1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar

    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


    User Avatar

    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


    User Avatar
    Staff Emeritus
    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: 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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?