$$$5 x^{x}$$$の導関数

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 $$$c = 5$$$ と $$$f{\left(x \right)} = x^{x}$$$ に対して適用します:

$${\color{red}\left(\frac{d}{dx} \left(5 x^{x}\right)\right)} = {\color{red}\left(5 \frac{d}{dx} \left(x^{x}\right)\right)}$$

$$$f{\left(x \right)} = x$$$ と $$$g{\left(x \right)} = x$$$ に対する $$$f^{g{\left(x \right)}}{\left(x \right)} = e^{g{\left(x \right)} \ln\left(f{\left(x \right)}\right)}$$$ を用いて、複雑な式を書き換えよ:

$$5 {\color{red}\left(\frac{d}{dx} \left(x^{x}\right)\right)} = 5 {\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)}$$

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

$$5 {\color{red}\left(\frac{d}{dx} \left(e^{x \ln\left(x\right)}\right)\right)} = 5 {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(x \ln\left(x\right)\right)\right)}$$

指数関数の微分は$$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$です:

$$5 {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right) = 5 {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(x \ln\left(x\right)\right)$$

元の変数に戻す:

$$5 e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = 5 e^{{\color{red}\left(x \ln\left(x\right)\right)}} \frac{d}{dx} \left(x \ln\left(x\right)\right) = 5 x^{x} \frac{d}{dx} \left(x \ln\left(x\right)\right)$$

積の微分法 $$$\frac{d}{dx} \left(f{\left(x \right)} g{\left(x \right)}\right) = \frac{d}{dx} \left(f{\left(x \right)}\right) g{\left(x \right)} + f{\left(x \right)} \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ を $$$f{\left(x \right)} = x$$$ と $$$g{\left(x \right)} = \ln\left(x\right)$$$ に適用する:

$$5 x^{x} {\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = 5 x^{x} {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$

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

$$5 x^{x} \left(x {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)\right) = 5 x^{x} \left(x {\color{red}\left(\frac{1}{x}\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)\right)$$

$$$n = 1$$$ 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 $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$5 x^{x} \left(\ln\left(x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + 1\right) = 5 x^{x} \left(\ln\left(x\right) {\color{red}\left(1\right)} + 1\right)$$

したがって、$$$\frac{d}{dx} \left(5 x^{x}\right) = 5 x^{x} \left(\ln\left(x\right) + 1\right)$$$。