Results 1 to 5 of 5

Math Help - Implicit differentiation

  1. #1
    Member
    Joined
    Oct 2006
    Posts
    88

    Implicit differentiation

    using the implicit differentiation, find y' (or dy/dx) of y^4 - 3xy + x = y^3

    4y^3 - 3y - 3x + 1 = 3y^2

    dy/dx(4y^3 - 3y - 3x) + 1 = 3y^2

    dy/dx(4y^3 - 3y - 3x) = 3y^2 - 1

    dy/dx = (3y^2 - 1)/(4y^3 - 3y - 3x)

    what did i do wrong? =/

    the answer is supposed to be (3y - 1)/(4y^3 - 3y^2 - 3x)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by jeph View Post
    using the implicit differentiation, find y' (or dy/dx) of y^4 - 3xy + x = y^3

    4y^3 - 3y - 3x + 1 = 3y^2

    dy/dx(4y^3 - 3y - 3x) + 1 = 3y^2

    dy/dx(4y^3 - 3y - 3x) = 3y^2 - 1

    dy/dx = (3y^2 - 1)/(4y^3 - 3y - 3x)

    what did i do wrong? =/

    the answer is supposed to be (3y - 1)/(4y^3 - 3y^2 - 3x)
    Hello,

    First you have to rearrange your equation:
    f(x,y)=y^4-y^3-3xy+x
    Then I use this formula:

    y'=-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}

    \frac{\partial f}{\partial x}=-3y+1. And
    \frac{\partial f}{\partial y}=4y^3-3y^2-3x

    Plug in these results in the formula mentioned above and you'll get the given result.

    EB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,939
    Thanks
    338
    Awards
    1
    Quote Originally Posted by jeph View Post
    using the implicit differentiation, find y' (or dy/dx) of y^4 - 3xy + x = y^3

    4y^3 - 3y - 3x + 1 = 3y^2

    dy/dx(4y^3 - 3y - 3x) + 1 = 3y^2

    dy/dx(4y^3 - 3y - 3x) = 3y^2 - 1

    dy/dx = (3y^2 - 1)/(4y^3 - 3y - 3x)

    what did i do wrong? =/

    the answer is supposed to be (3y - 1)/(4y^3 - 3y^2 - 3x)
    An alternative to Earboth's post is:
    y^4 - 3xy + x = y^3

    4y^3y' - 3y - 3xy' + 1 = 3y^2y'

    4y^3y' - 3y^2y' -3xy' = 3y - 1

    (4y^3 - 3y^2 -3x)y' = 3y - 1

    y' = \frac{3y - 1}{4y^3 - 3y^2 -3x}

    Remember, when you take differentiate a function (actually take the derivative) of a function f(y) of y you need to use the chain rule:
    f'(y) \cdot y'

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2006
    Posts
    88
    so when you differentiate 3xy you get 3y+3xy', but you don't change the y' to a 1? I think I got it...

    how do you gets write the math equations like that? It's so much easier to look at. Is it easier to type in the math as well that way?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by jeph View Post
    ...
    how do you gets write the math equations like that? It's so much easier to look at. Is it easier to type in the math as well that way?
    Hello, Jeph,

    have a look here: http://www.mathhelpforum.com/math-he...rial-latex.pdf

    EB
    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: May 1st 2010, 07:42 AM
  3. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 28th 2010, 03:19 PM
  4. Implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 26th 2010, 05:41 PM
  5. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 25th 2008, 07:33 PM

Search Tags


/mathhelpforum @mathhelpforum