$$$\sqrt{1 - x}$$$的导数

函数$$$\sqrt{1 - x}$$$是两个函数$$$f{\left(u \right)} = \sqrt{u}$$$和$$$g{\left(x \right)} = 1 - x$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

应用链式法则 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

$${\color{red}\left(\frac{d}{dx} \left(\sqrt{1 - x}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right) \frac{d}{dx} \left(1 - x\right)\right)}$$

应用幂次法则 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，其中 $$$n = \frac{1}{2}$$$:

$${\color{red}\left(\frac{d}{du} \left(\sqrt{u}\right)\right)} \frac{d}{dx} \left(1 - x\right) = {\color{red}\left(\frac{1}{2 \sqrt{u}}\right)} \frac{d}{dx} \left(1 - x\right)$$

返回到原变量：

$$\frac{\frac{d}{dx} \left(1 - x\right)}{2 \sqrt{{\color{red}\left(u\right)}}} = \frac{\frac{d}{dx} \left(1 - x\right)}{2 \sqrt{{\color{red}\left(1 - x\right)}}}$$

和/差的导数等于导数的和/差：

$$\frac{{\color{red}\left(\frac{d}{dx} \left(1 - x\right)\right)}}{2 \sqrt{1 - x}} = \frac{{\color{red}\left(\frac{d}{dx} \left(1\right) - \frac{d}{dx} \left(x\right)\right)}}{2 \sqrt{1 - x}}$$

常数的导数是$$$0$$$:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(1\right)\right)} - \frac{d}{dx} \left(x\right)}{2 \sqrt{1 - x}} = \frac{{\color{red}\left(0\right)} - \frac{d}{dx} \left(x\right)}{2 \sqrt{1 - x}}$$

应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：

$$- \frac{{\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{2 \sqrt{1 - x}} = - \frac{{\color{red}\left(1\right)}}{2 \sqrt{1 - x}}$$

因此，$$$\frac{d}{dx} \left(\sqrt{1 - x}\right) = - \frac{1}{2 \sqrt{1 - x}}$$$。