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Thread: need help dervative of hyperbolic

  1. #1
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    need help dervative of hyperbolic

    Find d/dx [arcsinh (x)]
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  2. #2
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    the derivative of archsinh(x) is $\displaystyle \dfrac{1} {\sqrt{x^{2}+1}}$
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  3. #3
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    Hello, gracy!

    Viet is correct . . . There is a formula for it.


    Find $\displaystyle \frac{d}{dx}\left[\text{arcsinh}(x)\right] $
    If we are expected to derive that formula, we need to know a few facts:

    . . $\displaystyle \frac{d}{dx}\left(\sinh(y)\right) \:=\:\cosh(y)$ [1]

    . . $\displaystyle \cosh^2(y) - \sinh^2(y)\:=\:1\quad\Rightarrow\quad\cosh(y) \:=\:\sqrt{1 + \sinh^2(y)} $ [2]


    We have: .$\displaystyle y \;=\;\text{arcsinh}(x)$

    Then: .$\displaystyle \sinh(y) \;=\;x$ [3]


    Using [1], differentiate implicitly: .$\displaystyle \cosh(y)\cdot\frac{dy}{dx} \;=\;1$

    . . Then we have: .$\displaystyle \frac{dy}{dx}\:=\:\frac{1}{\cosh(y)}$

    From [2], we have: .$\displaystyle \frac{dy}{dx}\:=\:\frac{1}{\sqrt{1 + \sinh^2(y)}} $

    From [3], we have: .$\displaystyle \frac{dy}{dx}\:=\:\frac{1}{\sqrt{1 + x^2}}$ . . . ta-DAA!

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