高等数学（第七版）同济大学习题9-2 个人解答

高等数学（第七版）同济大学习题9-2

1.求下列函数的偏导数：\begin{aligned}&1. \ 求下列函数的偏导数：&\end{aligned}1. 求下列函数的偏导数：

(1)z=x3y−y3x； (2)s=u2+v2uv；(3)z=ln(xy)； (4)z=sin(xy)+cos2(xy)；(5)z=lntanxy； (6)z=(1+xy)y；(7)u=xyz； (8)u=arctan(x−y)z.\begin{aligned} &\ \ (1)\ \ z=x^3y-y^3x；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ s=\frac{u^2+v^2}{uv}；\\\\ &\ \ (3)\ \ z=\sqrt{ln(xy)}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ z=sin(xy)+cos^2(xy)；\\\\ &\ \ (5)\ \ z=ln\ tan\ \frac{x}{y}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)\ \ z=(1+xy)^y；\\\\ &\ \ (7)\ \ u=x^{\frac{y}{z}}；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (8)\ \ u=arctan(x-y)^z. & \end{aligned} (1) z=x3y−y3x； (2) s=uvu2+v2； (3) z=ln(xy)； (4) z=sin(xy)+cos2(xy)； (5) z=ln tan yx； (6) z=(1+xy)y； (7) u=xzy； (8) u=arctan(x−y)z.

