$$$\ln\left(\sqrt{x} + 2\right)$$$ 的導數

函數 $$$\ln\left(\sqrt{x} + 2\right)$$$ 是兩個函數 $$$f{\left(u \right)} = \ln\left(u\right)$$$ 與 $$$g{\left(x \right)} = \sqrt{x} + 2$$$ 之複合 $$$f{\left(g{\left(x \right)} \right)}$$$。

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

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

自然對數的導數為 $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$：

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

返回原變數：

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

和/差的導數等於導數的和/差：

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

常數的導數為$$$0$$$：

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

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

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

化簡：

$$\frac{1}{2 \sqrt{x} \left(\sqrt{x} + 2\right)} = \frac{1}{2 \left(2 \sqrt{x} + x\right)}$$

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