$$$\ln\left(1 + \frac{3}{n}\right)$$$的导数

函数$$$\ln\left(1 + \frac{3}{n}\right)$$$是两个函数$$$f{\left(u \right)} = \ln\left(u\right)$$$和$$$g{\left(n \right)} = 1 + \frac{3}{n}$$$的复合$$$f{\left(g{\left(n \right)} \right)}$$$。

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

$${\color{red}\left(\frac{d}{dn} \left(\ln\left(1 + \frac{3}{n}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dn} \left(1 + \frac{3}{n}\right)\right)}$$

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

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dn} \left(1 + \frac{3}{n}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dn} \left(1 + \frac{3}{n}\right)$$

返回到原变量：

$$\frac{\frac{d}{dn} \left(1 + \frac{3}{n}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dn} \left(1 + \frac{3}{n}\right)}{{\color{red}\left(1 + \frac{3}{n}\right)}}$$

和/差的导数等于导数的和/差：

$$\frac{{\color{red}\left(\frac{d}{dn} \left(1 + \frac{3}{n}\right)\right)}}{1 + \frac{3}{n}} = \frac{{\color{red}\left(\frac{d}{dn} \left(1\right) + \frac{d}{dn} \left(\frac{3}{n}\right)\right)}}{1 + \frac{3}{n}}$$

对 $$$c = 3$$$ 和 $$$f{\left(n \right)} = \frac{1}{n}$$$ 应用常数倍法则 $$$\frac{d}{dn} \left(c f{\left(n \right)}\right) = c \frac{d}{dn} \left(f{\left(n \right)}\right)$$$：

$$\frac{{\color{red}\left(\frac{d}{dn} \left(\frac{3}{n}\right)\right)} + \frac{d}{dn} \left(1\right)}{1 + \frac{3}{n}} = \frac{{\color{red}\left(3 \frac{d}{dn} \left(\frac{1}{n}\right)\right)} + \frac{d}{dn} \left(1\right)}{1 + \frac{3}{n}}$$

常数的导数是$$$0$$$:

$$\frac{{\color{red}\left(\frac{d}{dn} \left(1\right)\right)} + 3 \frac{d}{dn} \left(\frac{1}{n}\right)}{1 + \frac{3}{n}} = \frac{{\color{red}\left(0\right)} + 3 \frac{d}{dn} \left(\frac{1}{n}\right)}{1 + \frac{3}{n}}$$

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

$$\frac{3 {\color{red}\left(\frac{d}{dn} \left(\frac{1}{n}\right)\right)}}{1 + \frac{3}{n}} = \frac{3 {\color{red}\left(- \frac{1}{n^{2}}\right)}}{1 + \frac{3}{n}}$$

化简：

$$- \frac{3}{n^{2} \left(1 + \frac{3}{n}\right)} = - \frac{3}{n \left(n + 3\right)}$$

因此，$$$\frac{d}{dn} \left(\ln\left(1 + \frac{3}{n}\right)\right) = - \frac{3}{n \left(n + 3\right)}$$$。