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Derivatives of Arctan and arcsin

  1. Nov 5, 2009 #1
    I am confused about how to find arctan and arcsin

    Specific Problem: y= arctan(4x/7) find derivative with respect to y

    I know that d/dx arctan is 1/(1+x^2) am stuck on what to do. Any help would be awesome thanks!


    *p.s. i am very new to this site and it looks awesome!! also not exactly sure where to post things so please any advice on posting would also be helpful
     
  2. jcsd
  3. Nov 5, 2009 #2
    if y=arctan(4x/7) what is the inverse of this?
     
  4. Nov 5, 2009 #3

    HallsofIvy

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    matt crouch is suggesting that you find x as a function of t, then find dx/dy and "invert". That allows you to use tan instead of arctan.

    But since you say you know the derivative of arctan so just use the chain rule.
    Since d(arctan(u))/du= 1/(1+u2), d(arctan(u))/dx= (1/(1+u^2))du/dx where u= 4x/7.
     
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