$$$r \cos{\left(\tanh{\left(\eta \right)} \right)}$$$ 對 $$$\eta$$$ 的導數
相關計算器: 對數微分計算器, 隱式微分計算器（附步驟）
您的輸入
求$$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)$$$。
解答
套用常數倍法則 $$$\frac{d}{d\eta} \left(c f{\left(\eta \right)}\right) = c \frac{d}{d\eta} \left(f{\left(\eta \right)}\right)$$$，使用 $$$c = r$$$ 與 $$$f{\left(\eta \right)} = \cos{\left(\tanh{\left(\eta \right)} \right)}$$$：
$${\color{red}\left(\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = {\color{red}\left(r \frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)}$$函數 $$$\cos{\left(\tanh{\left(\eta \right)} \right)}$$$ 是兩個函數 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ 與 $$$g{\left(\eta \right)} = \tanh{\left(\eta \right)}$$$ 之複合 $$$f{\left(g{\left(\eta \right)} \right)}$$$。
應用鏈式法則 $$$\frac{d}{d\eta} \left(f{\left(g{\left(\eta \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{d\eta} \left(g{\left(\eta \right)}\right)$$$：
$$r {\color{red}\left(\frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)}$$餘弦函數的導數為 $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$：
$$r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = r {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$返回原變數：
$$- r \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = - r \sin{\left({\color{red}\left(\tanh{\left(\eta \right)}\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$雙曲正切的導數為 $$$\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = \operatorname{sech}^{2}{\left(\eta \right)}$$$：
$$- r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)} = - r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\operatorname{sech}^{2}{\left(\eta \right)}\right)}$$因此，$$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}$$$。
答案
$$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}$$$A