′g1′.(4)方程组确定的两个二元隐函数u=u(x,y)，v=v(x,y)是已知函数的反函数，令F(x,y,u,v)=x−eu−usinv，G(x,y,u,v)=y−eu+ucosv，则Fx=1，Fy=0，Fu=−eu−sinv，Fv=−ucosv，Gx=0，Gy=1，Gu=−eu+cosv，Gv=−usinv，当J=∂(F,G)∂(u,v)=∣−eu−sinv−ucosv−eu+cosv−usinv∣=ueu(sinv−cosv)+u≠0时，由隐函数求导公式得∂u∂x=−∂(F,G)∂(x,v)J=−∣1−ucosv0−usinv∣J=sinveu(sinv−cosv)+1，∂u∂y=−∂(F,G)∂(y,v)J=−∣0−ucosv1−usinv∣J=−cosveu(sinv−cosv)+1，∂v∂x=−∂(F,G)∂(u,x)J=−∣−eu−sinv1−eu+cosv0∣J=cosv−euu[eu(sinv−cosv)+1]，∂v∂y=−∂(F,G)∂(u,y)J=−∣−eu−sinv0−eu+cosv1∣J=sinv+euu[eu(sinv−cosv)+1].\begin{aligned} &\ \ (1)\ 对两方程两端对x求导，得\begin{cases}\frac{dz}{dx}=2x+2y\frac{dy}{dx}，\\\\2x+4y\frac{dy}{dx}+6z\frac{dz}{dx}=0.\end{cases}，整理得\begin{cases}2y\frac{dy}{dx}-\frac{dz}{dx}=-2x，\\\\2y\frac{dy}{dx}+3z\frac{dz}{dx}=-x.\end{cases}，\\\\ &\ \ \ \ \ \ \ \ 当D=\left|\begin{array}{cccc}2y &-1\\2y &3z\end{array}\right|=6yz+2y \neq 0时，解方程组得\frac{dy}{dx}=\frac{\left|\begin{array}{cccc}-2x &-1\\-x &3z\end{array}\right|}{D}=\frac{-6xz-x}{6yz+2y}=\frac{-x(6z+1)}{2y(3z+1)}，\\\\ &\ \ \ \ \ \ \ \ \frac{dz}{dx}=\frac{\left|\begin{array}{cccc}2y &-2x\\2y &-x\end{array}\right|}{D}=\frac{2xy}{6yz+2y}=\frac{x}{3z+1}.\\\\ &\ \ (2)\ 方程组确定两个一元隐函数：x=x(z)和y=y(z)，对方程两端对z求导，整理得\begin{cases}\frac{dx}{dz}+\frac{dy}{dz}=-1，\\\\2x\frac{dx}{dz}+2y\frac{dy}{dz}=-2z.\end{cases}，\\\\ &\ \ \ \ \ \ \ \ 当D=\left|\begin{array}{cccc}1 &1\\2x &2y\end{array}\right|=2(y-x) \neq 0时，解方程组得\frac{dx}{dz}=\frac{\left|\begin{array}{cccc}-1 &1\\-2z &2y\end{array}\right|}{D}=\frac{-2y+2z}{2(y-x)}=\frac{y-z}{x-y}，\\\\ &\ \ \ \ \ \ \ \ \frac{dy}{dz}=\frac{\left|\begin{array}{cccc}1 &-1\\2x &-2z\end{array}\right|}{D}=\frac{-2z+2x}{2(y-x)}=\frac{z-x}{x-y}.\\\\ &\ \ (3)\ 方程组确定两个二元隐函数：u=u(x,\ y)，v=v(x, \ y)，分别对方程两端对x求偏导数，\\\\ &\ \ \ \ \ \ \ \ 得\begin{cases}\frac{\partial u}{\partial x}=f'_1\cdot \left(u+x\frac{\partial u}{\partial x}\right)+f'_2\cdot \frac{\partial v}{\partial x}，\\\\\frac{\partial v}{\partial x}=g'_1\cdot \left(\frac{\partial u}{\partial x}-1\right)+2g'_2yv\cdot \frac{\partial v}{\partial x}.