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Wierd implicit derivative

  1. Nov 10, 2008 #1
    when i took the derivative of X^2*Y^2-2x=6 i didn't get the same thing that i got when i took the derivative of y^2=(6+2x)/x^2 and all i did was rewrite the origional equation. why did rewriting the equation change my answer?
  2. jcsd
  3. Nov 10, 2008 #2


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    Staff: Mentor

    Could you show us your work? That would make it easier for us to help.
  4. Nov 10, 2008 #3
    not rewritting i gou
  5. Nov 10, 2008 #4
    when i rewrote it i got y^2=(3+2x)/(x^2)
    and my derivitive was (x-3)/(y(x^3))
  6. Nov 10, 2008 #5


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    Staff: Mentor

    I'm not tracking your math. Are you differentiating the equations with respect to x? It would help if you used LaTex to be explicit about your differentiations...

    [tex]\frac{dy}{dx} = etc.[/tex]


  7. Nov 10, 2008 #6


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    Science Advisor

    Okay, [itex]x^2y^2- 2x= 3 has derivative [itex]2xy^2+ 2x^2y y'- 2= 0 so [itex]2x^2y y'= 2- 2xy^2[/itex] [itex]y'= (2-2xy^2)/(2x^2y)[/itex], the "2"s cancel and y'= (1- xy^2)/(x^2y). You seem to have lost the "2" in the numerator.

    No, solving for y2 gives [itex]x^2 y^2=6+ 2x[/itex] so [itex]y^2= (6+ 2x)/x^2[/itex] The derivative of [itex]y^2[/itex], with respect to x, is 2y y' and the derivative of [itex](6+2x)/x^2[/itex] using the quotient rule is [itex][(2)x^2- (6+2x)(2x)]/(x^4)= [-2x^2- 12x]/(x^4)= -(2x^2+ 12x)/(x^4)= -2(x+ 6)/(x^3)[/itex]. [itex]2yy'= -2(x+6)/x^3[/itex] gives [itex]y'= -(x+1)/yx^3[/itex].

    Think those aren't the same? Replace the y2 in the numerator of the first derivative above with (2x+6)/x2 and simplify.
    Last edited by a moderator: Nov 10, 2008
  8. Nov 11, 2008 #7
    [itex]2yy'= -2(x+6)/x^3[/itex] gives [itex]y'= -(x+1)/yx^3[/itex].

    wouldn't that simplify to [itex]y'= -(x+6)/yx^3[/itex].
  9. Nov 14, 2008 #8
    When I rewrite this I get (-x-3)/(y*x^3). Could that be the problem
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