Results 1 to 4 of 4

Math Help - Second Order Implicit Partial Derivatives

  1. #1
    Newbie
    Joined
    May 2009
    From
    Perth, Western Australia
    Posts
    2

    Second Order Implicit Partial Derivatives

    This problem is doing my head in...

    If z is defined implicitly as a function of x and y by the equation

    xze^y + yz^3 = 1,

    find the values of dz/dx, dz/dy and d^2z/dxdy when x=1 and y=0.

    I've got dz/dy and dz/dx although these were worked out fairly quickly without double checking, but irregardless I can't for the life of me figure out how to get d^2z/dxdy, and no amount of googling it or reading the textbook seems to yield any more information.

    dz/dy = (-z^3-xze^y)/(3yz^2 + xe^y)

    dz/dx = -(ze^y)/(xe^y+3yz^2)

    Any help would be hugely appreciated, as this question is part of my exam preparation, and knowing my luck, a similar question is likely to come up in the actual exam.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    \frac{d^2z}{dxdy} means that you differentiate \frac{dz}{dy} with respect to x. (maybe for some people it's the other way round : diff dz/dx with respect to y...that will be depending on how you've been taught)

    Don't forget that z is still a function of x. So if you differentiate \frac{dz}{dy}, you'll have some terms containing \frac{dz}{dx}, which you will substitute by what you found.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    From
    Perth, Western Australia
    Posts
    2
    It shouldn't matter which way they're differentiated, [ie dz/dy with respect to x, or dz/dx with respect to y], since the answers would be equal, would they not?

    Either way, I have tried differentiating dz/dx with respect to y, and - as you mentioned - got an answer with a dz/dy term in it. So you think the way to go would be to just sub in the original answer for dz/dy? It sounds like a sensible idea, and one I had considered, but for some reason didn't think it'd work [although I had become quite frustrated with the problem by the time I got around to trying it, so maybe didn't give it a proper go].

    Cheers for the response - I'll try it and let you know how I go.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by wceagles10 View Post
    It shouldn't matter which way they're differentiated, [ie dz/dy with respect to x, or dz/dx with respect to y], since the answers would be equal, would they not?
    Oops, yes... I guess so. I don't exactly remember, but you must be right.

    Either way, I have tried differentiating dz/dx with respect to y, and - as you mentioned - got an answer with a dz/dy term in it. So you think the way to go would be to just sub in the original answer for dz/dy? It sounds like a sensible idea, and one I had considered, but for some reason didn't think it'd work [although I had become quite frustrated with the problem by the time I got around to trying it, so maybe didn't give it a proper go].
    Yes, you just have to sub.

    For the second derivative... It may be easier to work on e^y \left(z+x\frac{dz}{dy}\right)+3yz^2\frac{dz}{dy}+z  ^3=0, taking its derivative wrt x.
    In general, I prefer applying the product and addition rules of differentiation rather than the quotient rule...less mistakes. It's as you wish !
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 15th 2011, 08:54 PM
  2. Replies: 5
    Last Post: April 13th 2010, 03:07 PM
  3. Partial derivatives of implicit function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 22nd 2010, 11:58 AM
  4. Replies: 1
    Last Post: October 10th 2008, 08:29 AM
  5. Higher Order Implicit Derivatives
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 20th 2007, 02:56 PM

Search Tags


/mathhelpforum @mathhelpforum