$$$\left(x^{3} + 2\right)^{2} \left(x^{4} + 4\right)^{4}$$$の導関数

$$$H{\left(x \right)} = \left(x^{3} + 2\right)^{2} \left(x^{4} + 4\right)^{4}$$$ とする。

両辺の対数を取る: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\left(x^{3} + 2\right)^{2} \left(x^{4} + 4\right)^{4}\right)$$$.

対数の性質を用いて右辺を書き換えよ: $$$\ln\left(H{\left(x \right)}\right) = 2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)$$$。

方程式の両辺をそれぞれ微分せよ: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)\right)$$$

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:

$${\color{red}\left(\frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right)\right) + \frac{d}{dx} \left(4 \ln\left(x^{4} + 4\right)\right)\right)}$$

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 $$$c = 2$$$ と $$$f{\left(x \right)} = \ln\left(x^{3} + 2\right)$$$ に対して適用します:

$${\color{red}\left(\frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right)\right)\right)} + \frac{d}{dx} \left(4 \ln\left(x^{4} + 4\right)\right) = {\color{red}\left(2 \frac{d}{dx} \left(\ln\left(x^{3} + 2\right)\right)\right)} + \frac{d}{dx} \left(4 \ln\left(x^{4} + 4\right)\right)$$

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 $$$c = 4$$$ と $$$f{\left(x \right)} = \ln\left(x^{4} + 4\right)$$$ に対して適用します:

$${\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x^{4} + 4\right)\right)\right)} + 2 \frac{d}{dx} \left(\ln\left(x^{3} + 2\right)\right) = {\color{red}\left(4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right)\right)} + 2 \frac{d}{dx} \left(\ln\left(x^{3} + 2\right)\right)$$

関数$$$\ln\left(x^{3} + 2\right)$$$は、2つの関数$$$f{\left(u \right)} = \ln\left(u\right)$$$と$$$g{\left(x \right)} = x^{3} + 2$$$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:

$$2 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{3} + 2\right)\right)\right)} + 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right) = 2 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{3} + 2\right)\right)} + 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right)$$

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

$$2 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{3} + 2\right) + 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right) = 2 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{3} + 2\right) + 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right)$$

元の変数に戻す:

$$4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right) + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{{\color{red}\left(u\right)}} = 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right) + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{{\color{red}\left(x^{3} + 2\right)}}$$

関数$$$\ln\left(x^{4} + 4\right)$$$は、2つの関数$$$f{\left(u \right)} = \ln\left(u\right)$$$と$$$g{\left(x \right)} = x^{4} + 4$$$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:

$$4 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right)\right)} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = 4 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{4} + 4\right)\right)} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

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

$$4 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{4} + 4\right) + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = 4 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{4} + 4\right) + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

元の変数に戻す:

$$\frac{4 \frac{d}{dx} \left(x^{4} + 4\right)}{{\color{red}\left(u\right)}} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = \frac{4 \frac{d}{dx} \left(x^{4} + 4\right)}{{\color{red}\left(x^{4} + 4\right)}} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

和/差の導関数は、導関数の和/差である:

$$\frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{4} + 4\right)\right)}}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = \frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{4}\right) + \frac{d}{dx} \left(4\right)\right)}}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

定数の導数は$$$0$$$です:

$$\frac{4 \left({\color{red}\left(\frac{d}{dx} \left(4\right)\right)} + \frac{d}{dx} \left(x^{4}\right)\right)}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = \frac{4 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{4}\right)\right)}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 4$$$ に対して適用する:

$$\frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{4}\right)\right)}}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = \frac{4 {\color{red}\left(4 x^{3}\right)}}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

和/差の導関数は、導関数の和/差である:

$$\frac{16 x^{3}}{x^{4} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{3} + 2\right)\right)}}{x^{3} + 2} = \frac{16 x^{3}}{x^{4} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) + \frac{d}{dx} \left(2\right)\right)}}{x^{3} + 2}$$

定数の導数は$$$0$$$です:

$$\frac{16 x^{3}}{x^{4} + 4} + \frac{2 \left({\color{red}\left(\frac{d}{dx} \left(2\right)\right)} + \frac{d}{dx} \left(x^{3}\right)\right)}{x^{3} + 2} = \frac{16 x^{3}}{x^{4} + 4} + \frac{2 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{3}\right)\right)}{x^{3} + 2}$$

冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = 3$$$ に対して適用する:

$$\frac{16 x^{3}}{x^{4} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)}}{x^{3} + 2} = \frac{16 x^{3}}{x^{4} + 4} + \frac{2 {\color{red}\left(3 x^{2}\right)}}{x^{3} + 2}$$

したがって、$$$\frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)\right) = \frac{16 x^{3}}{x^{4} + 4} + \frac{6 x^{2}}{x^{3} + 2}$$$。

したがって、$$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = \frac{16 x^{3}}{x^{4} + 4} + \frac{6 x^{2}}{x^{3} + 2}$$$。

したがって、$$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(\frac{16 x^{3}}{x^{4} + 4} + \frac{6 x^{2}}{x^{3} + 2}\right) H{\left(x \right)} = 2 x^{2} \left(x^{3} + 2\right) \left(x^{4} + 4\right)^{3} \left(3 x^{4} + 8 x \left(x^{3} + 2\right) + 12\right)$$$。