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$$$x \ln\left(x\right)$$$の導関数
入力内容
$$$\frac{d}{dx} \left(x \ln\left(x\right)\right)$$$ を求めよ。
解答
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:
$${\color{red}\left(\frac{d}{dx} \left(x \ln\left(x\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) \ln\left(x\right) + x \frac{d}{dx} \left(\ln\left(x\right)\right)\right)}$$自然対数の導関数は $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:
$$x {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right) = x {\color{red}\left(\frac{1}{x}\right)} + \ln\left(x\right) \frac{d}{dx} \left(x\right)$$$$$n = 1$$$ 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 $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$\ln\left(x\right) {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + 1 = \ln\left(x\right) {\color{red}\left(1\right)} + 1$$したがって、$$$\frac{d}{dx} \left(x \ln\left(x\right)\right) = \ln\left(x\right) + 1$$$。
解答
$$$\frac{d}{dx} \left(x \ln\left(x\right)\right) = \ln\left(x\right) + 1$$$A
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