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Derivative of inverse tangent function

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Find derivative of tan[itex]^{-1}[/itex]([itex]\frac{3sinx}{4+5cosx}[/itex])

    2. Relevant equations

    deriviative of tan[itex]^{-1}[/itex]=[itex]\frac{U'}{1+U^{2}}[/itex]



    3. The attempt at a solution

    I found U'= [itex]\frac{12cosx+15}{(4+5cosx)^{2}}[/itex]

    1+U[itex]^{2}[/itex]=1+[itex]\frac{9sin^{2}x}{(4+5cosx)^{2}}[/itex]


    I think my components are correct but my answer is still incorrect. Are these basic components even correct (if so I can proceed on my own from here)?

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Mar 8, 2012
  2. jcsd
  3. Mar 8, 2012 #2

    tiny-tim

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    hi biochem850! :smile:

    looks ok so far
     
  4. Mar 8, 2012 #3
    How do you then simplify:

    [itex]\frac{U'}{1+U^{2}}[/itex]?

    I've tried to simplify this but with no luck.
     
  5. Mar 9, 2012 #4
    Can someone please help me?
     
  6. Mar 9, 2012 #5

    tiny-tim

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    hi biochem850! :wink:

    (just got up :zzz:)

    well, the (4 + 5cosx)2 should cancel and disappear …

    show us your full calculations, and then we'll know how to help! :smile:
     
  7. Mar 9, 2012 #6
    [itex]\frac{12cosx+15}{(4+5cosx)^{2}}[/itex]*1+[itex]\frac{(4+5cosx)^{2}}{9sin^{2}x}[/itex]=

    [itex]\frac{12cosx+15(9sin^{2}x)+(4+5cosx)^{2}(4+5cosx)^{2}}{(4+5cosx)^{2}(9sin^{2}x)}[/itex]

    I'm not sure this is correct.
     
    Last edited: Mar 9, 2012
  8. Mar 9, 2012 #7

    tiny-tim

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    hmm :rolleyes: … let's use tex instead of itex, to make it bigger :wink:
    no, that first line is wrong,

    the bracket is 1 + 1/U2, it should be 1/(1 + U2)
     
  9. Mar 9, 2012 #8
    I changed my work just before you posted
     
  10. Mar 9, 2012 #9

    tiny-tim

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    but you didn't change the bit i said was wrong :confused:
     
  11. Mar 9, 2012 #10
    [itex]\frac{12cosx+15}{(4+5cosx)^{2}}[/itex][itex]\frac{(4+5cosx)^{2}+1}{9sin^{2}x+1}[/itex]=


    [itex]\frac{12cosx+15}{9sin^{2}x+1}[/itex]
    I'm not sure this is correct. I really don't understand what you asked me to change.
     
  12. Mar 9, 2012 #11

    tiny-tim

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    change your original …
    … to something with (4 + 5cosx)2 on the bottom
     
  13. Mar 9, 2012 #12


    [itex]\frac{(4+5cosx)^{2}+9sin^{2}x}{(4+5cosx)^{2}}[/itex]?
     
  14. Mar 9, 2012 #13

    tiny-tim

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    yup! :smile:

    now turn that upside-down, and multiply, and the (4 + 5cosx)2 should cancel :wink:
     
  15. Mar 9, 2012 #14
    [itex]\frac{12cosx+15}{(4+5cosx)^{2}+9sin^{2}x}[/itex]

    In terms of finding the derivative I know I've found it but this can be simplified further.

    I don't know if I'm tired or what because I cannot seem to do these very simple problems. This does not bode well for my calculus mark. Would you factor out a 3 in the numerator and then see what will simplify?
     
  16. Mar 9, 2012 #15

    tiny-tim

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    no, and i'd be very surprised if this simplifies

    (except that you could get rid of the sin2 by using cos2 + sin2 = 1 :wink:)

    get some sleep! :zzz:​
     
  17. Mar 9, 2012 #16
    Through some weird algebraic manipulation I simplified down to [itex]\frac{3}{5+4cosx}[/itex]. I beleive this is correct. I'm going to sleep.:smile:

    Thanks so much!
     
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