Results 1 to 6 of 6

Math Help - implicit differentiation

  1. #1
    Member
    Joined
    Feb 2010
    Posts
    93

    implicit differentiation

    Hi:

    If 8y(x+y)=4x, find an expression for dy/dx

    8y(x+y)=4x
    u=8y
    v=(x+y)

    du/dx=8
    dv/dx=1

    dy/dx=8y.1dy/dx+(x+y).1=9

    dy/dx=8ydy/dx+(x+y)=9

    I'm not sure if my method is solid here. Would somebody please check this for me. Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    8y(x+y) = 4x

    8xy + 8y^2 = 4x

    Using implicit differentiation and the product rule:

    8xy' + 8y + 16yy' = 4

    8xy' + 16yy' = 4 - 8y

    y'(8x + 16y) = 4 - 8y

    y' = \frac{4 - 8y}{8x + 16y} = \frac{1 - 2y}{2(x + 2y)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by stealthmaths View Post
    Hi:

    If 8y(x+y)=4x, find an expression for dy/dx

    8y(x+y)=4x uv=8y(x+y)=4x
    u=8y
    v=(x+y)

    du/dx=8 typo....du/dy=8
    dv/dx=1 dv/dx=1+dy/dx

    dy/dx=8y.1dy/dx+(x+y).1=9 no!

    dy/dx=8ydy/dx+(x+y)=9

    I'm not sure if my method is solid here. Would somebody please check this for me. Thanks
    Hi stealthmaths,

    you need to reconsider this...

    8y(x+y)=4x

    Both sides are equal.
    Differentiate both sides, the derivatives are equal...

    \frac{d}{dx}[8y(x+y)]=\frac{d}{dx}[4x]

    v\frac{du}{dx}+u\frac{dv}{dx}=4

    u=8y,\ \frac{du}{dx}=\frac{d}{dx}[8y]=8\frac{dy}{dx}

    v=x+y,\ \frac{dv}{dx}=\frac{d}{dx}x+\frac{d}{dx}y=1+\frac{  dy}{dx}

    (x+y)8\frac{dy}{dx}+8y\left(1+\frac{dy}{dx}\right)  =4

    \frac{dy}{dx}\left(8(x+y)+8y\right)+8y=4

    \frac{dy}{dx}\left(8(x+y)+8y\right)=4-8y

    \frac{dy}{dx}=\frac{4-8y}{8x+16y}=\frac{1-2y}{2x+4y}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Feb 2010
    Posts
    93
    I'm really missed some fundamentals from my lectures on this. I have a video tutorial I think I will watch before I fire some questions back, if that's ok.

    In the meantime: I just bought mathtype so I can put my workings in Latex for easy comprehension, and managed to create the problem in the program. But how do I get it into this window without uploading an image file?
    Attached Thumbnails Attached Thumbnails implicit differentiation-.gif  
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,659
    Thanks
    600
    Hello, stealthmaths!

    Sorry, I can't follow your work . . .


    If 8y(x+y)\:=\:4x,\ find an expression for \frac{dy}{dx}

    First, I would divide by 4: . 2y(x+y) \:=\:x

    Then I would expand: . 2xy + 2y^2 \:=\:x


    Differentiate implicitly: . 2x\frac{dy}{dx} + 2y + 4y\frac{dy}{dx} \:=\:1

    . . . .Rearrange terms: . . . . 2x\frac{dy}{dx} + 4y\frac{dy}{dx} \;=\;1 - 2y

    . . . . . . . . . . .Factor: . . . . . 2(x+2y)\frac{dy}{dx} \;=\;1-2y

    . . . . . . . . Therefore: . . . . . . . . . . . \frac{dy}{dx} \;=\;\frac{1-2y}{2(x+2y)}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2010
    Posts
    93
    Great!
    I have studied and can now understand.
    Thanks to Archie Mead for the full version, as this is how we are being taught.
    I have also enjoyed to learn the shorter versions thanks to Soroban and Icemanfan.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 05:24 PM
  2. implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 10th 2010, 05:58 PM
  3. implicit differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 20th 2009, 10:34 AM
  4. implicit differentiation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 12th 2009, 03:15 PM
  5. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 21st 2008, 08:12 PM

Search Tags


/mathhelpforum @mathhelpforum