Derivative of $$$\frac{1}{x + 1}$$$

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

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

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

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

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

Return to the old variable:

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

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

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

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

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

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

Thus, $$$\frac{d}{dx} \left(\frac{1}{x + 1}\right) = - \frac{1}{\left(x + 1\right)^{2}}$$$.