# Hyperbolic Functions

1. Jun 23, 2008

### goaliejoe35

The problem statement, all variables and given/known data

Find the derivative of $$y=ln(cosh(2x^3))$$

The attempt at a solution

is this the same as saying (1/(cosh(2x^3)) ?

The correct answer is 6x^2 - ((12x^2)/(e^4x^3 + 1))...... how do you derive this?

I am really stuck on this question I didn't learn about hyperbolic functions in Calc 1 and now my Calc 2 teacher expects me to know it. Any help is greatly appreciated!!

Last edited: Jun 23, 2008
2. Jun 23, 2008

### rock.freak667

$$y=ln(cosh(2x^3))$$

Let $t=2x^3$

So what you now have is y=ln(cosh(t))

and let u=cosh(t)

and it simplifies to y=ln(u)

$$\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dt} \times \frac{dt}{dx}$$

3. Jun 23, 2008

### dirk_mec1

No! Use the chain rule and note that

$$\cosh(x) = \frac{e^x +e^{-x}}{2}$$

edit: rockfreak was faster.

4. Jun 23, 2008

### goaliejoe35

Ok I did all that but I still get a wacky answer....

I got ((-3x^2 tanh(2x^3))/2)

I don't understand how they get the answer 6x^2-((12x^2)/(e^(4x^3)+1))

Last edited: Jun 23, 2008
5. Jun 23, 2008

### tiny-tim

Hi goaliejoe35!

Hint: tanhx = sinhx/coshx

$$= \frac{e^x\,-\,e^{-x}}{e^x\,+\,e^{-x}}$$

$$= \frac{e^{2x}\,-\,1}{e^{2x}\,+\,1}$$

6. Jun 23, 2008