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Math Help - deriving the formula

  1. #1
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    deriving the formula

    derive the formula d(sinh^-1 z)/dz=1/(1+z^2)^1/2 also i know there needs to be certain conditions that make this possible
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  2. #2
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    I was just doing these for class last unit. You have to use the formula for deriving Inverse functions 1/f ' (g(x)) where f(x) is the inverse of g(x). In this case f (x)=sinch x and g(x)=arcsinch x (I don't know if it is legal to call it arcsinch, but I don't know how to write the math formulas on the computer) Anyway after this use the identity
    cosh x= (1-(sinch x)^2)^1/2). try it from there and keep in mind it is similar to deriving inverse trig functions.
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  3. #3
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    Quote Originally Posted by Hongosh View Post
    derive the formula d(sinh^-1 z)/dz=1/(1+z^2)^1/2 also i know there needs to be certain conditions that make this possible
    y  = \sinh^{-1} z \Rightarrow z = \sinh y.

    Therefore \frac{dz}{dy} = \cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1 + z^2}.

    You should think about why it's \sqrt{1 + \sinh^2 y} and not -\sqrt{1 + \sinh^2 y}.

    Therefore \frac{dy}{dz} = \frac{1}{\sqrt{1 + z^2}}.
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  4. #4
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    Hello, Hongosh!

    A variation of Mr. F's solution . . .


    Derive the formula: . \frac{d}{dx}(\sinh^{-1}\!z) \;=\;\frac{1}{\sqrt{1+z^2}}

    We have: . y \:=\:\sin^{-1}\!z \quad\Rightarrow \quad \sinh y \:=\:z .[1]

    Differentiate implicitly: . \cosh y\cdot\frac{dy}{dz} \:=\:1 \quad\Rightarrow\quad \frac{dy}{dz} \:=\:\frac{1}{\cosh y} .[2]


    \text{Since }\:\cosh^2\!y - \sin^2\!y \:=\:1\quad\Rightarrow\quad \cosh y\:=\:\sqrt{1 + \sinh^2\!y}

    Substitute [1]: . \cosh y \:=\:\sqrt{1+z^2}


    Substitute into [2]: . \frac{dy}{dx} \;=\;\frac{1}{\sqrt{1+z^2}}

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