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Homework Help: Could i get some help working out dy/dx of arccoth(x)

  1. Aug 26, 2010 #1
    1. The problem statement, all variables and given/known data
    y=arccoth(x), find the derivative



    2. Relevant equations
    cothx - coshx/sinhx


    3. The attempt at a solution
    y = arccoth(x)
    x = coth(y)
    = cosh(y)/sinh(y)
    quotient rule = (sinh(y)*sinh(y))-(cosh(y)*cosh(y))/(sinh(y))^2
    = sinh^2(y) - cosh^2(y)/sinh^2(y)
    x= -1/sinh^2(y)

    this is what i got which is the derivative of coth, but no the inverse. Is there a way to go from here and rearrange it back to y =
     
  2. jcsd
  3. Aug 26, 2010 #2

    hunt_mat

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    Homework Helper

    You're almost there! If y=argcoth(x) then x=costh(y), then from the chain rule:
    [tex]
    1=\frac{dy}{dx}\frac{d}{dy}(\coth x)
    [/tex]
    Re-arrange to get dy/dx
     
  4. Aug 26, 2010 #3
    sorry but i'm still not getting it, can you give me another hint or show me please
     
  5. Aug 26, 2010 #4

    hunt_mat

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    Homework Helper

    ok, you have:
    [tex]
    y=\coth^{-1}x
    [/tex]
    Then as you have stated
    [tex]
    x=\coth y
    [/tex]
    Differentiate this w.r.t. x and use the chain rule:
    [tex]
    1=\frac{d}{dx}(\coth y)=\frac{d}{dy}(\coth y)\frac{dy}{dx}
    [/tex]
    You know how to compute the derivative of coth x, so...
     
  6. Aug 26, 2010 #5
    does that mean 1 = -1/sinh^2(y) * (coth(y))
     
  7. Aug 26, 2010 #6

    hunt_mat

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    Homework Helper

    No it means that:
    [tex]
    1=-\sinh^{2}y\frac{dy}{dx}
    [/tex]
     
  8. Aug 26, 2010 #7
    You differentiated the right side with respect to x, but not the left side. You should have
    [tex]x = \frac{\cosh(y)}{\sinh(y)} \Rightarrow \frac{d}{dx}x = \frac{d}{dx}\frac{\cosh(y)}{\sinh(y)} \Rightarrow 1 = -\frac{1}{\sinh^2(y)}\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = -\sinh^2(y)[/tex]

    Since y = arccoth(x), we have
    [tex]-\sinh^2y = -\sinh^2(arccoth(x)) = \frac{1}{-csch^2(arccoth(x))}[/tex]

    Then using the identity coth2x = 1 + csch2x ==> -csch2x = 1 - coth2x, we have

    [tex]\frac{dy}{dx} = \frac{1}{-csch^2(arccoth(x))} = \frac{1}{1 -coth^2(arccoth(x))} = ~?[/tex]
     
  9. Aug 26, 2010 #8
    thanks for that I have the answer now
     
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