Results 1 to 5 of 5

Math Help - does d^2y / dx^2 mean second derivative?

  1. #1
    Newbie
    Joined
    Mar 2007
    Posts
    8

    does d^2y / dx^2 mean second derivative?

    does d^2y / dx^2 mean second derivative?

    and how do u take the derivative of something like this

    3y^2 - 4x=16 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by juiicycouture View Post
    does d^2y / dx^2 mean second derivative?
    Yes


    and how do u take the derivative of something like this

    3y^2 - 4x=16 ?
    i suppose the derivative you are talking about is dy/dx, you can find this two ways, explicitly or implicity.

    implicity:

    3y^2 - 4x = 16
    => 6y y' - 4 = 0
    => y' = 4/6y = 2/3y
    and if you want, you can solve for y and plug it in


    explicitly:
    we solve for y first and then take the derivative normally

    3y^2 - 4x = 16
    => 3y^2 = 16 + 4x
    => y^2 = (16 + 4x)/3
    => y = sqrt[(16 + 4x)/3] = [(16 + 4x)/3]^(1/2) = (16/3)^(1/2) + (4x/3)^(1/2) .................incorrect, see post below
    => y' = (1/2)(4x/3)^(-1/2) * (4/3) = (2/3)(4x/3)^(-1/2)
    Last edited by Jhevon; April 5th 2007 at 10:44 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by Jhevon View Post
    ...
    explicitly:
    we solve for y first and then take the derivative normally

    3y^2 - 4x = 16
    => 3y^2 = 16 + 4x
    => y^2 = (16 + 4x)/3
    => y = sqrt[(16 + 4x)/3] = [(16 + 4x)/3]^(1/2) = (16/3)^(1/2) + (4x/3)^(1/2)
    => y' = (1/2)(4x/3)^(-1/2) * (4/3) = (2/3)(4x/3)^(-1/2)
    Hello, Jhevon,

    I will not comment this kind of transformation - but it is extremly wrong:

    Repair:

    y = sqrt[(16 + 4x)/3] = [(16 + 4x)/3]^(1/2) ===>

    y' = * ((1/3)(16+4x))^(- )*4 = 2/(sqrt((1/3)(16+4x))

    EB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by earboth View Post
    Hello, Jhevon,

    I will not comment this kind of transformation - but it is extremly wrong:

    Repair:

    y = sqrt[(16 + 4x)/3] = [(16 + 4x)/3]^(1/2) ===>

    y' = * ((1/3)(16+4x))^(- )*4 = 2/(sqrt((1/3)(16+4x))

    EB
    Ah yes, i see my mistake. my bad, i'm half asleep right now
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,802
    Thanks
    692
    Hello, juiicycouture!

    How do u take the derivative of something like this? .3y - 4x .= .16
    I'll assume you want the second derivative . . .


    Differentiate implicitly: .6y(dy/dx) - 4 .= .0

    . . Then: .dy/dx .= .2/(3y) .= .(2/3)y^
    (-1) .[1]


    Differentiate again: .dy/dx .= .-(2/3)y^
    (-2) (dy/dx)

    Substitute [1]: .dy/dx .= .-(2/3)y^
    (-2) (2/3)y^(-1)


    . . Therefore: . dy/dx .= .-4/(9y)

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: January 24th 2011, 11:40 AM
  3. Replies: 2
    Last Post: November 6th 2009, 02:51 PM
  4. Replies: 1
    Last Post: January 7th 2009, 01:59 PM
  5. Frchet derivative and Gteaux derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 23rd 2008, 04:40 PM

Search Tags


/mathhelpforum @mathhelpforum