\end{cases}，整理得\begin{cases}(xf'_1-1)\frac{\partial u}{\partial x}+f'_2\frac{\partial v}{\partial x}=-uf'_1，\\\\g'_1\frac{\partial u}{\partial x}+(2yvg'_2-1)\frac{\partial v}{\partial x}=g'_1.\end{cases}，\\\\ &\ \ \ \ \ \ \ \ 当D=\left|\begin{array}{cccc}xf'_1-1 &f'_2\\g'_1 &2yvg'_2-1\end{array}\right|=(xf'_1-1)(2yvg'_2-1)-f'_2g'_1 \neq 0时，解方程组得\\\\ &\ \ \ \ \ \ \ \ \frac{\partial u}{\partial x}=\frac{\left|\begin{array}{cccc}-uf'_1 &f'_2\\g'_1 &2yvg'_2-1\end{array}\right|}{D}=\frac{-uf'_1(2yvg'_2-1)-f'_2g'_1}{(xf'_1-1)(2yvg'_2-1)-f'_2g'_1}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial v}{\partial x}=\frac{\left|\begin{array}{cccc}xf'_1-1 &-uf'_1\\g'_1 &g'_1\end{array}\right|}{D}=\frac{g'_1(xf'_1+uf'_1-1)}{(xf'_1-1)(2yvg'_2-1)-f'_2g'_1}.\\\\ &\ \ (4)\ 方程组确定的两个二元隐函数u=u(x, \ y)，v=v(x, \ y)是已知函数的反函数，\\\\ &\ \ \ \ \ \ \ \ 令F(x, \ y, \ u, \ v)=x-e^u-usin\ v，G(x, \ y, \ u, \ v)=y-e^u+ucos\ v，\\\\ &\ \ \ \ \ \ \ \ 则F_x=1，F_y=0，F_u=-e^u-sin\ v，F_v=-ucos\ v，G_x=0，G_y=1，G_u=-e^u+cos\ v，G_v=-usin\ v，\\\\ &\ \ \ \ \ \ \ \ 当J=\frac{\partial(F, \ G)}{\partial(u, \ v)}=\left|\begin{array}{cccc}-e^u-sin\ v &-ucos\ v\\-e^u+cos\ v &-usin\ v\end{array}\right|=ue^u(sin\ v-cos\ v)+u \neq 0时，由隐函数求导公式得\\\\ &\ \ \ \ \ \ \ \ \frac{\partial u}{\partial x}=-\frac{\frac{\partial(F, \ G)}{\partial(x, \ v)}}{J}=-\frac{\left|\begin{array}{cccc}1 &-ucos\ v\\0 &-usin\ v\end{array}\right|}{J}=\frac{sin\ v}{e^u(sin\ v-cos\ v)+1}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial u}{\partial y}=-\frac{\frac{\partial(F, \ G)}{\partial(y, \ v)}}{J}=-\frac{\left|\begin{array}{cccc}0 &-ucos\ v\\1 &-usin\ v\end{array}\right|}{J}=\frac{-cos\ v}{e^u(sin\ v-cos\ v)+1}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial v}{\partial x}=-\frac{\frac{\partial(F, \ G)}{\partial(u, \ x)}}{J}=-\frac{\left|\begin{array}{cccc}-e^u-sin\ v &1\\-e^u+cos\ v &0\end{array}\right|}{J}=\frac{cos\ v-e^u}{u[e^u(sin\ v-cos\ v)+1]}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial v}{\partial y}=-\frac{\frac{\partial(F, \ G)}{\partial(u, \ y)}}{J}=-\frac{\left|\begin{array}{cccc}-e^u-sin\ v &0\\-e^u+cos\ v &1\end{array}\right|}{J}=\frac{sin\ v+e^u}{u[e^u(sin\ v-cos\ v)+1]}. & \end{aligned} (1) 对两方程两端对x求导，得⎩⎨⎧dxdz=2x+2ydxdy，2x+4ydxdy+6zdxdz=0.