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Math Help - Help with the derivative of arcsinh

  1. #1
    Super Member craig's Avatar
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    Help with the derivative of arcsinh

    Hi, I'm a little stuck on this problem.

    Differentiate with respect to x, \arcsin{(\frac{1}{x})}.

    I know the derivative of \arcsin is \frac{1}{\sqrt{1-x^2}}, but I'm not too sure how to approach this problem.

    Any pointers would be greatly appreciated

    Thanks in advance
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  2. #2
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    Krizalid's Avatar
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    ya know what the chain rule is?
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  3. #3
    Super Member Random Variable's Avatar
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     y = \sin^{-1}\frac{1}{x}

     \sin y = \frac{1}{x}

     \cos y \ \frac{dy}{dx} = \text{-} \frac {1}{x^{2}}

     \frac {dy}{dx} = \frac{\text{-1}}{x^{2} \cos y} = \frac{-1}{x^{2} \sqrt{x^2-1}}
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  4. #4
    Super Member craig's Avatar
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    Quote Originally Posted by Random Variable View Post
     y = \sin^{-1}\frac{1}{x}

     \sin y = \frac{1}{x}

     \cos y \ \frac{dy}{dx} = \text{-} \frac {1}{x^{2}}

     \frac {dy}{dx} = \frac{\text{-1}}{x^{2} \cos y} = \frac{-1}{x^{2} \sqrt{x^2-1}}
    Thanks, I never though to do it though implicit differentiation.
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  5. #5
    Super Member craig's Avatar
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    Quote Originally Posted by Krizalid View Post
    ya know what the chain rule is?
    Yep I know what the chain rule is, how would you use it in this situation?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by craig View Post
    Yep I know what the chain rule is, how would you use it in this situation?
    I think Kriz is referring to this: \frac{\,d}{\,dx}\left[\sin^{-1}u\right]=\frac{1}{\sqrt{1-u^2}}\cdot\frac{\,du}{\,dx}. As you see here, chain rule is necessary.
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  7. #7
    Super Member Random Variable's Avatar
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    Quote Originally Posted by craig View Post
    Thanks, I never though to do it though implicit differentiation.
    You can use implict differentiation to get the derivatives of all the basic inverse trigonometric functions.

    For this problem it's not necessary if you already know the derivative of  \sin ^{-1} u
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