解：

(1)∂z∂x=3yx2−y3，∂z∂y=x3−3xy2.(2)∂s∂u=1v−vu2，∂z∂v=−uv2+1u，因为u和v分别看作常数，可以直接约掉。(3)∂z∂x=12ln(xy)⋅1xy⋅y=12xln(xy)，∂z∂y=12ln(xy)⋅1xy⋅x=12yln(xy)(4)∂z∂x=ycos(xy)+2cos(xy)⋅[−sin(xy)]⋅y=ycos(xy)−ysin(2xy)，∂z∂y=xcos(xy)+2cos(xy)⋅[−sin(xy)]⋅x=xcos(xy)−xsin(2xy).(5)∂z∂x=1tanxy⋅sec2xy⋅1y=2ysin2xy，∂z∂y=1tanxy⋅sec2xy⋅(−xy2)=−2xy2sin2xy.(6)∂z∂x=y(1+xy)y−1⋅y=y2(1+xy)y−1，∂z∂y=(1+xy)y[ln(1+xy)+xy1+xy].(7)∂u∂x=yzxyz−1，∂u∂y=1zxyzlnx，∂u∂z=−yz2xyzlnx.(8)∂u∂x=z(x−y)z−11+(x−y)2z，∂u∂y=−z(x−y)z−11+(x−y)2z，∂u∂z=(x−y)zln(x−y)1+(x−y)2z.\begin{aligned} &\ \ (1)\ \frac{\partial z}{\partial x}=3yx^2-y3，\frac{\partial z}{\partial y}=x^3-3xy^2.\\\\ &\ \ (2)\ \frac{\partial s}{\partial u}=\frac{1}{v}-\frac{v}{u^2}，\frac{\partial z}{\partial v}=-\frac{u}{v^2}+\frac{1}{u}，因为u和v分别看作常数，可以直接约掉。\\\\ &\ \ (3)\ \frac{\partial z}{\partial x}=\frac{1}{2\sqrt{ln(xy)}}\cdot \frac{1}{xy}\cdot y=\frac{1}{2x\sqrt{ln(xy)}}，\frac{\partial z}{\partial y}=\frac{1}{2\sqrt{ln(xy)}}\cdot \frac{1}{xy}\cdot x=\frac{1}{2y\sqrt{ln(xy)}}\\\\ &\ \ (4)\ \frac{\partial z}{\partial x}=ycos(xy)+2cos(xy)\cdot [-sin(xy)]\cdot y=ycos(xy)-ysin(2xy)，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial z}{\partial y}=xcos(xy)+2cos(xy)\cdot [-sin(xy)]\cdot x=xcos(xy)-xsin(2xy).\\\\ &\ \ (5)\ \frac{\partial z}{\partial x}=\frac{1}{tan\ \frac{x}{y}}\cdot sec^2\ \frac{x}{y}\cdot \frac{1}{y}=\frac{2}{ysin\ \frac{2x}{y}}，\frac{\partial z}{\partial y}=\frac{1}{tan\ \frac{x}{y}}\cdot sec^2\ \frac{x}{y}\cdot \left(-\frac{x}{y^2}\right)=-\frac{2x}{y^2sin\ \frac{2x}{y}}.\\\\ &\ \ (6)\ \frac{\partial z}{\partial x}=y(1+xy)^{y-1}\cdot y=y^2(1+xy)^{y-1}，\frac{\partial z}{\partial y}=(1+xy)^y\left[ln(1+xy)+\frac{xy}{1+xy}\right].\\\\ &\ \ (7)\ \frac{\partial u}{\partial x}=\frac{y}{z}x^{\frac{y}{z}-1}，\frac{\partial u}{\partial y}=\frac{1}{z}x^{\frac{y}{z}}ln\ x，\frac{\partial u}{\partial z}=-\frac{y}{z^2}x^{\frac{y}{z}}ln\ x.\\\\ &\ \ (8)\ \frac{\partial u}{\partial x}=\frac{z(x-y)^{z-1}}{1+(x-y)^{2z}}，\frac{\partial u}{\partial y}=-\frac{z(x-y)^{z-1}}{1+(x-y)^{2z}}，\frac{\partial u}{\partial z}=\frac{(x-y)^zln(x-y)}{1+(x-y)^{2z}}. & \end{aligned} (1) ∂x∂z=3yx2−y3，∂y∂z=x3−3xy2. (2) ∂u∂s=v1−u2v，∂v∂z=−v2u+u1，因为u和v分别看作常数，可以直接约掉。 (3) ∂x∂z=2ln(xy)1⋅xy1⋅y=2xln(xy)1，∂y∂z=2ln(xy)1⋅xy1⋅x=2yln(xy)1 (4) ∂x∂z=ycos(xy)+2cos(xy)⋅[−sin(xy)]⋅y=ycos(xy)−ysin(2xy)， ∂y∂z=xcos(xy)+2cos(xy)⋅[−sin(xy)]⋅x=xcos(xy)−xsin(2xy). (5) ∂x∂z=tan yx1⋅sec2 yx⋅y1=ysin y2x2，∂y∂z=tan yx1⋅sec2 yx⋅(−y2x)=−y2sin y2x2x. (6) ∂x∂z=y(1+xy)y−1⋅y=y2(1+xy)y−1，∂y∂z=(1+xy)y[ln(1+xy)+1+xyxy]. (7) ∂x∂u=zyxzy−1，∂y∂u=z1xzyln x，∂z∂u=−z2yxzyln x. (8) ∂x∂u=1+(x−y)2zz(x−y)z−1，∂y∂u=−1+(x−y)2zz(x−y)z−1，∂z∂u=1+(x−y)2z(x−y)zln(x−y).

2.设T=2πlg，求证l∂T∂l+g∂T∂g=0.\begin{aligned}&2. \ 设T=2\pi \sqrt{\frac{l}{g}}，求证l\frac{\partial T}{\partial l}+g\frac{\partial T}{\partial g}=0.&\end{aligned}2. 设T=2πgl，求证l∂l∂T+g∂g∂T=0.

