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$$$\cos{\left(e^{t} \right)}$$$の導関数
入力内容
$$$\frac{d}{dt} \left(\cos{\left(e^{t} \right)}\right)$$$ を求めよ。
解答
関数$$$\cos{\left(e^{t} \right)}$$$は、2つの関数$$$f{\left(u \right)} = \cos{\left(u \right)}$$$と$$$g{\left(t \right)} = e^{t}$$$の合成$$$f{\left(g{\left(t \right)} \right)}$$$である。
連鎖律 $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$ を適用する:
$${\color{red}\left(\frac{d}{dt} \left(\cos{\left(e^{t} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{dt} \left(e^{t}\right)\right)}$$余弦関数の導関数は$$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:
$${\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{dt} \left(e^{t}\right) = {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{dt} \left(e^{t}\right)$$元の変数に戻す:
$$- \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{dt} \left(e^{t}\right) = - \sin{\left({\color{red}\left(e^{t}\right)} \right)} \frac{d}{dt} \left(e^{t}\right)$$指数関数の微分は$$$\frac{d}{dt} \left(e^{t}\right) = e^{t}$$$です:
$$- \sin{\left(e^{t} \right)} {\color{red}\left(\frac{d}{dt} \left(e^{t}\right)\right)} = - \sin{\left(e^{t} \right)} {\color{red}\left(e^{t}\right)}$$したがって、$$$\frac{d}{dt} \left(\cos{\left(e^{t} \right)}\right) = - e^{t} \sin{\left(e^{t} \right)}$$$。
解答
$$$\frac{d}{dt} \left(\cos{\left(e^{t} \right)}\right) = - e^{t} \sin{\left(e^{t} \right)}$$$A
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