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$$$e^{\frac{1}{x}}$$$の導関数
入力内容
$$$\frac{d}{dx} \left(e^{\frac{1}{x}}\right)$$$ を求めよ。
解答
関数$$$e^{\frac{1}{x}}$$$は、2つの関数$$$f{\left(u \right)} = e^{u}$$$と$$$g{\left(x \right)} = \frac{1}{x}$$$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:
$${\color{red}\left(\frac{d}{dx} \left(e^{\frac{1}{x}}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(\frac{1}{x}\right)\right)}$$指数関数の微分は$$$\frac{d}{du} \left(e^{u}\right) = e^{u}$$$です:
$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(\frac{1}{x}\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(\frac{1}{x}\right)$$元の変数に戻す:
$$e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(\frac{1}{x}\right) = e^{{\color{red}\left(\frac{1}{x}\right)}} \frac{d}{dx} \left(\frac{1}{x}\right)$$冪法則 $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ を $$$n = -1$$$ に対して適用する:
$$e^{\frac{1}{x}} {\color{red}\left(\frac{d}{dx} \left(\frac{1}{x}\right)\right)} = e^{\frac{1}{x}} {\color{red}\left(- \frac{1}{x^{2}}\right)}$$したがって、$$$\frac{d}{dx} \left(e^{\frac{1}{x}}\right) = - \frac{e^{\frac{1}{x}}}{x^{2}}$$$。
解答
$$$\frac{d}{dx} \left(e^{\frac{1}{x}}\right) = - \frac{e^{\frac{1}{x}}}{x^{2}}$$$A
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