解：

因为∂T∂l=2π⋅12lg⋅1g=πgl，∂T∂g=2π⋅12lg⋅(−lg2)=−πglg，所以l∂T∂l+g∂T∂g=πlg−πlg=0.\begin{aligned} &\ \ 因为\frac{\partial T}{\partial l}=2\pi \cdot \frac{1}{2\sqrt{\frac{l}{g}}}\cdot \frac{1}{g}=\frac{\pi}{\sqrt{gl}}，\frac{\partial T}{\partial g}=2\pi \cdot \frac{1}{2\sqrt{\frac{l}{g}}}\cdot \left(-\frac{l}{g^2}\right)=-\frac{\pi}{g}\sqrt{\frac{l}{g}}，所以l\frac{\partial T}{\partial l}+g\frac{\partial T}{\partial g}=\pi \sqrt{\frac{l}{g}}-\pi \sqrt{\frac{l}{g}}=0. & \end{aligned} 因为∂l∂T=2π⋅2gl1⋅g1=glπ，∂g∂T=2π⋅2gl1⋅(−g2l)=−gπgl，所以l∂l∂T+g∂g∂T=πgl−πgl=0.

3.设z=e−(1x+1y)，求证x2∂z∂x+y2∂z∂y=2z.\begin{aligned}&3. \ 设z=e^{-(\frac{1}{x}+\frac{1}{y})}，求证x^2\frac{\partial z}{\partial x}+y^2\frac{\partial z}{\partial y}=2z.&\end{aligned}3. 设z=e−(x1+y1)，求证x2∂x∂z+y2∂y∂z=2z.

解：

因为∂z∂x=1x2e−(1x+1y)，∂z∂y=1y2e−(1x+1y)，所以x2∂z∂x+y2∂z∂y=2e−(1x+1y)=2z.\begin{aligned} &\ \ 因为\frac{\partial z}{\partial x}=\frac{1}{x^2}e^{-(\frac{1}{x}+\frac{1}{y})}，\frac{\partial z}{\partial y}=\frac{1}{y^2}e^{-(\frac{1}{x}+\frac{1}{y})}，所以x^2\frac{\partial z}{\partial x}+y^2\frac{\partial z}{\partial y}=2e^{-(\frac{1}{x}+\frac{1}{y})}=2z. & \end{aligned} 因为∂x∂z=x21e−(x1+y1)，∂y∂z=y21e−(x1+y1)，所以x2∂x∂z+y2∂y∂z=2e−(x1+y1)=2z.

4.设f(x,y)=x+(y−1)arcsinxy，求fx(x,1).\begin{aligned}&4. \ 设f(x, \ y)=x+(y-1)arcsin\ \sqrt{\frac{x}{y}}，求f_x(x, \ 1).&\end{aligned}4. 设f(x, y)=x+(y−1)arcsin yx，求fx(x, 1).

解：

fx(x,y)=1+y−11−xy⋅12xy⋅1y，fx(x,1)=1.\begin{aligned} &\ \ f_x(x, \ y)=1+\frac{y-1}{\sqrt{1-\frac{x}{y}}}\cdot \frac{1}{2\sqrt{\frac{x}{y}}}\cdot \frac{1}{y}，f_x(x, \ 1)=1. & \end{aligned} fx(x, y)=1+1−yxy−1⋅2yx1⋅y1，fx(x, 1)=1.

5.曲线{z=x2+y24，y=4在点(2,4,5)处的切线对于x轴的倾角是多少？\begin{aligned}&5. \ 曲线\begin{cases}z=\frac{x^2+y^2}{4}，\\\\y=4\end{cases}在点(2, \ 4, \ 5)处的切线对于x轴的倾角是多少？&\end{aligned}5. 曲线⎩⎨⎧z=4x2+y2，y=4在点(2, 4, 5)处的切线对于x轴的倾角是多少？

解：

设z=f(x,y)，根据偏导数的几何意义，fx(2,4)是曲线在点(2,4,5)处的切线对于x轴的斜率，而fx(2,4)=12x∣x=2=1，即k=tanα=1，得倾角α=π4.\begin{aligned} &\ \ 设z=f(x, \ y)，根据偏导数的几何意义，f_x(2, \ 4)是曲线在点(2, \ 4, \ 5)处的切线对于x轴的斜率，\\\\ &\ \ 而f_x(2, \ 4)=\frac{1}{2}x\bigg|_{x=2}=1，即k=tan\ \alpha=1，得倾角\alpha=\frac{\pi}{4}. & \end{aligned} 设z=f(x, y)，根据偏导数的几何意义，fx(2, 4)是曲线在点(2, 4, 5)处的切线对于x轴的斜率，而fx(2, 4)=21x∣∣x=2=1，即k=tan α=1，得倾角α=4π.

