$$$\sqrt{1 - x}$$$ 的導數

函數 $$$\sqrt{1 - x}$$$ 是兩個函數 $$$f{\left(u \right)} = \sqrt{u}$$$ 與 $$$g{\left(x \right)} = 1 - x$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。

應用鏈式法則 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$：

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

套用冪次法則 $$$\frac{d}{du} \left(u^{n}\right) = n u^{n - 1}$$$，取 $$$n = \frac{1}{2}$$$：

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

返回原變數：

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

和/差的導數等於導數的和/差：

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

常數的導數為$$$0$$$：

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

套用冪次法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是 $$$\frac{d}{dx} \left(x\right) = 1$$$：

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

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