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krypton

inverse and inverse trignometric functions find d/dx(arctan^2(sqrt(x))

  • one year ago
  • one year ago

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  1. hsmt
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    hi krypton.. you still there? I think i might be able to help if you are

    • one year ago
  2. krypton
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    yea dear

    • one year ago
  3. hsmt
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    consider the derivative of arctan^2(u) where u=sqrt x and use the chain rule...

    • one year ago
  4. hsmt
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    sorry. let's start with the derivative of (tan-1u)^2. what is the derivative of this little beastie. lol there we go...

    • one year ago
  5. krypton
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    u mean?

    • one year ago
  6. hsmt
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    what is the derivative of (f(x))^2?

    • one year ago
  7. hsmt
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    it would be something like 2(f(x))times f prime of x

    • one year ago
  8. hsmt
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    in this case, f(u)=tan-1u, where u=sqrt of x. does this make sense?

    • one year ago
  9. hsmt
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    questions are good here. the more you ask, the more i can help... :)

    • one year ago
  10. krypton
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    yea

    • one year ago
  11. hsmt
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    so what is the derivative of tan-1u?

    • one year ago
  12. hsmt
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    you can give me your best guess if you like

    • one year ago
  13. krypton
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    hmm actually i wasnt in the class when the topic was taught, but i thinkis =1/srt 1-x^2?

    • one year ago
  14. krypton
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    sorry 1/1+x^2

    • one year ago
  15. hsmt
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    that is totally correct. :) now replace the x with u so that we can use the chain rule where u=1/sqrtx.

    • one year ago
  16. krypton
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    so thats the answer?

    • one year ago
  17. hsmt
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    no, we have a power of tan-1. so the answer is 2(tan-1(sqrtx))(1/(1+x))1/2sqrtx

    • one year ago
  18. hsmt
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    is that making sense to you?

    • one year ago
  19. hsmt
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    still there?

    • one year ago
  20. krypton
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    ok

    • one year ago
  21. krypton
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    so what will be my final answer?

    • one year ago
  22. hsmt
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    it is above

    • one year ago
  23. krypton
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    can u help me with one other questionÉ

    • one year ago
  24. hsmt
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    if i can, i would be happy to

    • one year ago
  25. krypton
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    find \[find d/dx (\sqrt{1-x ^{2}}+1/2\cos ^{-1}(x) and verify whether or \not this is equal \to (2x+1)d/dx \sin ^{-1}(-x)\]

    • one year ago
  26. hsmt
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    ok. any ideas where to start?

    • one year ago
  27. hsmt
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    oh... and i didnt get anything beyond the arrow... sorry... but i think i have enough to figure it out. :)

    • one year ago
  28. krypton
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    hmm i dont have an idea

    • one year ago
  29. hsmt
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    ok, well, it is a rather large function, but it is a quotient (hint). So where might we start with this lovely little thing?

    • one year ago
  30. krypton
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    hmm..no idea lol

    • one year ago
  31. krypton
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    dont know anytin about the topic yet

    • one year ago
  32. hsmt
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    it is ok... remember the quotient rule d(f/g)/dx=[(fprime)g-(gprime)f]/g^2?

    • one year ago
  33. hsmt
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    in this case the numerator is f=sqrt(1-x^2) +1 and g=2cos-1(x)

    • one year ago
  34. hsmt
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    do you know the product rule?

    • one year ago
  35. krypton
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    yes

    • one year ago
  36. hsmt
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    phew...!! i am so glad. i thought for a sec we were outta options. lol

    • one year ago
  37. hsmt
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    ok

    • one year ago
  38. hsmt
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    look at f(x)=sqrt(1-x^2)+1 and g(x)=(2cos-1(x))^-1. does this make sense?

    • one year ago
  39. krypton
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    hahaha i can still remember my calculus 1, this calculus 2

    • one year ago
  40. krypton
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    yes it does

    • one year ago
  41. hsmt
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    glad to hear it... calc I is totally needed here... good, it makes sense. now we need fprime and gprime and i think you are pretty much homefree. so what is fprime and what is gprime?

    • one year ago
  42. hsmt
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    just algebra after you get the derivatives.

    • one year ago
  43. krypton
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    so i should get the derivative of each of them separatelyÉ

    • one year ago
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