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Implicit Differentiation

  1. Mar 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

    x^2+y^2=(2x^2+2y^2-x)^2 @ (0,(1/2))

    2. Relevant equations



    3. The attempt at a solution

    So, I don't have a problem differentiating this.

    Which I get 2x+2y(dy/dx)=2(2x^2+2y^2-x)(4x+4y(dy/dx)-1)

    I am having trouble getting (dy/dx) alone. I was wondering if I could plug in the coordinates and treat (dy/dx) like a variable (e.g. being added and subtracted to other dy/dx like 1(dy/dx)-2(dy/dx)=-1?)

    So, could I just plug in (0,(1/2))? Also, I have a test tomorrow and I was wondering in your expert opinions (I know you don't know my professor), but do you think it would be possible that he wants me to simplify this on a test instead (I mean is that more shall I say proper)?

    If that is the case when I have 2 large polynomials is it true that I can do something like:
    using (2x^2+2y^2-x)(4x+4y(dy/dx)-1) take out 4y(dy/dx) and multiply it by the other polynomial? by that I mean

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

    or like (2x^2+2y^2-x)(4x+4y(dy/dx)-1) is equivalent to (2x^2+2y^2)((4x+4y(dy/dx)-1)-x(4x+4y(dy/dx)-1)?
     
  2. jcsd
  3. Mar 18, 2013 #2

    LCKurtz

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    You should probably bone up on your algebra skills then, eh?
    Yes, that is a very good idea to simplify it quickly, especially when time is at a premium.
     
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