6.求下列函数的∂2z∂x2，∂2z∂y2，∂2z∂x∂y；\begin{aligned}&6. \ 求下列函数的\frac{\partial^2 z}{\partial x^2}，\frac{\partial^2 z}{\partial y^2}，\frac{\partial^2 z}{\partial x \partial y}；&\end{aligned}6. 求下列函数的∂x2∂2z，∂y2∂2z，∂x∂y∂2z；

(1)z=x4+y4−4x2y2；(2)z=arctanyx；(3)z=yx.\begin{aligned} &\ \ (1)\ \ z=x^4+y^4-4x^2y^2；\\\\ &\ \ (2)\ \ z=arctan \frac{y}{x}；\\\\ &\ \ (3)\ \ z=y^x. & \end{aligned} (1) z=x4+y4−4x2y2； (2) z=arctanxy； (3) z=yx.

解：

(1)∂z∂x=4x3−8xy2，∂2z∂x2=12x2−8y2，∂z∂y=4y3−8x2y，∂2z∂y2=12y2−8x2，∂2z∂x∂y=∂∂y(4x3−8xy2)=−16xy.(2)∂z∂x=11+(yx)2⋅(−yx2)=−yx2+y2，∂2z∂x2=2xy(x2+y2)2，∂z∂y=11+(yx)2⋅1x=xx2+y2，∂2z∂y2=−2xy(x2+y2)2，∂2z∂x∂y=∂∂y(−yx2+y2)=−(x2+y2)−y⋅2y(x2+y2)2=y2−x2(x2+y2)2.(3)∂z∂x=yxlny，∂2z∂x2=yxln2y，∂z∂y=xyx−1，∂2z∂y2=x(x−1)yx−2，∂2z∂x∂y=∂∂y(yxlny)=yx−1(1+xlny).\begin{aligned} &\ \ (1)\ \frac{\partial z}{\partial x}=4x^3-8xy^2，\frac{\partial^2 z}{\partial x^2}=12x^2-8y^2，\frac{\partial z}{\partial y}=4y^3-8x^2y，\frac{\partial^2 z}{\partial y^2}=12y^2-8x^2，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial}{\partial y}(4x^3-8xy^2)=-16xy.\\\\ &\ \ (2)\ \frac{\partial z}{\partial x}=\frac{1}{1+\left(\frac{y}{x}\right)^2}\cdot \left(-\frac{y}{x^2}\right)=-\frac{y}{x^2+y^2}，\frac{\partial^2 z}{\partial x^2}=\frac{2xy}{(x^2+y^2)^2}，\frac{\partial z}{\partial y}=\frac{1}{1+\left(\frac{y}{x}\right)^2}\cdot \frac{1}{x}=\frac{x}{x^2+y^2}，\frac{\partial^2 z}{\partial y^2}=-\frac{2xy}{(x^2+y^2)^2}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(-\frac{y}{x^2+y^2}\right)=-\frac{(x^2+y^2)-y\cdot 2y}{(x^2+y^2)^2}=\frac{y^2-x^2}{(x^2+y^2)^2}.\\\\ &\ \ (3)\ \frac{\partial z}{\partial x}=y^xln\ y，\frac{\partial^2 z}{\partial x^2}=y^xln^2\ y，\frac{\partial z}{\partial y}=xy^{x-1}，\frac{\partial^2 z}{\partial y^2}=x(x-1)y^{x-2}，\frac{\partial^2 z}{\partial x \partial y}=\frac{\partial}{\partial y}(y^xln\ y)=y^{x-1}(1+xln\ y). & \end{aligned} (1) ∂x∂z=4x3−8xy2，∂x2∂2z=12x2−8y2，∂y∂z=4y3−8x2y，∂y2∂2z=12y2−8x2， ∂x∂y∂2z=∂y∂(4x3−8xy2)=−16xy. (2) ∂x∂z=1+(xy)21⋅(−x2y)=−x2+y2y，∂x2∂2z=(x2+y2)22xy，∂y∂z=1+(xy)21⋅x1=x2+y2x，∂y2∂2z=−(x2+y2)22xy， ∂x∂y∂2z=∂y∂(−x2+y2y)=−(x2+y2)2(x2+y2)−y⋅2y=(x2+y2)2y2−x2. (3) ∂x∂z=yxln y，∂x2∂2z=yxln2 y，∂y∂z=xyx−1，∂y2∂2z=x(x−1)yx−2，∂x∂y∂2z=∂y∂(yxln y)=yx−1(1+xln y).

