Advertisement
If you have a new account but are having problems posting or verifying your account, please email us on hello@boards.ie for help. Thanks :)
Hello all! Please ensure that you are posting a new thread or question in the appropriate forum. The Feedback forum is overwhelmed with questions that are having to be moved elsewhere. If you need help to verify your account contact hello@boards.ie
Hi there,
There is an issue with role permissions that is being worked on at the moment.
If you are having trouble with access or permissions on regional forums please post here to get access: https://www.boards.ie/discussion/2058365403/you-do-not-have-permission-for-that#latest

Partial Derivatives

  • 19-05-2004 6:08pm
    #1
    Closed Accounts Posts: 2,922 ✭✭✭


    Need a hand with the explanation of this. Sorry about the threads, it's exam time. Anyway here's the question.

    Find all the first order and second order partial derivatives for f(x, y) = sin(xy)


Comments

  • Closed Accounts Posts: 14,483 ✭✭✭✭daveirl


    This post has been deleted.


  • Registered Users, Registered Users 2 Posts: 1,372 ✭✭✭silverside


    you missed the 2 mixed partial derivatives
    :D , which are equal (function is twice ctsly diff'able)

    d2f/dydx = d2f/dxdy = cos xy - (xy sin xy)


  • Registered Users, Registered Users 2 Posts: 5,523 ✭✭✭ApeXaviour


    lets say
    df/du = df/dx + df/dy

    then what would the answer to

    d2f/du2 = ?


    How do you mess around with these things to get what you want?


  • Registered Users, Registered Users 2 Posts: 1,372 ✭✭✭silverside


    d2f/du2 = (d/du) of (df/du). so basically differentiate each term in the expression of df/du again. Sorry if this is not clear - you may need to look at your notes and/or an analysis book for further explantaion / examples.


  • Registered Users, Registered Users 2 Posts: 5,523 ✭✭✭ApeXaviour


    Ok but I don't have the derivatives of them.

    If I want to get it in terms of df/du, df/dx, df/dy etc.

    How do I do that d'ya reckon?


  • Advertisement
  • Registered Users, Registered Users 2 Posts: 5,523 ✭✭✭ApeXaviour


    Apologies I wan't giving enough information. I worked it out now.

    Assuming f is a continious function then:

    since df/du = df/dx + df/dy

    then also as you said

    d2f/du2 = d/du (df/dx + df/dy)

    and since

    d/du = d/dx + d/dy

    The next line is:

    d2f/du2 = (d/dx + d/dy) (df/dx + df/dy)

    which when done out boils down to:

    d2f/du2 = d2f/dx2 + d2f/dxdy +d2f/dydx + d2f/dx2


    That's what I was looking for, but thanks, ya sent me on the right road.


  • Closed Accounts Posts: 2,922 ✭✭✭Dave


    Cheers lads. The exam isn't on till friday, and that question is worth 4% to me.


Advertisement