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fp3 hypobolic functions,

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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Reply 1

ram8_

Differentiate using quotient rule so you can see more clearly where the negative comes from

Reply 2

Hedwigeeeee

thanks but why I cannot apply the osborn's rule here? is this method not available or my working has mistakes?

(edited 3 years ago)

Reply 3

Hedwigeeeee

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

Reply 4

ram8_

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

Reply 5

mqb2766

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.

Reply 6

Hedwigeeeee

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

Reply 7

Hedwigeeeee

Original post by mqb2766

You can't use that rule using calculus. Stuff starts to become imaginary.

Reply 8

ghostwalker

@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.

Reply 9

Hedwigeeeee

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.

Reply 10

ram8_

I think you are getting confused about Osborn's rule

Reply 11

Hedwigeeeee

Original post by ram8_

I think you are getting confused about Osborn's rule

could you tell me what is wrong in my method

Reply 12

ram8_

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.

Reply 13

davros

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 :smile:

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 :smile: