anonymous
  • anonymous
what is y' in Arc csc squaroot of 2x
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[y=\csc^{-1}(\sqrt{2x})\]?
anonymous
  • anonymous
if i recall correctly the derivative of \(\csc^{-1}(x)\) is \(-\frac{1}{x\sqrt{x^2-1}}\)
anonymous
  • anonymous
yeah i got it

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anonymous
  • anonymous
the y'
anonymous
  • anonymous
so you want \(\frac{d}{dx}[\csc^{-1}(\sqrt{2x})]\)
anonymous
  • anonymous
what you need is the derivative of \(\csc^{-1}(x)\) and the chain rule
anonymous
  • anonymous
\[\frac{d}{dx}[\csc^{-1}(x)=-\frac{1}{x\sqrt{x^2-1}}\] 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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