Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Whatever1

  • one year ago

what is y' in Arc csc squaroot of 2x

  • This Question is Closed
  1. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[y=\csc^{-1}(\sqrt{2x})\]?

  2. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if i recall correctly the derivative of \(\csc^{-1}(x)\) is \(-\frac{1}{x\sqrt{x^2-1}}\)

  3. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah i got it

  4. Whatever1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the y'

  5. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so you want \(\frac{d}{dx}[\csc^{-1}(\sqrt{2x})]\)

  6. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what you need is the derivative of \(\csc^{-1}(x)\) and the chain rule

  7. satellite73
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\frac{d}{dx}[\csc^{-1}(x)=-\frac{1}{x\sqrt{x^2-1}}\] 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

  8. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.