Derivative of $$$\ln\left(1 + \frac{3}{n}\right)$$$

The function $$$\ln\left(1 + \frac{3}{n}\right)$$$ is the composition $$$f{\left(g{\left(n \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(n \right)} = 1 + \frac{3}{n}$$$.

Apply the chain rule $$$\frac{d}{dn} \left(f{\left(g{\left(n \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dn} \left(g{\left(n \right)}\right)$$$:

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

The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Return to the old variable:

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

The derivative of a sum/difference is the sum/difference of derivatives:

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

Apply the constant multiple rule $$$\frac{d}{dn} \left(c f{\left(n \right)}\right) = c \frac{d}{dn} \left(f{\left(n \right)}\right)$$$ with $$$c = 3$$$ and $$$f{\left(n \right)} = \frac{1}{n}$$$:

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

The derivative of a constant is $$$0$$$:

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

Apply the power rule $$$\frac{d}{dn} \left(n^{m}\right) = m n^{m - 1}$$$ with $$$m = -1$$$:

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

Simplify:

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

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