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Homework Help: Implicit Differentiation

  1. Jan 8, 2005 #1
    Hello all

    Given: x^2 + xy + y^2 - 7 = 0, solve for y using the quadratic forumula. Then find dy/dx at P(1,2) from a function of the form f(x).

    My solution:

    y = -x (+/-) sqrt( x^2 - 28) / 2.

    I am not sure if this is correct. After solving for y, do you have to implicitly take the derivative of y?
     
  2. jcsd
  3. Jan 8, 2005 #2
    you have to calculate the derivative with repsect to x of this given function. So just look at y as some function dependent on x ; so you have dy/dx...and dy²/dx = 2ydy/dx and so on...

    2x + y + xdy/dx + 2ydy/dx = 0

    you get this too ???


    regards
    marlon
     
  4. Jan 8, 2005 #3

    dextercioby

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    [tex] y^{2}+xy+x^{2}-7=0 [/tex]
    Use the quadratic formula to find:
    [tex] y_{1,2}(x) =\frac{-x\pm\sqrt{28-3x^{2}}}{2} [/tex]
    (1)
    The condition that the point (1,2) should be on the graph of "y" yields:
    [tex] y_{1}(1)=\frac{-1+\sqrt{28-3}}{2}=\frac{-1+5}{2}=2 [/tex](2)

    [tex] y_{2}(1)=\frac{-1-5}{2}=-3 [/tex] (3)


    So you need to chose the "+" sign from the explicitation.
    [tex] y(x)=\frac{1}{2}(\sqrt{28-3x^{2}}-x) [/tex]

    Compute its derivative and make "x=1" in the result.

    Daniel.
     
    Last edited: Jan 8, 2005
  5. Jan 8, 2005 #4
    GUYS GUYS what are you doing...

    Your solution is NOT an implicit derivation. You don't need the quadratic formula at all. Besides the motivation, dextercioby for taken the + value is not correct.

    What you need to do is look at y as y(x) and calculate the derivative of the given formula with respect to x. This is the IMPLICIT part...the derivavtive that is asked is in a formula itself

    x²+xy+y²-7=0
    derivative with respect to x yields

    2x + y + x(dy/dx) + 2y(dy/dx) = 0

    or

    [tex]\frac{dy}{dx} (x + 2y) = -(2x+y)[/tex]
    [tex]\frac{dy}{dx} = \frac{-2x-y}{x+2y}[/tex]

    Then fill in the given x and y values in the right hand side and all is done...

    regards
    marlon
     
  6. Jan 8, 2005 #5
    how did you get 3x^2 for the term in the quadratic forumula dextercioby?

    Thanks
     
  7. Jan 8, 2005 #6

    dextercioby

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    The discriminant is [itex] \Delta=b^{2}-4ac [/tex]
    Compute it,with [itex] a=1;b=x;c=x^{2}-7 [/itex]

    Daniel.
     
  8. Jan 8, 2005 #7
    ok thanks alot
     
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