$$$\ln\left(3 x\right)$$$の導関数
入力内容
$$$\frac{d}{dx} \left(\ln\left(3 x\right)\right)$$$ を求めよ。
解答
関数$$$\ln\left(3 x\right)$$$は、2つの関数$$$f{\left(u \right)} = \ln\left(u\right)$$$と$$$g{\left(x \right)} = 3 x$$$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:
$${\color{red}\left(\frac{d}{dx} \left(\ln\left(3 x\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(3 x\right)\right)}$$自然対数の導関数は $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(3 x\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(3 x\right)$$元の変数に戻す:
$$\frac{\frac{d}{dx} \left(3 x\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(3 x\right)}{{\color{red}\left(3 x\right)}}$$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 $$$c = 3$$$ と $$$f{\left(x \right)} = x$$$ に対して適用します:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(3 x\right)\right)}}{3 x} = \frac{{\color{red}\left(3 \frac{d}{dx} \left(x\right)\right)}}{3 x}$$$$$n = 1$$$ 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 $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{x} = \frac{{\color{red}\left(1\right)}}{x}$$したがって、$$$\frac{d}{dx} \left(\ln\left(3 x\right)\right) = \frac{1}{x}$$$。
解答
$$$\frac{d}{dx} \left(\ln\left(3 x\right)\right) = \frac{1}{x}$$$A