$$$\frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}$$$的导数

设$$$H{\left(x \right)} = \frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}$$$。

对等式两边取对数：$$$\ln\left(H{\left(x \right)}\right) = \ln\left(\frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}\right)$$$。

利用对数的性质改写等式右边：$$$\ln\left(H{\left(x \right)}\right) = 6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)$$$。

分别对方程两边求导：$$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right)$$$。

对方程的左边求导。

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

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

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

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

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

返回到原变量：

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

因此，$$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}}$$$。

对等式右边求导。

和/差的导数等于导数的和/差：

$${\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{d}{dx} \left(6 \ln\left(x^{2} + 8\right)\right)\right)}$$

对 $$$c = 6$$$ 和 $$$f{\left(x \right)} = \ln\left(x^{2} + 8\right)$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

$$- {\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x^{2} + 8\right)\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) = - {\color{red}\left(6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)$$

函数$$$\ln\left(x^{2} + 8\right)$$$是两个函数$$$f{\left(u \right)} = \ln\left(u\right)$$$和$$$g{\left(x \right)} = x^{2} + 8$$$的复合$$$f{\left(g{\left(x \right)} \right)}$$$。

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

$$- 6 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) = - 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{2} + 8\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)$$

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

$$- 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{2} + 8\right) + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) = - 6 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{2} + 8\right) + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)$$

返回到原变量：

$$\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 \frac{d}{dx} \left(x^{2} + 8\right)}{{\color{red}\left(u\right)}} = \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 \frac{d}{dx} \left(x^{2} + 8\right)}{{\color{red}\left(x^{2} + 8\right)}}$$

和/差的导数等于导数的和/差：

$$\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 {\color{red}\left(\frac{d}{dx} \left(x^{2} + 8\right)\right)}}{x^{2} + 8} = \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(8\right)\right)}}{x^{2} + 8}$$

常数的导数是$$$0$$$:

$$\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 \left({\color{red}\left(\frac{d}{dx} \left(8\right)\right)} + \frac{d}{dx} \left(x^{2}\right)\right)}{x^{2} + 8} = \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)\right)}{x^{2} + 8}$$

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

$$\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}}{x^{2} + 8} = \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{6 {\color{red}\left(2 x\right)}}{x^{2} + 8}$$

对 $$$c = 6$$$ 和 $$$f{\left(x \right)} = \ln\left(x + 1\right)$$$ 应用常数倍法则 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$：

$$- \frac{12 x}{x^{2} + 8} + {\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)\right)} = - \frac{12 x}{x^{2} + 8} + {\color{red}\left(6 \frac{d}{dx} \left(\ln\left(x + 1\right)\right)\right)}$$

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

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

$$- \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x + 1\right)\right)\right)} = - \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x + 1\right)\right)}$$

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

$$- \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x + 1\right) = - \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x + 1\right)$$

返回到原变量：

$$- \frac{12 x}{x^{2} + 8} + \frac{6 \frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(u\right)}} = - \frac{12 x}{x^{2} + 8} + \frac{6 \frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(x + 1\right)}}$$

和/差的导数等于导数的和/差：

$$- \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x + 1\right)\right)}}{x + 1} = - \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x\right) + \frac{d}{dx} \left(1\right)\right)}}{x + 1}$$

常数的导数是$$$0$$$:

$$- \frac{12 x}{x^{2} + 8} + \frac{6 \left({\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x\right)\right)}{x + 1} = - \frac{12 x}{x^{2} + 8} + \frac{6 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x\right)\right)}{x + 1}$$

应用幂法则 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$，取 $$$n = 1$$$，也就是说，$$$\frac{d}{dx} \left(x\right) = 1$$$：

$$- \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{x + 1} = - \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(1\right)}}{x + 1}$$

因此，$$$\frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right) = - \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}$$$。

因此，$$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = - \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}$$$。

因此，$$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(- \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}\right) H{\left(x \right)} = - \frac{6 \left(x - 2\right) \left(x + 1\right)^{5} \left(x + 4\right)}{\left(x^{2} + 8\right)^{7}}$$$。