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Math Help - Inverse Hyperbolic Trig Functions?

  1. #1
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    Inverse Hyperbolic Trig Functions?

    Hi,
    Can somebody please help me with finding the derivative of this function? I can find the derivative if it is a normal function but the inverse function has me stumped at the moment.

    Thanks

    Elbarto
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by elbarto View Post
    Hi,
    Can somebody please help me with finding the derivative of this function? I can find the derivative if it is a normal function but the inverse function has me stumped at the moment.

    Thanks

    Elbarto
    Are you looking for the derivative of ar~sinh(2x) or the derivative of its inverse?
    y^{\prime} = ar~cosh(2x) \cdot 2 = 2ar~cosh(2x)
    by the chain rule
    and note that \frac{d}{dx}sinh^{-1}(x) = \frac{1}{\sqrt{x^2 + 1}}
    so
    y = \frac{1}{2}sinh^{-1} \left ( \frac{x}{ar} \right )
    is the inverse function and has a derivative
    y^{\prime} = \frac{1}{2} \cdot \frac{1}{\sqrt{ \left ( \frac{x}{ar} \right )^2 + 1}} \cdot \frac{1}{ar} = \frac{1}{2\sqrt{x^2 + a^2r^2}}
    after some simplification.

    -Dan
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  3. #3
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    Hi dan,

    Im just trying to find the derivative of y. is it as simple as
    y^{\prime} = ar~cosh(2x) \cdot 2 = 2ar~cosh(2x)?

    Thanks for your help

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    Hello, Elbarto!

    There is a formula for this problem.
    If you don't know it, we can derive it.


    y \;=\;\sinh^{-1}(4x)

    Take the \sinh of both sides: . \sinh y \;=\;4x .[1]

    Differentiate implicitly: . \cosh y\cdot\frac{dy}{dx}\;=\;4\quad\Rightarrow\quad\fra  c{dy}{dx}\;=\;\frac{4}{\cosh y}<br />
.[2]


    From the identity: . \cosh^2u - \sinh^2u \:=\:1, we have: . \cosh u \:=\:\sqrt{1 + \sinh^2u}

    Then [2] becomes: . \frac{dy}{dx} \;=\;\frac{4}{\sqrt{1+\sinh^2y}}

    Substitute [1] and we have: . \frac{dy}{dx}\;=\;\frac{4}{\sqrt{1 + (4x)^2}} \;=\;\frac{4}{\sqrt{1+16x^2}}

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    Forum Admin topsquark's Avatar
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    Oh! That "ar" on the outside of the sinh(x), that was supposed to mean "arcsinh(x)"? Sorry about that.

    -Dan
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