纯粹拿来练手的题目:《数学分析,高等教育出版社》例14.1.6,练习14.1,例14.2.1,例14.2.4

 (1) $u(x,y)=xy$,(2) $u(x,y)=xe^x\sin y$,求二阶偏导数.

解:(1)$\frac{\partial u}{\partial x}=y$,$\frac{\partial u}{\partial y}=x$.因此 $\frac{\partial^2 u}{\partial y^2}=0$,$\frac{\partial^{2}u}{\partial x^{2}}=0$,$\frac{\partial^2u}{\partial y\partial x}=1$,$\frac{\partial^2u}{\partial x\partial y}=1$.

(2)$\frac{\partial u}{\partial x}=\sin y(e^x+xe^x)$.$\frac{\partial u}{\partial y}=xe^x\cos y$,$\frac{\partial^2u}{\partial y\partial x}=\cos y(e^x+xe^x)$.$\frac{\partial^2u}{\partial x\partial y}=\cos y(e^x+xe^x)$,$\frac{\partial^2u}{\partial y^2}=-xe^x\sin y$,$\frac{\partial^2u}{\partial x^2}=\sin y(2e^x+xe^x)$.

 

 

 

 

练习14.1:求下列函数的偏导数.

(1)$z=x^2\ln (x^2+y^2)$.

$$\frac{\partial z}{\partial x}=2x\ln (x^2+y^2)+
\frac{2x^3}{x^2+y^2}.$$
$$
\frac{\partial z}{\partial y}=\frac{2x^2y}{x^2+y^2}.
$$



(2)$u=e^{xy}$.


$$
\frac{\partial u}{\partial x}=ye^{xy}.
$$
$$
\frac{\partial u}{\partial y}=xe^{xy}.
$$

(3)$z=xy+\frac{x}{y}$.
$$
\frac{\partial z}{\partial x}=y+\frac{1}{y}.
$$
$$
\frac{\partial z}{\partial y}=x-x \frac{1}{y^2}.
$$
(4)$u=\arctan \frac{y}{x}$.
$$
\frac{\partial u}{\partial x}=\frac{-1}{x^{2}+y^2}.
$$
$$
\frac{\partial u}{\partial y}=\frac{1}{x+\frac{y^2}{x}}.
$$
(5)$u=x^2+y^2+z^2+2xy+2yz+2zx$.
$$
\frac{\partial u}{\partial x}=2x+2y+2z.
$$
至于其余两个,根据对称性易得也是上面的答案.

(6)$u=e^{\phi-\theta}\cos(\theta +\phi)$.
$$
\frac{\partial u}{\partial\theta}=-e^{\phi-\theta}\cos(\theta+\phi)-e^{\phi-\theta}\sin(\theta+\phi).
$$
$$
\frac{\partial u}{\partial \phi}=e^{\phi-\theta}\cos(\theta+\phi)-e^{\phi-\theta}\sin(\theta+\phi).
$$

 

例14.2.1:设 $u=e^x\sin y,x=2st,y=t+s^2$,求 $u_s,u_t$.

解:$s$ 变化导致 $x$ 变化,同时导致 $y$ 变化,而 $x,y$ 变化导致 $u$ 变化.从$s$ 到 $(x,y)$ 是一个函数 $f:\mathbf{R}\to \mathbf{R}^2$,$f$ 的导数是
$$
\begin{pmatrix}
  2t\\
2s\\
\end{pmatrix}.
$$
从 $x,y$ 到 $u$ 是一个函数 $g:\mathbf{R}^2\to \mathbf{R}$.它的导数是
$$
\begin{pmatrix}
  \sin y e^x&e^x\cos y
\end{pmatrix}.
$$
因此
$$
u_s=\begin{pmatrix}
  e^x\sin y&e^x\cos y
\end{pmatrix}\begin{pmatrix}
  2t\\
2s\\
\end{pmatrix}=2t\sin ye^x+2se^x\cos y.
$$
从 $t$ 到 $(x,y)$ 是一个函数 $f':\mathbf{R}\to \mathbf{R}^2$,$f'$ 的导数是
$$
\begin{pmatrix}
  2s\\
1\\
\end{pmatrix},
$$
因此
$$
u_t=\begin{pmatrix}
  \sin ye^x&e^x\cos y
\end{pmatrix}\begin{pmatrix}
  2s\\
1\\
\end{pmatrix}=2s\sin ye^x+e^x\cos y.
$$

 

例14.2.4:已知 $u=u(x,y)$,在极坐标 $x=r\cos\theta$,$y=r\sin\theta$ 变换下,证明
$$
(\frac{\partial u}{\partial r})^2+\frac{1}{r^2}(\frac{\partial u}{\partial
  \theta})^2=(\frac{\partial u}{\partial x})^2+(\frac{\partial u}{\partial y})^2.
$$
证明:函数 $f:\mathbf{R}^2\to \mathbf{R}^2$,具体规则如下:
$$
f:(\theta,r)\to (x,y).
$$
函数 $g:\mathbf{R}^2\to \mathbf{R}$,具体规则如下:
$$
g:(x,y)\to u.
$$
根据链法则易得
$$
\begin{pmatrix}
  \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}
\end{pmatrix}\begin{pmatrix}
  -r\sin\theta&\cos\theta\\
r\cos\theta&\sin\theta\\
\end{pmatrix}=\begin{pmatrix}
  \frac{\partial u}{\partial \theta}&\frac{\partial u}{\partial r}
\end{pmatrix}.
$$
也就是
$$
\begin{cases}
  \frac{\partial u}{\partial \theta}=-r\sin\theta\frac{\partial u}{\partial x}+r\cos\theta
  \frac{\partial u}{\partial y}\\
\frac{\partial u}{\partial r}=\cos\theta \frac{\partial u}{\partial x}+\sin\theta
\frac{\partial u}{\partial y}.
\end{cases}
$$
所以易得命题成立.