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Another tough differentiation question

  1. Oct 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Find y' given
    tan-1(xy) = 1 + x2y

    2. Relevant equations



    3. The attempt at a solution
    I know I'll have to start with implicit differentiation, and I can differentiate the RHS to be:
    0 + 2xy + x2(dy/dx)

    And I know tan-1 x = 1/(1+x2), but I don't know how to differentiate tan-1 (xy) implicitly!
     
  2. jcsd
  3. Oct 25, 2010 #2

    Mark44

    Staff: Mentor

    Then you know something that isn't true. What is true, though, is that
    d/dx(tan-1(x)) = 1/(1 + x2).
    All you really need to do is to use the chain rule and the product rule.

    d/dx(tan-1 (xy)) = 1/(1 + (xy)2)) * d/dx(xy)

    Can you finish this?
     
    Last edited: Oct 25, 2010
  4. Oct 25, 2010 #3
    Yes, I had a typo. I meant d/dx tan-1 x = 1 / (1 + x2)

    Ok I think I can now.
     
  5. Oct 25, 2010 #4
    I got:

    dy/dx = (2xy(1 + x2y2) / ( 1 - 2x)

    Am I correct?
     
  6. Oct 25, 2010 #5

    Mark44

    Staff: Mentor

    That's not what I get. Can you show your work?
     
  7. Oct 25, 2010 #6
    d/dx tan-1 xy = d/dx ( 1 + x2 y)

    dy/dx (xy) (1/(1 + x2y2) = 0 + 2xy2 + dy/dx (2y) x2

    Rearrange to make dy/dx subject of formula, I got:

    dy/dx (xy / (1 + x2y2 - 2yx2) = 2x2y2

    dy/dx = 2x2y2 / ((xy)/(1+x2y2) - 2yx2)

    dy/dx = 2xy / ( (1-2x) / (1+x2y2)

    dy/dx = 2xy( 1 + x2y2) / ( 1 -2x)
     
  8. Oct 25, 2010 #7

    Mark44

    Staff: Mentor

    You went astray right off the bat. d/dx(tan-1(u)) = 1/(1 + u2) * du/dx

    In this case, u = xy, so you need to use the product rule to get d/dx(xy).
     
  9. Oct 26, 2010 #8
    So,
    d/dx tan-1 u = 1/(1 + u2) (du/dx)


    u = yx
    du/dx = x(dy/dx) + y

    Thus I get:
    d/dx tan-1 (xy) = 1/(1 + x2y2) (y + x (dy/dx)

    Correct so far?
     
  10. Oct 26, 2010 #9

    Mark44

    Staff: Mentor

    Yes.
     
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