# How useful are hyperbolic functions

## Derive hyperbolic functions

Hello my dears!

Once again I am plagued by ignorance. Nothing new as far as a student. ;-)
This time I would be interested in how to derive the hyperbolic functions ...

1.) sinh´ (x) = cosh (x)
2.) cosh´ (x) = sinh (x) and
3) cosh ^ 2 (x) - sinh ^ 2 (x) = 1 ... determined.

Have a nice weekend,
Sven

PS: What do you need it for in "practice"? Google doesn't really know either.

For everyone who wants to help me (automatically generated by OnlineMathe):
"I want to create the solution in collaboration with others."

QPhma

11:12 p.m., December 18, 2010

Hello mathabolica,

the hyperbolic functions are defined as follows:
de.wikipedia.org/wiki/Hyperbelffunktion

With this you can trace the derivation of the hyperbolic functions back to the derivation of the exponential function and also prove the third equation.

greeting

QPhma
I've already looked at the definition.
But as I said ... I really don't get any further.

hmm :-(
sinh '(x) =

and the rest is analog ... you know how to balance exponential functions, right?

anonymous

11:21 p.m., December 18, 2010

As QPhma wrote, think of:

Then applies to

It just stops when it is diverted. results in derived and results in derived because the chain rule has to be used to differentiate:

and now I'll leave it to you. Unless there are still questions.

The derivatives are. useful for optimizations, since the zeros of the first derivative can have relative highs or lows.
And I know that at least the graph of a cosh function is also called a catenoid (chain line), since it describes the slack of a chain that is suspended at two points and is influenced by gravity.

Edit: Unfortunately you swapped the definitions of and with each other, Mathinator90.
That's a strange function. I now understand the "how" but not the "why".

a) Why does the 1/2 stay like this when deriving? Isn't that actually by definition = 0 ?!
b) Nothing changes on the e ^ x either. is that because x = e ^ (ln x) ?!
c) Just as an excursus: How could I show continuity / growth in an exam?
-> Isn't that feasible with the intermediate value theorem?

@ Mathinator90: Had not even close to the log in mind, because the function is so "messed up".

Thank you Mi.St. for the application example. I'll do some more research tomorrow.
I belong to the learner types: seen applying, understanding, then playing around with it =)

Thanks again for the first!

QPhma

10:27 p.m., December 19, 2010

Hello mathabolica,

there is a derivation rule for deriving when a function is preceded by a constant factor. www.mathe-online.at/mathint/diff1/i_ableitungen.html
You have to learn these rules by heart. Sometimes it is not so easy to "understand" the rules. It takes exact mathematical evidence to make it clear that the rules are correct.

is considered an elementary function. You also have to learn the derivatives of the elementary functions by heart, since you don't want to use the definition of the derivative with the limit value again and again.
is the only function where the derivative is equal to the function itself. In a sense, this is the personal touch of the e-function.

First of all: continuity and growth (you probably mean monotony) are two very different properties.
Both can be demonstrated for a given function by applying the respective definition. With continuity you can make your work easier in some cases if you use the fact that the sum and the product of two continuous functions are again a continuous function, and the concatenation of continuous functions is again continuous.
http://www.iag.uni-hannover.de/~hulek/Skripten/AnaA/Kapitel2.pdf
I cannot imagine how one should use the intermediate value theorem to prove continuity. This theorem assumes that a function is continuous and then makes a statement about the possible function values ​​of the function.

greeting

QPhma

Thank you QPhma!

You're right. I should take the derivation rules to heart again. Perhaps I will also find a nice proof for one or the other derivation.

I wish you all a Merry Christmas!

Greeting,
Sven