Results 1 to 5 of 5

Math Help - another partial derivative

  1. #1
    Member
    Joined
    Jul 2009
    Posts
    149

    another partial derivative

     3x^2\ln(1-xy)

    i want to differentiate with respect to y.

    So.
    [(0)(\ln(1-xy)]+[(3x^2)(-x)]

    which is -3x^3


    Is that correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2009
    Posts
    112
    again you're running into this same problem where you want to do a partial derivative in terms of y, but you're not holding x as a constant.

    what type of function of y are you needing to differentiate (ie. polynomial, quotient etc.)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by el123 View Post
     3x^2\ln(1-xy)

    i want to differentiate with respect to y.

    So.
    [(0)(\ln(1-xy)]+[(3x^2)(-x)]

    which is -3x^3


    Is that correct?
    Just treat x like a number. 3x^2 and x are just constants.

    If you were derivating c_1\ln(1-c_2 y), what would you get? You'd get \frac{c_1}{1-c_2 y}\cdot -c_2. Just extend that notion to this problem.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2009
    Posts
    149
    so its  \frac{3x^3}{1-xy}

    Is that what your saying?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    You have to multiply that by \frac{d}{dx}[1-xy]=-x, to complete the chain rule, so the answer is \frac{-3x^3}{1-xy}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Partial derivative
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 27th 2011, 02:18 PM
  2. Please help with partial derivative.
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 12th 2010, 04:54 PM
  3. Derivative of arctan in a partial derivative
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 12th 2010, 02:52 PM
  4. help me!!! (partial derivative)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 5th 2008, 11:57 PM
  5. Partial Derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 24th 2007, 05:02 PM

Search Tags


/mathhelpforum @mathhelpforum