7.设f(x,y,z)=xy2+yz2+zx2，求fxx(0,0,1)，fxz(1,0,2)，fyz(0,−1,0)及fzzx(2,0,1).\begin{aligned}&7. \ 设f(x, \ y, \ z)=xy^2+yz^2+zx^2，求f_{xx}(0, \ 0 , \ 1)，f_{xz}(1, \ 0, \ 2)，f_{yz}(0, \ -1, \ 0)及f_{zzx}(2, \ 0, \ 1).&\end{aligned}7. 设f(x, y, z)=xy2+yz2+zx2，求fxx(0, 0, 1)，fxz(1, 0, 2)，fyz(0, −1, 0)及fzzx(2, 0, 1).

解：

因为fx=y2+2xz，fxx=2z，fxz=2x，fy=2xy+z2，fyz=2z，fz=2yz+x2，fzz=2y，fzzx=0，所以fxx(0,0,1)=2，fxz(1,0,2)=2，fyz(0,−1,0)=2，fzzx(2,0,1)=0.\begin{aligned} &\ \ 因为f_x=y^2+2xz，f_{xx}=2z，f_{xz}=2x，f_y=2xy+z^2，f_{yz}=2z，f_z=2yz+x^2，f_{zz}=2y，f_{zzx}=0，\\\\ &\ \ 所以f_{xx}(0, \ 0 , \ 1)=2，f_{xz}(1, \ 0, \ 2)=2，f_{yz}(0, \ -1, \ 0)=2，f_{zzx}(2, \ 0, \ 1)=0. & \end{aligned} 因为fx=y2+2xz，fxx=2z，fxz=2x，fy=2xy+z2，fyz=2z，fz=2yz+x2，fzz=2y，fzzx=0，所以fxx(0, 0, 1)=2，fxz(1, 0, 2)=2，fyz(0, −1, 0)=2，fzzx(2, 0, 1)=0.

8.设z=xln(xy)，求∂3z∂x2∂y及∂3z∂x∂y2.\begin{aligned}&8. \ 设z=xln(xy)，求\frac{\partial^3 z}{\partial x^2 \partial y}及\frac{\partial^3 z}{\partial x \partial y^2}.&\end{aligned}8. 设z=xln(xy)，求∂x2∂y∂3z及∂x∂y2∂3z.

解：

∂z∂x=ln(xy)+x⋅yxy=ln(xy)+1，∂2z∂x2=yxy=1x，∂3z∂x2∂y=0，∂2z∂x∂y=xxy=1y，∂3z∂x∂y2=−1y2.\begin{aligned} &\ \ \frac{\partial z}{\partial x}=ln(xy)+x\cdot \frac{y}{xy}=ln(xy)+1，\frac{\partial^2 z}{\partial x^2}=\frac{y}{xy}=\frac{1}{x}，\frac{\partial^3 z}{\partial x^2 \partial y}=0，\frac{\partial^2 z}{\partial x \partial y}=\frac{x}{xy}=\frac{1}{y}，\frac{\partial^3 z}{\partial x \partial y^2}=-\frac{1}{y^2}. & \end{aligned} ∂x∂z=ln(xy)+x⋅xyy=ln(xy)+1，∂x2∂2z=xyy=x1，∂x2∂y∂3z=0，∂x∂y∂2z=xyx=y1，∂x∂y2∂3z=−y21.

