Hyperbolic Functions

  • #1
Homework Statement

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!!
 
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Answers and Replies

  • #2
rock.freak667
Homework Helper
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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]
 
  • #3
761
13
is this the same as saying (1/(cosh(2x^3)) ?
No! Use the chain rule and note that

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

edit: rockfreak was faster.
 
  • #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))
 
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  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
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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! :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:
 
  • #6
29
1
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.
 

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