Derivative of y=ln(cosh(2x^3)): Calc 2 Help Needed

[goaliejoe35]

Homework Statement

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!

 

[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)

Now using your chain rule:

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

 

[dirk_mec1]

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.

 

[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))

 

[tiny-tim]

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}

 

[shortydeb]

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))

I got (6x^2)(tanh(2x^3)) instead.

anyway, that goes to: (6x^2) * ((e^(4x^3)-1)/(e^(4x^3)+1)).
multiply (6x^2) by the numerator of the 2nd term. u should get: ((6x^2)(e^4x^3) - (6x^2))/(e^(4x^3)+1).
If you separate the numerator into: (6x^2)(e^4x^3) + (6x^2) - (12x^2) ------> (6x^2)((e^4x^3) + 1) - (12x^2), (this is the numerator only), you should be able to get correct answer you posted.