$$$\eta$$$ に関する $$$r \cos{\left(\tanh{\left(\eta \right)} \right)}$$$ の導関数
関連する計算機: 対数微分計算機, 陰関数微分計算機（手順付き）
入力内容
$$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)$$$ を求めよ。
解答
定数倍の法則 $$$\frac{d}{d\eta} \left(c f{\left(\eta \right)}\right) = c \frac{d}{d\eta} \left(f{\left(\eta \right)}\right)$$$ を $$$c = r$$$ と $$$f{\left(\eta \right)} = \cos{\left(\tanh{\left(\eta \right)} \right)}$$$ に対して適用します:
$${\color{red}\left(\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = {\color{red}\left(r \frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)}$$関数$$$\cos{\left(\tanh{\left(\eta \right)} \right)}$$$は、2つの関数$$$f{\left(u \right)} = \cos{\left(u \right)}$$$と$$$g{\left(\eta \right)} = \tanh{\left(\eta \right)}$$$の合成$$$f{\left(g{\left(\eta \right)} \right)}$$$である。
連鎖律 $$$\frac{d}{d\eta} \left(f{\left(g{\left(\eta \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{d\eta} \left(g{\left(\eta \right)}\right)$$$ を適用する:
$$r {\color{red}\left(\frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)}$$余弦関数の導関数は$$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:
$$r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = r {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$元の変数に戻す:
$$- r \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = - r \sin{\left({\color{red}\left(\tanh{\left(\eta \right)}\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$双曲線正接関数の導関数は$$$\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = \operatorname{sech}^{2}{\left(\eta \right)}$$$です:
$$- r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)} = - r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\operatorname{sech}^{2}{\left(\eta \right)}\right)}$$したがって、$$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}$$$。
解答
$$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}$$$A