$$$\frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}$$$の導関数

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

両辺の対数を取る: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}\right)$$$.

対数の性質を用いて右辺を書き換えよ: $$$\ln\left(H{\left(x \right)}\right) = 6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)$$$。

方程式の両辺をそれぞれ微分せよ: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right)$$$

方程式の左辺を微分せよ。

関数$$$\ln\left(H{\left(x \right)}\right)$$$は、2つの関数$$$f{\left(u \right)} = \ln\left(u\right)$$$と$$$g{\left(x \right)} = H{\left(x \right)}$$$の合成$$$f{\left(g{\left(x \right)} \right)}$$$である。

連鎖律 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ を適用する:

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

自然対数の導関数は $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

元の変数に戻す:

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

したがって、$$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}}$$$。

方程式の右辺を微分する。

和/差の導関数は、導関数の和/差である:

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

定数倍の法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ を $$$c = 6$$$ と $$$f{\left(x \right)} = \ln\left(x^{2} + 8\right)$$$ に対して適用します:

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

関数$$$\ln\left(x^{2} + 8\right)$$$は、2つの関数$$$f{\left(u \right)} = \ln\left(u\right)$$$と$$$g{\left(x \right)} = x^{2} + 8$$$の合成$$$f{\left(g{\left(x \right)} \right)}$$$である。

連鎖律 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ を適用する:

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

自然対数の導関数は $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

元の変数に戻す:

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

和/差の導関数は、導関数の和/差である:

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

定数の導数は$$$0$$$です:

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

冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 2$$$ に対して適用する:

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

定数倍の法則 $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ を $$$c = 6$$$ と $$$f{\left(x \right)} = \ln\left(x + 1\right)$$$ に対して適用します:

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

関数$$$\ln\left(x + 1\right)$$$は、2つの関数$$$f{\left(u \right)} = \ln\left(u\right)$$$と$$$g{\left(x \right)} = x + 1$$$の合成$$$f{\left(g{\left(x \right)} \right)}$$$である。

連鎖律 $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ を適用する:

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

自然対数の導関数は $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

元の変数に戻す:

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

和/差の導関数は、導関数の和/差である:

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

定数の導数は$$$0$$$です:

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

$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dx} \left(x\right) = 1$$$:

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

したがって、$$$\frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right) = - \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}$$$。

したがって、$$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = - \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}$$$。

したがって、$$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(- \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}\right) H{\left(x \right)} = - \frac{6 \left(x - 2\right) \left(x + 1\right)^{5} \left(x + 4\right)}{\left(x^{2} + 8\right)^{7}}$$$。