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Simplifying a Derivative

  1. Jul 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of arctan[(1 - x) / (1 + x)].


    2. Relevant equations
    Everything in the "Show Steps" section:
    http://www.wolframalpha.com/input/?i=derivative+arctan[(1+-+x)+/+(1+++x)]

    My problem is that I don't know how Wolfram manages to simplify:

    - 2 / [1 + ((1 - x) / (1 + x))^2](1 + x)^2

    ^ which is also what I've managed to get,

    to get ,

    - 1 / (1+x^2)


    3. The attempt at a solution
    Everything you see Wolfram does pretty much.
     
  2. jcsd
  3. Jul 16, 2012 #2
    It multiplies through by the (1+x)^2 on the right and then expands the squares.

    But really, it might be more instructive to use the chain rule and do this by hand?
     
  4. Jul 16, 2012 #3
    try multiplying everything out in the denominator and see what happens
     
  5. Jul 16, 2012 #4
    Just multiply out your denominator. You get -2/{(1+x)^2+(1-x)^2}=-2/{2+2x^2}=....
     
  6. Jul 16, 2012 #5
    Tell me what to do next or what I've done wrong here:

    just showing the denominator:

    = 1 + 2x + x^2 + [(1 - 2x + x^2)(1 + 2x + x^2)(1 + 2x + x^2) / (1 + 2x + x^2)]

    = 1 + 2x + x^2 + (1 - 2x + x^2)(1 + 2x + x^2)

    = 1 + 2x + x^2 + 1 + 2x + x^2 - 2x - 4x^2 - 2x^3 + x^2 + 2x^3 + x^4

    = 2 + x^4 - 2x^2 + 2x
     
  7. Jul 16, 2012 #6
    The denominator you gave was
    [1 + ((1 - x) / (1 + x))^2](1 + x)^2

    take the (1+x)^2 inside the brackets
    [(1 + x)^2 + (1 + x)^2((1 - x) / (1 + x))^2]

    Go from there, I'm not sure how you started off with
    "= 1 + 2x + x^2 + [(1 - 2x + x^2)(1 + 2x + x^2)(1 + 2x + x^2) / (1 + 2x + x^2)]"
     
  8. Jul 16, 2012 #7
    Ah, never mind, I was thinking I'm meant to multiply the fraction in the denominator as if they were being added/subtracted, trying to multiply the numerators by the denominators of both the fraction and (1 + x)^2. I get it now. Thanks, I appreciate your patience : D
     
    Last edited: Jul 16, 2012
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