I am slightly confused when it comes to Osborne's rule when you take derivatives of hyperbolic functions. For example. The derivative of $\cot x$ is $-\mathrm{cosec}^2 x$, so there is a product of sines. So should then derivative of $\mathrm{coth}\, x$ not be $\mathrm{cosech}^2x$ because of the sign changing rule? Instead it is $-\mathrm{cosech}^2x$...

Thank you in advance for any help.

EDIT: I just realised that it is not Osborne's rule that is not working out, but there is something going on with the transitions of derivatives from trigonometric to hyperbolic functions. It seems to be that when you go from trig to hyperbolic and consider the derivatives, the derivatives always use the same function (although in hyperbolic terms rather than trig terms). But the derivatives of the reciprocal functions have the sign changed. For instance, derivative of $\sec x$ is $\sec x \tan x$, and of $\mathrm{sech}\, x$ is $-\mathrm{sech}\, x \tanh x$. Derivative of $\mathrm{cosec}\, x$ is $-\cot x\, \mathrm{cosec}\, x$, and of $\mathrm{cosech}\, x$ is $-\mathrm{coth}\, x\, \mathrm {cosech}\, x$, so there seems to have been one sign change when transitioning from trig to hyperbolic derivatives, and another one due to Osborne's rule (repeated sine term) so that the sign is unchanged. Is there a reason why there is this sign change when going from derivatives of trig to hyperbolic functions?

  • $\begingroup$ It's negative because when taking the differential using the quotient rule, your numerator is $\sinh^{2}x - \cosh^{2}x = - (\cosh^{2}x - \sinh^{2}x) = -1$ $\endgroup$ – Mattos Apr 12 '15 at 9:54
  • $\begingroup$ In response to your edit.. It's probably because $(\sinh x)' = \cosh x$ and (notice the sign here) $(\cosh x)' = + \sinh x$ $\endgroup$ – Mattos Apr 12 '15 at 10:05
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    $\begingroup$ For the curious, Osborne's Rule is described here, and it shows how to translate an identity from (circular) trigonometry into its counterpart in hyperbolic trigonometry. Now, since derivative formulas are not trig identities, there's no reason to expect Osborne's Rule to apply. $\endgroup$ – Blue Apr 12 '15 at 10:06
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    $\begingroup$ Why would anyone make up a rule for that, instead of simply using $\sin(ix)=i \sinh(x)$ and $\cos(ix)=\cosh(x)$...? $\endgroup$ – Hans Lundmark Apr 12 '15 at 11:54

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