Derivative of $$$r \cos{\left(\tanh{\left(\eta \right)} \right)}$$$ with respect to $$$\eta$$$
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Find $$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)$$$.
Solution
Apply the constant multiple rule $$$\frac{d}{d\eta} \left(c f{\left(\eta \right)}\right) = c \frac{d}{d\eta} \left(f{\left(\eta \right)}\right)$$$ with $$$c = r$$$ and $$$f{\left(\eta \right)} = \cos{\left(\tanh{\left(\eta \right)} \right)}$$$:
$${\color{red}\left(\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = {\color{red}\left(r \frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)}$$The function $$$\cos{\left(\tanh{\left(\eta \right)} \right)}$$$ is the composition $$$f{\left(g{\left(\eta \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ and $$$g{\left(\eta \right)} = \tanh{\left(\eta \right)}$$$.
Apply the chain rule $$$\frac{d}{d\eta} \left(f{\left(g{\left(\eta \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{d\eta} \left(g{\left(\eta \right)}\right)$$$:
$$r {\color{red}\left(\frac{d}{d\eta} \left(\cos{\left(\tanh{\left(\eta \right)} \right)}\right)\right)} = r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right) \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)}$$The derivative of the cosine is $$$\frac{d}{du} \left(\cos{\left(u \right)}\right) = - \sin{\left(u \right)}$$$:
$$r {\color{red}\left(\frac{d}{du} \left(\cos{\left(u \right)}\right)\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = r {\color{red}\left(- \sin{\left(u \right)}\right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$Return to the old variable:
$$- r \sin{\left({\color{red}\left(u\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = - r \sin{\left({\color{red}\left(\tanh{\left(\eta \right)}\right)} \right)} \frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)$$The derivative of the hyperbolic tangent is $$$\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right) = \operatorname{sech}^{2}{\left(\eta \right)}$$$:
$$- r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\frac{d}{d\eta} \left(\tanh{\left(\eta \right)}\right)\right)} = - r \sin{\left(\tanh{\left(\eta \right)} \right)} {\color{red}\left(\operatorname{sech}^{2}{\left(\eta \right)}\right)}$$Thus, $$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}.$$$
Answer
$$$\frac{d}{d\eta} \left(r \cos{\left(\tanh{\left(\eta \right)} \right)}\right) = - r \sin{\left(\tanh{\left(\eta \right)} \right)} \operatorname{sech}^{2}{\left(\eta \right)}$$$A