fp3 hypobolic functions,

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Hedwigeeeee
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#1
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#1
I think the differentiation in the first line should be positive, d(cosecx)/dx=-cosecxcotx, as there are 2 sine and according to the Osborn's rule, so I think d(cosechx)/dx =cosechxcothx, but the answer is negative, why is that?
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ram8_
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#2
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#2
Differentiate using quotient rule so you can see more clearly where the negative comes from
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Hedwigeeeee
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#3
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thanks but why I cannot apply the osborn's rule here? is this method not available or my working has mistakes?
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Hedwigeeeee
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#4
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(Original post by ram8_)
Differentiate using quotient rule so you can see more clearly where the negative comes from
because if I use the osborn rule it seems that as there are two sines, so we should change the sign to positive
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ram8_
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(Original post by Hedwigeeeee)
because if I use the osborn rule it seems that as there are two sines, so we should change the sign to positive
You don't need to use that rule? It's only for trying to work out an equivalent identity for hyperbolics
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mqb2766
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#6
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(Original post by Hedwigeeeee)
thanks but why I cannot apply the osborn's rule here? is this method not available or my working has mistakes?
You can't use that rule using calculus. Stuff starts to become imaginary.
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Hedwigeeeee
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#7
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(Original post by ram8_)
You don't need to use that rule? It's only for trying to work out an equivalent identity for hyperbolics
:yes:
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Hedwigeeeee
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#8
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(Original post by mqb2766)
You can't use that rule using calculus. Stuff starts to become imaginary.
:yep:
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ghostwalker
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#9
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Hedwigeeeee


If you want convincing Osborn's rule doesn't work here:

\frac{d}{dx}\cos x = -\sin x

\frac{d}{dx}\cosh x = \sinh x

And not a sin^2 in sight.
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Hedwigeeeee
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#10
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#10
I mean there are two sines in cosechx(1/sinhx) and cothx(coshx/ sinhx), so it is sinh^2, so I think we should change the sign. though it is true that if we do not use the Osborn's rule, the sign will not change.
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ram8_
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#11
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I think you are getting confused about Osborn's rule
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Hedwigeeeee
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#12
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(Original post by ram8_)
I think you are getting confused about Osborn's rule
could you tell me what is wrong in my method
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ram8_
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#13
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It's for converting a non-hyperbolic trigonometric identity into an equivalent hyperbolic one. So it would swap signs for when there's a product of two sines, so the sinh version with be the opposite sign, e.g. cos^2(x) - sin^2(x) will be cosh^2(x) + sinh^2(x). You wouldn't need to use it for this problem as it is just differentiating. You can also confirm this by multiplying two sinh(x)'s in its exponential form.
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davros
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#14
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#14
(Original post by Hedwigeeeee)
could you tell me what is wrong in my method
It's not remotely relevant to what you're trying to do

Osborn's rule allows you to convert a trigonometric identity into an equivalent hyperbolic identity. It has nothing to do with differentiation!

Seriously, forget all about it! I think I saw Osborn's rule once when I first learnt hyperbolics and then never used it again - it's as easy to derive the identities directly as it is to remember the correct conditions for applying the rule
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