# Hyperbolic functions

1. Sep 21, 2007

### Oblio

The hyperbolic functions are defined as follows:

coshz = e$$^{z}$$ + e$$^{-z}$$ /2

sinhz = e$$^{z}$$ - e$$^{-z}$$ /2

a.)Show that coshz = cos (iz). What is the corresponding relationship for sinhz?
b.)What are the derivatives of coshz and sinhz? What about their integrals?
c.)Show that cosh^2z - sin^2 =1
d.)Show that the integral of dx/sqrt[1+x^2 = arcsin x.

Hint : substitution x = sinhz.

I'd LOVE starters on showing this, we're told to assume z is real.
I get the idea that theres an imaginery aspect to hyperbolic functions, since coshz = cos(iz) ?

2. Sep 21, 2007

### learningphysics

can you use $$e^{iz} = cos(z) + isin(z)$$

3. Sep 21, 2007

### Oblio

I'm not sure.
Beforehand these values were sketched over a range of real values of z.
I don't know if that answers whether 'were allowed' ...

4. Sep 21, 2007

### Oblio

Correction: part a.) isnt necessary. My bad.

5. Sep 21, 2007

### Oblio

I have really no idea how to derive hyperbolic functions.
Is d/dx of cosh = -sinh?

6. Sep 21, 2007

### learningphysics

No. Get the derivative of $$\frac{e^z + e^{-z}}{2}$$ with respect to z.

7. Sep 21, 2007

### Oblio

=2e^z + 2e^-z / 4

?

8. Sep 21, 2007

### learningphysics

9. Sep 21, 2007

### Oblio

is d/dx of e^z + e^-z the same thing or not?

10. Sep 21, 2007

### Oblio

nm reading now, i know thats wrong

11. Sep 21, 2007

### Oblio

is d/dx 0?

12. Sep 21, 2007

### Oblio

orrr..

e^z - e^-z / 4
?

13. Sep 21, 2007

### learningphysics

14. Sep 21, 2007

### Oblio

e$$^{z}$$ + e$$^{-z}$$ / 2

d/dx e$$^{z}$$ = e$$^{z}$$ * d/dx (z)
z=real number, d/dx = 1

d/dx e$$^{-z}$$ = e$$^{-z}$$ * d/dx (-z)
z=real number, d/dx = -1

Denominator ^2 by quotient rule...

=e$$^{z}$$ - e$$^{-z}$$ / 4

?

15. Sep 21, 2007

### learningphysics

just do d/dz...

it's just (1/2)(e^z + e^-z)

taking d/dz you get $$\frac{1}{2}(\frac{d}{dz}e^z + \frac{d}{dz}e^{-z})$$

so the answer is just (e^z - e^-z)/2

if you do it using the quotient rule... you need to do derivative of the numerator by the denominator, minus the numerator*deriavative of the denominator divided by the denominator squared so...

$$\frac{(e^z - e^{-z})2 - (e^z + e^{-z})(0)}{2^2}$$

and you get the same result.

16. Sep 21, 2007

### Oblio

I dont know the rule of putting a half there...

17. Sep 21, 2007

### learningphysics

It's just taking out the constant.

if z = A*y

then taking the derivative of both sides with respect to x...

dz/dx = A*(dy/dx)

For example... the derivative of 5e^(2x) = 5*d/dx(e^(2x)) = 5*2e^(2x) = 10e^(2x)

18. Sep 21, 2007

### Oblio

alrighty, i think i get it.
basically what i get is that d/dx of cosh is sinh and vice versa right?

19. Sep 21, 2007

### Oblio

I forget integrating quotients.. and e...

20. Sep 21, 2007

Yeah.