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