Dérivée de $$$r \cos{\left(\tanh{\left(\eta \right)} \right)}$$$ par rapport à $$$\eta$$$
Calculatrices associées: Calculatrice de dérivation logarithmique, Calculatrice de dérivation implicite pas à pas
Votre saisie
Déterminez $$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)$$$.
Solution
Appliquez la règle du facteur constant $$$\frac{d}{d\eta} \left(c f{\left(\eta \right)}\right) = c \frac{d}{d\eta} \left(f{\left(\eta \right)}\right)$$$ avec $$$c = r$$$ et $$$f{\left(\eta \right)} = \cos{\left(\tanh{\left(\eta \right)} \right)}$$$:
$${\color{red}\left(\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = {\color{red}\left(r \frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)}$$La fonction $$$\cos{\left(\tanh{\left(\eta \right)} \right)}$$$ est la composée $$$f{\left(g{\left(\eta \right)} \right)}$$$ de deux fonctions $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ et $$$g{\left(\eta \right)} = \tanh{\left(\eta \right)}$$$.
Appliquez la règle de la chaîne $$$\frac{d}{d\eta} \left(f{\left(g{\left(\eta \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{d\eta} \left(g{\left(\eta \right)}\right)$$$:
$$r {\color{red}\left(\frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)}$$La dérivée du cosinus est $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$ :
$$r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = r {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$Revenir à la variable initiale:
$$- r \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = - r \sin{\left({\color{red}\left(\tanh{\left(\eta \right)}\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$La dérivée de la tangente hyperbolique est $$$\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = \operatorname{sech}^{2}{\left(\eta \right)}$$$ :
$$- r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)} = - r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\operatorname{sech}^{2}{\left(\eta \right)}\right)}$$Ainsi, $$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}.$$$
Réponse
$$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}$$$A