stumped on a seemingly simple math problem..

Discussion in 'Options' started by cdcaveman, Feb 5, 2013.

  1. i figure this relates so much to options .. some of you must know this..

    find the dervative.. of


    i thought you bring factor so you don't have fractions, put everything to a power.. like sqrt(x) is x^1/2 i'm very good at much more complex questions relative to this..
    a simple one is differentiate f(x) = x(x^2+3)
    f'(x) =3x^2+3
    thats simple..

    so question is.. how to find the derivative of 3/(2x) i'm using parenthesis here so you don't think its three halfs of x .. its three over two times x..
  2. hft_boy


    It's like (3/2) * x^-1. Use the power rule and you get -1 * (3/2) * x^-2.
  3. addchild


    dx/dy for a quotient = ((derivitive of the numerator)(denominator)-(numerator)(derivitive of the denominator))/ (denominator)^2
  4. because 3/(2x) = 3/2 *1/x

    there for lifting the x from the denominator you get


    then you differentiate

    -1 * (3/2) ... as the first part..
    then subtract 1 from the exponent of x.. putting it at X^-2
    so you end up with


    that sure seems right.. and follows what you are saying..
  5. Eyez


    -(3x^-2)/2 or -3/(2x^2)
  6. parenthesis to be explicit..

    i got negative three halfs X to the negative second..

  7. Eyez


  8. now.. (3/2)*((x^-2)/1) is the same as (3/2)*(1/(x^2))

    yes moving the exponent up makes it negative and the reverse respectively..
    such that negative three over 2x^2 is the same thing as 3/2( x ^-2)
  9. Eyez


    I guess im being a syntax nazi lol... :D
  10. at least we aren't actually arguing about the answer being right haha
    #10     Feb 5, 2013