，整理得⎩⎨⎧2ydxdy−dxdz=−2x，2ydxdy+3zdxdz=−x.，当D=∣∣2y2y−13z∣∣=6yz+2y=0时，解方程组得dxdy=D∣∣−2x−x−13z∣∣=6yz+2y−6xz−x=2y(3z+1)−x(6z+1)， dxdz=D∣∣2y2y−2x−x∣∣=6yz+2y2xy=3z+1x. (2) 方程组确定两个一元隐函数：x=x(z)和y=y(z)，对方程两端对z求导，整理得⎩⎨⎧dzdx+dzdy=−1，2xdzdx+2ydzdy=−2z.，当D=∣∣12x12y∣∣=2(y−x)=0时，解方程组得dzdx=D∣∣−1−2z12y∣∣=2(y−x)−2y+2z=x−yy−z， dzdy=D∣∣12x−1−2z∣∣=2(y−x)−2z+2x=x−yz−x. (3) 方程组确定两个二元隐函数：u=u(x, y)，v=v(x, y)，分别对方程两端对x求偏导数，得⎩⎨⎧∂x∂u=f1′⋅(u+x∂x∂u)+f2′⋅∂x∂v，∂x∂v=g1′⋅(∂x∂u−1)+2g2′yv⋅∂x∂v.，整理得⎩⎨⎧(xf1′−1)∂x∂u+f2′∂x∂v=−uf1′，g1′∂x∂u+(2yvg2′−1)∂x∂v=g1′.，当D=∣∣xf1′−1g1′f2′2yvg2′−1∣∣=(xf1′−1)(2yvg2′−1)−f2′g1′=0时，解方程组得 ∂x∂u=D∣∣−uf1′g1′f2′2yvg2′−1∣∣=(xf1′−1)(2yvg2′−1)−f2′g1′−uf1′(2yvg2′−1)−f2′g1′， ∂x∂v=D∣∣xf1′−1g1′−uf1′g1′∣∣=(xf1′−1)(2yvg2′−1)−f2′g1′g1′(xf1′+uf1′−1). (4) 方程组确定的两个二元隐函数u=u(x, y)，v=v(x, y)是已知函数的反函数，令F(x, y, u, v)=x−eu−usin v，G(x, y, u, v)=y−eu+ucos v，则Fx=1，Fy=0，Fu=−eu−sin v，Fv=−ucos v，Gx=0，Gy=1，Gu=−eu+cos v，Gv=−usin v，当J=∂(u, v)∂(F, G)=∣∣−eu−sin v−eu+cos v−ucos v−usin v∣∣=ueu(sin v−cos v)+u=0时，由隐函数求导公式得 ∂x∂u=−J∂(x, v)∂(F, G)=−J∣∣10−ucos v−usin v∣∣=eu(sin v−cos v)+1sin v， ∂y∂u=−J∂(y, v)∂(F, G)=−J∣∣01−ucos v−usin v∣∣=eu(sin v−cos v)+1−cos v， ∂x∂v=−J∂(u, x)∂(F, G)=−J∣∣−eu−sin v−eu+cos v10∣∣=u[eu(sin v−cos v)+1]cos v−eu， ∂y∂v=−J∂(u, y)∂(F, G)=−J∣∣−eu−sin v−eu+cos v01∣∣=u[eu(sin v−cos v)+1]sin v+eu.

11.设y=f(x,t)，而t=t(x,y)是由方程F(x,y,t)=0所确定的函数，其中f，F都具有一阶连续偏导数，试证明dydx=∂f∂x∂F∂t−∂f∂t∂F∂x∂f∂t∂F∂y+∂F∂t.\begin{aligned}&11. \ 设y=f(x, \ t)，而t=t(x, \ y)是由方程F(x, \ y, \ t)=0所确定的函数，其中f，F都具有一阶连续偏导数，\\\\&\ \ \ \ \ \ 试证明\frac{dy}{dx}=\frac{\frac{\partial f}{\partial x}\frac{\partial F}{\partial t}-\frac{\partial f}{\partial t}\frac{\partial F}{\partial x}}{\frac{\partial f}{\partial t}\frac{\partial F}{\partial y}+\frac{\partial F}{\partial t}}.&\end{aligned}11. 设y=f(x, t)，而t=t(x, y)是由方程F(x, y, t)=0所确定的函数，其中f，F都具有一阶连续偏导数，试证明dxdy=∂t∂f∂y∂F+∂t∂F∂x∂f∂t∂F−∂t∂f∂x∂F.

