Afgeleide van $$$r \cos{\left(\tanh{\left(\eta \right)} \right)}$$$ naar $$$\eta$$$
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Uw invoer
Bepaal $$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)$$$.
Oplossing
Pas de regel van de constante factor $$$\frac{d}{d\eta} \left(c f{\left(\eta \right)}\right) = c \frac{d}{d\eta} \left(f{\left(\eta \right)}\right)$$$ toe met $$$c = r$$$ en $$$f{\left(\eta \right)} = \cos{\left(\tanh{\left(\eta \right)} \right)}$$$:
$${\color{red}\left(\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = {\color{red}\left(r \frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)}$$De functie $$$\cos{\left(\tanh{\left(\eta \right)} \right)}$$$ is de samenstelling $$$f{\left(g{\left(\eta \right)} \right)}$$$ van twee functies $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ en $$$g{\left(\eta \right)} = \tanh{\left(\eta \right)}$$$.
Pas de kettingregel $$$\frac{d}{d\eta} \left(f{\left(g{\left(\eta \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{d\eta} \left(g{\left(\eta \right)}\right)$$$ toe:
$$r {\color{red}\left(\frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)}$$De afgeleide van de cosinus is $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:
$$r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = r {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$Keer terug naar de oorspronkelijke variabele:
$$- r \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = - r \sin{\left({\color{red}\left(\tanh{\left(\eta \right)}\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$De afgeleide van de hyperbolische tangens is $$$\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = \operatorname{sech}^{2}{\left(\eta \right)}$$$:
$$- r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)} = - r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\operatorname{sech}^{2}{\left(\eta \right)}\right)}$$Dus, $$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}.$$$
Antwoord
$$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}$$$A