## Hyperbolic Functions

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!!
 Recognitions: Homework Help $$y=ln(cosh(2x^3))$$ 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) Now using your chain rule: $$\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dt} \times \frac{dt}{dx}$$

 Quote by goaliejoe35 is this the same as saying (1/(cosh(2x^3)) ?
No! Use the chain rule and note that

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

edit: rockfreak was faster.

## Hyperbolic Functions

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))

Blog Entries: 27
Recognitions:
Gold Member
Homework Help
 Quote by 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))
Hi goaliejoe35!

Hint: tanhx = sinhx/coshx

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

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

 Quote by 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))