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$$$\sin{\left(4 t \right)}$$$の導関数
入力内容
$$$\frac{d}{dt} \left(\sin{\left(4 t \right)}\right)$$$ を求めよ。
解答
関数$$$\sin{\left(4 t \right)}$$$は、2つの関数$$$f{\left(u \right)} = \sin{\left(u \right)}$$$と$$$g{\left(t \right)} = 4 t$$$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:
$${\color{red}\left(\frac{d}{dt} \left(\sin{\left(4 t \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right) \frac{d}{dt} \left(4 t\right)\right)}$$正弦関数の導関数は$$$\frac{d}{du} \left(\sin{\left(u \right)}\right) = \cos{\left(u \right)}$$$:
$${\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right)\right)} \frac{d}{dt} \left(4 t\right) = {\color{red}\left(\cos{\left(u \right)}\right)} \frac{d}{dt} \left(4 t\right)$$元の変数に戻す:
$$\cos{\left({\color{red}\left(u\right)} \right)} \frac{d}{dt} \left(4 t\right) = \cos{\left({\color{red}\left(4 t\right)} \right)} \frac{d}{dt} \left(4 t\right)$$定数倍の法則 $$$\frac{d}{dt} \left(c f{\left(t \right)}\right) = c \frac{d}{dt} \left(f{\left(t \right)}\right)$$$ を $$$c = 4$$$ と $$$f{\left(t \right)} = t$$$ に対して適用します:
$$\cos{\left(4 t \right)} {\color{red}\left(\frac{d}{dt} \left(4 t\right)\right)} = \cos{\left(4 t \right)} {\color{red}\left(4 \frac{d}{dt} \left(t\right)\right)}$$$$$n = 1$$$ を用いて冪法則 $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ を適用すると、すなわち $$$\frac{d}{dt} \left(t\right) = 1$$$:
$$4 \cos{\left(4 t \right)} {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} = 4 \cos{\left(4 t \right)} {\color{red}\left(1\right)}$$したがって、$$$\frac{d}{dt} \left(\sin{\left(4 t \right)}\right) = 4 \cos{\left(4 t \right)}$$$。
解答
$$$\frac{d}{dt} \left(\sin{\left(4 t \right)}\right) = 4 \cos{\left(4 t \right)}$$$A