解：

由方程组{y=f(x,t)，F(x,y,t)=0可确定两个一元隐函数y=y(x)，t=t(x)，分别对两个方程两端对x求导，得{dydx=∂f∂x+∂f∂t⋅dtdx，∂F∂x+∂F∂y⋅dydx+∂F∂t⋅dtdx=0.，整理得{dydx−∂f∂t⋅dtdx=∂f∂x，∂F∂y⋅dydx+∂F∂t⋅dtdx=−∂F∂x.，当D=∣1−∂f∂t∂F∂y∂F∂t∣=∂F∂t+∂f∂t⋅∂F∂y≠0时，解方程组得dydx=∣∂f∂x−∂f∂t−∂F∂x∂F∂t∣D=∂f∂x⋅∂F∂t−∂f∂t⋅∂F∂x∂F∂t+∂f∂t⋅∂F∂y\begin{aligned} &\ \ 由方程组\begin{cases}y=f(x, \ t)，\\\\F(x, \ y, \ t)=0\end{cases}可确定两个一元隐函数y=y(x)，t=t(x)，分别对两个方程两端对x求导，\\\\ &\ \ 得\begin{cases}\frac{dy}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}\cdot \frac{dt}{dx}，\\\\\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\cdot \frac{dy}{dx}+\frac{\partial F}{\partial t}\cdot \frac{dt}{dx}=0.\end{cases}，整理得\begin{cases}\frac{dy}{dx}-\frac{\partial f}{\partial t}\cdot \frac{dt}{dx}=\frac{\partial f}{\partial x}，\\\\\frac{\partial F}{\partial y}\cdot \frac{dy}{dx}+\frac{\partial F}{\partial t}\cdot \frac{dt}{dx}=-\frac{\partial F}{\partial x}.\end{cases}，\\\\ &\ \ 当D=\left|\begin{array}{cccc}1 &-\frac{\partial f}{\partial t}\\ \\\frac{\partial F}{\partial y} &\frac{\partial F}{\partial t}\end{array}\right|=\frac{\partial F}{\partial t}+\frac{\partial f}{\partial t}\cdot \frac{\partial F}{\partial y} \neq 0时，解方程组得\\\\ &\ \ \frac{dy}{dx}=\frac{\left|\begin{array}{cccc}\frac{\partial f}{\partial x} &-\frac{\partial f}{\partial t}\\ \\-\frac{\partial F}{\partial x} &\frac{\partial F}{\partial t}\end{array}\right|}{D}=\frac{\frac{\partial f}{\partial x}\cdot \frac{\partial F}{\partial t}-\frac{\partial f}{\partial t}\cdot \frac{\partial F}{\partial x}}{\frac{\partial F}{\partial t}+\frac{\partial f}{\partial t}\cdot \frac{\partial F}{\partial y}} & \end{aligned} 由方程组⎩⎨⎧y=f(x, t)，F(x, y, t)=0可确定两个一元隐函数y=y(x)，t=t(x)，分别对两个方程两端对x求导，得⎩⎨⎧dxdy=∂x∂f+∂t∂f⋅dxdt，∂x∂F+∂y∂F⋅dxdy+∂t∂F⋅dxdt=0.，整理得⎩⎨⎧dxdy−∂t∂f⋅dxdt=∂x∂f，∂y∂F⋅dxdy+∂t∂F⋅dxdt=−∂x∂F.，当D=∣∣1∂y∂F−∂t∂f∂t∂F∣∣=∂t∂F+∂t∂f⋅∂y∂F=0时，解方程组得 dxdy=D∣∣∂x∂f−∂x∂F−∂t∂f∂t∂F∣∣=∂t∂F+∂t∂f⋅∂y∂F∂x∂f⋅∂t∂F−∂t∂f⋅∂x∂F