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$$$x e^{x}$$$の導関数
入力内容
$$$\frac{d}{dx} \left(x e^{x}\right)$$$ 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 $$$g{\left(x \right)} = e^{x}$$$ に適用する:
$${\color{red}\left(\frac{d}{dx} \left(x e^{x}\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(x\right) e^{x} + x \frac{d}{dx} \left(e^{x}\right)\right)}$$$$$n = 1$$$ 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 $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$x \frac{d}{dx} \left(e^{x}\right) + e^{x} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = x \frac{d}{dx} \left(e^{x}\right) + e^{x} {\color{red}\left(1\right)}$$指数関数の微分は$$$\frac{d}{dx} \left(e^{x}\right) = e^{x}$$$です:
$$x {\color{red}\left(\frac{d}{dx} \left(e^{x}\right)\right)} + e^{x} = x {\color{red}\left(e^{x}\right)} + e^{x}$$簡単化せよ:
$$x e^{x} + e^{x} = \left(x + 1\right) e^{x}$$したがって、$$$\frac{d}{dx} \left(x e^{x}\right) = \left(x + 1\right) e^{x}$$$。
解答
$$$\frac{d}{dx} \left(x e^{x}\right) = \left(x + 1\right) e^{x}$$$A
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