Results 1 to 2 of 2

Math Help - Partial derivative with exponential & product rule

  1. #1
    Junior Member
    Joined
    Nov 2007
    Posts
    53

    Partial derivative with exponential & product rule

    So I've been trying to solve this partial derivative with respect to either x or y but my answer (using the product rule) is way off of what the solutions manual is.

    The original function is f(x,y) = (x^2 + 4y^2) * e^(1-x^2-y^2)

    The solutions manual states the partial derivative with respect to x is 2xe^(1-x^2-y^2) * (1-x^2-4y^2) and the partial derivative with respect to y is 2ye^(1-x^2-y^2) * (4-x^2-4y^2).

    The partial derivative with respect to xx is 2e^(1-x^2-y^2) * (1-5x^2+2x^4-4y^2+8x^2y^2) and respect to yy is 2e^(1-x^2-Y^2) * (4-x^2-20y^2+8y^4+2x^2y^2) and with respect to xy is -4xye^(1-x^2-y^2) * (5-x^2-4y^2).

    Is there a special way to using the product rule (and quotient rule for that matter) as far as partial derivatives go outside of holding the variable constant that you're not taking the derivative with respect to?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    652
    Thanks
    2
    Awards
    1
    When you treat it the same as always, the product rule holds in Partial Derivatives:

    We have:

    z = f(x,y) = (x^2 + 4y^2)e^{(1 - x^2 - y^2)}

    We need to find:

    \frac{\partial z}{\partial x} = 2xe^{(1 - x^2 - y^2)} - 2xe^{(1 - x^2 - y^2)}(x^2 + 4y^2)

    Factor out a 2xe^{(1 - x^2 - y^2)}:

    \frac{\partial z}{\partial x} = 2xe^{(1 - x^2 - y^2)}(1 - (x^2 + 4y^2))

    \frac{\partial z}{\partial x} = 2xe^{(1 - x^2 - y^2)}(1 - x^2 - 4y^2)

    And it's the same method for \frac{\partial z}{\partial y}.

    Hope that helped.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivative by using product rule
    Posted in the Calculus Forum
    Replies: 7
    Last Post: October 13th 2011, 02:54 PM
  2. Replies: 5
    Last Post: October 19th 2009, 01:04 PM
  3. Derivative using the product rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 1st 2009, 12:11 AM
  4. product rule derivative check!
    Posted in the Calculus Forum
    Replies: 9
    Last Post: September 8th 2008, 02:36 AM
  5. Find derivative using product rule....
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 1st 2008, 07:03 PM

Search Tags


/mathhelpforum @mathhelpforum