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Whatever1

  • 3 years ago

what is y' in Arc csc squaroot of 2x

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  1. anonymous
    • 3 years ago
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    \[y=\csc^{-1}(\sqrt{2x})\]?

  2. anonymous
    • 3 years ago
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    if i recall correctly the derivative of \(\csc^{-1}(x)\) is \(-\frac{1}{x\sqrt{x^2-1}}\)

  3. anonymous
    • 3 years ago
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    yeah i got it

  4. Whatever1
    • 3 years ago
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    the y'

  5. anonymous
    • 3 years ago
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    so you want \(\frac{d}{dx}[\csc^{-1}(\sqrt{2x})]\)

  6. anonymous
    • 3 years ago
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    what you need is the derivative of \(\csc^{-1}(x)\) and the chain rule

  7. anonymous
    • 3 years ago
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    \[\frac{d}{dx}[\csc^{-1}(x)=-\frac{1}{x\sqrt{x^2-1}}\] i know because i just looked it up replace \(x\) by \(\sqrt{2x}\) and then multiply by the derivative of \(\sqrt{2x}\) which is \(\frac{1}{\sqrt{2x}}\) then probably some algebra to clean it up

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