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krypton

  • 2 years ago

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

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

  2. krypton
    • 2 years ago
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    yea dear

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

  4. hsmt
    • 2 years ago
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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...

  5. krypton
    • 2 years ago
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    u mean?

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

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

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

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

  10. krypton
    • 2 years ago
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    yea

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

  12. hsmt
    • 2 years ago
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    you can give me your best guess if you like

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

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

  15. hsmt
    • 2 years ago
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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.

  16. krypton
    • 2 years ago
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    so thats the answer?

  17. hsmt
    • 2 years ago
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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

  18. hsmt
    • 2 years ago
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    is that making sense to you?

  19. hsmt
    • 2 years ago
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    still there?

  20. krypton
    • 2 years ago
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    ok

  21. krypton
    • 2 years ago
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    so what will be my final answer?

  22. hsmt
    • 2 years ago
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    it is above

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

  24. hsmt
    • 2 years ago
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    if i can, i would be happy to

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

  26. hsmt
    • 2 years ago
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    ok. any ideas where to start?

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

  28. krypton
    • 2 years ago
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    hmm i dont have an idea

  29. hsmt
    • 2 years ago
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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?

  30. krypton
    • 2 years ago
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    hmm..no idea lol

  31. krypton
    • 2 years ago
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    dont know anytin about the topic yet

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

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

  34. hsmt
    • 2 years ago
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    do you know the product rule?

  35. krypton
    • 2 years ago
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    yes

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

  37. hsmt
    • 2 years ago
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    ok

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

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

  40. krypton
    • 2 years ago
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    yes it does

  41. hsmt
    • 2 years ago
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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?

  42. hsmt
    • 2 years ago
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    just algebra after you get the derivatives.

  43. krypton
    • 2 years ago
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    so i should get the derivative of each of them separatelyÉ

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