9.验证：\begin{aligned}&9. \ 验证：&\end{aligned}9. 验证：

(1)y=e−kn2tsinnx满足∂y∂t=k∂2y∂x2；(2)r=x2+y2+z2满足∂2r∂x2+∂2r∂y2+∂2r∂z2=2r.\begin{aligned} &\ \ (1)\ \ y=e^{-kn^2t}sin\ nx满足\frac{\partial y}{\partial t}=k\frac{\partial^2 y}{\partial x^2}；\\\\ &\ \ (2)\ \ r=\sqrt{x^2+y^2+z^2}满足\frac{\partial^2 r}{\partial x^2}+\frac{\partial^2 r}{\partial y^2}+\frac{\partial^2 r}{\partial z^2}=\frac{2}{r}. & \end{aligned} (1) y=e−kn2tsin nx满足∂t∂y=k∂x2∂2y； (2) r=x2+y2+z2满足∂x2∂2r+∂y2∂2r+∂z2∂2r=r2.

解：

(1)因为∂y∂t=−kn2e−kn2tsinnx，∂y∂x=ne−kn2tcosnx，∂2y∂x2=∂∂x(ne−kn2tcosnx)=−n2e−kn2tsinnx，所以∂y∂t=k(−n2e−kn2tsinnx)=k∂2y∂x2.(2)因为∂r∂x=xx2+y2+r2=xr，∂2r∂x2=∂∂x(xr)=1r−xr2⋅xr=r2−x2r3，根据函数关于自变量得对称性得∂2r∂y2=r2−y2r3，∂2r∂z2=r2−z2r3，所以∂2r∂x2+∂2r∂y2+∂2r∂z2=r2−x2r3+r2−y2r3+r2−z2r3=2r.\begin{aligned} &\ \ (1)\ 因为\frac{\partial y}{\partial t}=-kn^2e^{-kn^2t}sin\ nx，\frac{\partial y}{\partial x}=ne^{-kn^2t}cos\ nx，\frac{\partial^2 y}{\partial x^2}=\frac{\partial}{\partial x}(ne^{-kn^2t}cos\ nx)=-n^2e^{-kn^2t}sin\ nx，\\\\ &\ \ \ \ \ \ \ \ 所以\frac{\partial y}{\partial t}=k(-n^2e^{-kn^2t}sin\ nx)=k\frac{\partial^2 y}{\partial x^2}.\\\\ &\ \ (2)\ 因为\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2+r^2}}=\frac{x}{r}，\frac{\partial^2 r}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{x}{r}\right)=\frac{1}{r}-\frac{x}{r^2}\cdot \frac{x}{r}=\frac{r^2-x^2}{r^3}，根据函数关于自变量得对称性得\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 r}{\partial y^2}=\frac{r^2-y^2}{r^3}，\frac{\partial^2 r}{\partial z^2}=\frac{r^2-z^2}{r^3}，所以\frac{\partial^2 r}{\partial x^2}+\frac{\partial^2 r}{\partial y^2}+\frac{\partial^2 r}{\partial z^2}=\frac{r^2-x^2}{r^3}+\frac{r^2-y^2}{r^3}+\frac{r^2-z^2}{r^3}=\frac{2}{r}. & \end{aligned} (1) 因为∂t∂y=−kn2e−kn2tsin nx，∂x∂y=ne−kn2tcos nx，∂x2∂2y=∂x∂(ne−kn2tcos nx)=−n2e−kn2tsin nx，所以∂t∂y=k(−n2e−kn2tsin nx)=k∂x2∂2y. (2) 因为∂x∂r=x2+y2+r2x=rx，∂x2∂2r=∂x∂(rx)=r1−r2x⋅rx=r3r2−x2，根据函数关于自变量得对称性得 ∂y2∂2r=r3r2−y2，∂z2∂2r=r3r2−z2，所以∂x2∂2r+∂y2∂2r+∂z2∂2r=r3r2−x2+r3r2−y2+r3r2−z2=r2.