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Derivative of inverse hyperbolic trigonometric functions

  1. Aug 14, 2006 #1


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    I'm working on a pre-freshman year math packet for college, and at one point it asks for the derivative of sinh-1(x), followed up by the derivative of ln( x + sqrt(1+x2) ). In high school, we never really covered hyperbolic trigonometry, but I have previously derived that the inverse of sinh is ln( x + sqrt(1+x2) ), and used that fact to solve for the derivative of sinh-1' (by taking the derivative of the ln term)

    The trouble i'm having is that it seems obvious that you weren't supposed use the ln term to solve for the derivative, otherwise they wouldn't have posed both questions next to each other. So my guess is that there's some alternative method through which you can solve for this that I may need to know for the school year. If anyone has any ideas, that would be great.
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  3. Aug 14, 2006 #2


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    The usual way to find the derivative of inverse functions is to use implicit differentiation. For example, for sin-1 x, we have:

    [tex] y= \sin^{-1} x [/tex]

    [tex] \sin y =x [/tex]

    [tex] \cos y \frac{dy}{dx} = 1 [/tex]

    [tex] \frac{dy}{dx} = \frac{1}{\cos y} [/tex]

    You can get up to this point for any inverse function, but the tricky part is now to solve for the RHS in terms of x. In this case, as well as in yours, it isn't too hard, and you can write:

    [tex] \frac{1}{\cos y}=\frac{1}{\sqrt{1- \sin^2 y}} = \frac{1}{\sqrt{1- x^2}} [/tex]

    And so:

    [tex]\frac{dy}{dx} = \frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1- x^2}} [/tex]
  4. Aug 14, 2006 #3
    there is an error in the calculations of SatusX.

    The basic relation with hyperbolic function is
    cosh^2 x - sinh^2 x = 1.

    Now if
    y = sinh-1 x
    x = sinh y
    dx/dy = coshy
    but cosh^2 y = 1+sinh^2y =1+x^2.
    dy/dx =1/sqrt(1+x^2)
    where sqrt means square root.
  5. Aug 14, 2006 #4


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    There is no error. StatusX used sin x instead of sinh x as an example.
  6. Aug 14, 2006 #5


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    I'm retarded. I know how to do it like that (essentially, it looks a little different), it's just that I used a right triangle relationship between cosine and sin, and couldn't figure out the relationship between sinh and cosh (maybe if I draw a curved triangle.... :rofl: ). Anyways, I can't believe I didn't think of cosh^2 x - sinh^2 x = 1
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