求解 $z$ $=$ $\frac{y}{\sqrt{x^{2} + y^{2}}}$ 的全微分

一、题目

求解下面这个函数的全微分：

$$
z = \frac{y}{\sqrt{x^{2} + y^{2}}}
$$

难度评级：

二、解析

已知：

$$
z = \frac{y}{\sqrt{x^{2} + y^{2}}} = \frac{y}{(x^{2} + y^{2})^{\frac{1}{2}}} \Rightarrow
$$

$$
z = y \cdot (x^{2} + y^{2})^{-\frac{1}{2}}
$$

Next

则：

$$
\frac{\partial z}{\partial x} = -\frac{1}{2} y \cdot 2x (x^{2} + y^{2})^{-\frac{3}{2}} \Rightarrow
$$

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

Next

继续：

$$
\frac{\partial z}{\partial y} = \frac{\sqrt{x^{2} + y^{2}} – y \cdot \frac{1}{2} \cdot 2y (x^{2} + y^{2})^{-\frac{1}{2}}}{x^{2} + y^{2}} \Rightarrow
$$

$$
\frac{\partial z}{\partial y} = \frac{\sqrt{x^{2} + y^{2}} – \frac{y^{2}}{\sqrt{x^{2} + y^{2}}}}{x^{2} + y^{2}} \Rightarrow
$$

$$
\frac{\partial z}{\partial y} = \frac{\sqrt{x^{2} + y^{2}} \big(\sqrt{x^{2} + y^{2}} – \frac{y^{2}}{\sqrt{x^{2} + y^{2}}} \big)}{\sqrt{x^{2} + y^{2}}(x^{2} + y^{2})} \Rightarrow
$$

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

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

Next

于是：

$$
\mathrm{d} z = \frac{\partial z}{\partial x} \mathrm{d} x + \frac{\partial z}{\partial y} \mathrm{d} y \Rightarrow
$$

$$
\mathrm{d} z = \frac{-xy}{(x^{2} + y^{2})^{\frac{3}{2}}} \mathrm{d} x + \frac{x^{2}}{(x^{2} + y^{2})^{\frac{3}{2}}} \mathrm{d} y \Rightarrow
$$

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