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Hyperbolic Functions

  1. Jun 23, 2008 #1
    The problem statement, all variables and given/known data

    Find the derivative of [tex]y=ln(cosh(2x^3))[/tex]

    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. jcsd
  3. Jun 23, 2008 #2

    rock.freak667

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    [tex]y=ln(cosh(2x^3))[/tex]

    Let [itex]t=2x^3[/itex]

    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:

    [tex]\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dt} \times \frac{dt}{dx}[/tex]
     
  4. Jun 23, 2008 #3
    No! Use the chain rule and note that

    [tex]\cosh(x) = \frac{e^x +e^{-x}}{2} [/tex]

    edit: rockfreak was faster.
     
  5. Jun 23, 2008 #4
    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
  6. Jun 23, 2008 #5

    tiny-tim

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    Hi goaliejoe35! :smile:

    Hint: tanhx = sinhx/coshx

    [tex]= \frac{e^x\,-\,e^{-x}}{e^x\,+\,e^{-x}}[/tex]

    [tex]= \frac{e^{2x}\,-\,1}{e^{2x}\,+\,1}[/tex] :smile:
     
  7. Jun 23, 2008 #6
    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.
     
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