Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By chiro

Math Help - Second Derivative in Terms of X and Y

  1. #1
    Junior Member
    Joined
    Oct 2012
    From
    Houston
    Posts
    51

    Second Derivative in Terms of X and Y

    So, i have no problem finding the first derivative. But then i cant get my answer to match the books answer for the second derivative. How are they getting the highlighted area? It doesnt look like they used the Quotient Rule. Have they moved the denominator up and then power rule?

    Thanks for the help.
    Attached Thumbnails Attached Thumbnails Second Derivative in Terms of X and Y-screen-shot-2012-10-11-7.33.03-pm-1.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,337
    Thanks
    1247

    Re: Second Derivative in Terms of X and Y

    Quote Originally Posted by petenice View Post
    So, i have no problem finding the first derivative. But then i cant get my answer to match the books answer for the second derivative. How are they getting the highlighted area? It doesnt look like they used the Quotient Rule. Have they moved the denominator up and then power rule?

    Thanks for the help.
    They multiplied both sides by \displaystyle \begin{align*} x^2y \end{align*} and then moved all non-constant terms to the LHS before trying to evaluate the second derivative.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,607
    Thanks
    591

    Re: Second Derivative in Terms of X and Y

    Hey petenice.

    They applied implicit differentiation to the expression in line 3. In other words they did d/dx [x^2*y*y' + xy^2 - 9] = d/dx(0)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2012
    From
    Houston
    Posts
    51

    Re: Second Derivative in Terms of X and Y

    Quote Originally Posted by chiro View Post
    Hey petenice.

    They applied implicit differentiation to the expression in line 3. In other words they did d/dx [x^2*y*y' + xy^2 - 9] = d/dx(0)
    Ok so now i can get past that part. However i get stuck at this next part. I keep going and keep getting the same answer but its incorrect. :/ Im pretty sure im doing everything right up to this point. But now what?

    Thanks for any help!
    Attached Thumbnails Attached Thumbnails Second Derivative in Terms of X and Y-photo.jpg  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,607
    Thanks
    591

    Re: Second Derivative in Terms of X and Y

    The differentiation of the first term is not correct: you differentiating x^2*y correctly but not the other one.

    So d/dx[x^2*y*y']
    = 2xyy' + x^2[d/dx](yy')
    = 2xyy' + x^2[y'y' + y*y'']
    = 2xyy' + x^2[y']^2 + x^2[yy'']

    Try using this and see how you go.
    Thanks from yvonnehr
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: June 6th 2012, 08:51 AM
  2. Getting Y in terms of x
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 19th 2011, 04:28 PM
  3. Replies: 0
    Last Post: January 24th 2011, 11:40 AM
  4. Replies: 5
    Last Post: September 3rd 2010, 01:45 AM
  5. like terms
    Posted in the Algebra Forum
    Replies: 3
    Last Post: February 1st 2007, 05:35 AM

Search Tags


/mathhelpforum @mathhelpforum