命题 设函数$f:\mathbb R^2\to\mathbb R$可微，则对于$r:=\sqrt{x^2+y^2}$和$\theta:=\arctan\dfrac yx$成立
$$
\begin{cases}
\mathrm dr=\dfrac{x\mathrm dx+y\mathrm dy}r\\
\mathrm d\theta=\dfrac{x\mathrm dy-y\mathrm dx}{r^2}
\end{cases}.
$$
陈纪修习题12.2.13 设$z=u^v$，$u=\ln\sqrt{x^2+y^2}$，$v=\arctan \dfrac yx$，求$\mathrm dz$。
解
$$
\mathrm dz=\frac{\partial z}{\partial u}\mathrm du+\frac{\partial z}{\partial v}\mathrm dv=vu^{v-1}\frac1r\frac{x\mathrm dx+y\mathrm dy}r+u^v\ln u\cdot\frac{x\mathrm dy-y\mathrm dx}{r^2}.\ \square
$$
陈纪修习题12.2.14 设$z=(x^2+y^2)e^{-\arctan\frac yx}$，求$\mathrm dz$。
解
$$
\begin{align*}
\mathrm dz &= \mathrm d(r^2e^{-\theta})=r^2(-e^{-\theta})\mathrm d\theta+e^{-\theta}2r\mathrm dr\\
&=e^{-\theta}(2r\mathrm dr-r^2\mathrm d\theta)=e^{-\theta}(2x\mathrm dx+2y\mathrm dy-x\mathrm dy+y\mathrm dx).\ \square
\end{align*}
$$
陈纪修习题12.2.17 设$z=f(x,y)$在全平面上有定义，具有连续的偏导数，且满足方程
$$
xf_x(x,y)+yf_y(x,y)=0,
$$
证明：$f(x,y)$为常数。
证
$$
f_x(x,y) = \frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}=\frac{\partial f}{\partial r}\frac xr-\frac{\partial f}{\partial \theta}\frac y{r^2},\\
f_y(x,y) = \frac{\partial f}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial y}=\frac{\partial f}{\partial r}\frac yr+\frac{\partial f}{\partial \theta}\frac x{r^2}.
$$
于是
$$
0=xf_x(x,y)+yf_y(x,y)=r\frac{\partial f}{\partial r},
$$
故$\dfrac{\partial f}{\partial r}\equiv0$，进而$\dfrac{\partial f}{\partial \theta}\equiv0$。$\square$