$$$\ln\left(\sqrt{10} \sqrt{x}\right)$$$的导数

函数$$$\ln\left(\sqrt{10} \sqrt{x}\right)$$$是两个函数$$$f{\left(u \right)} = \ln\left(u\right)$$$和$$$g{\left(x \right)} = \sqrt{10} \sqrt{x}$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

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

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\sqrt{10} \sqrt{x}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(\sqrt{10} \sqrt{x}\right)\right)}$$

自然对数的导数为 $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$：

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(\sqrt{10} \sqrt{x}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(\sqrt{10} \sqrt{x}\right)$$

返回到原变量：

$$\frac{\frac{d}{dx} \left(\sqrt{10} \sqrt{x}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(\sqrt{10} \sqrt{x}\right)}{{\color{red}\left(\sqrt{10} \sqrt{x}\right)}}$$

对 $$$c = \sqrt{10}$$$ 和 $$$f{\left(x \right)} = \sqrt{x}$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

$$\frac{\sqrt{10} {\color{red}\left(\frac{d}{dx} \left(\sqrt{10} \sqrt{x}\right)\right)}}{10 \sqrt{x}} = \frac{\sqrt{10} {\color{red}\left(\sqrt{10} \frac{d}{dx} \left(\sqrt{x}\right)\right)}}{10 \sqrt{x}}$$

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

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

因此，$$$\frac{d}{dx} \left(\ln\left(\sqrt{10} \sqrt{x}\right)\right) = \frac{1}{2 x}$$$。