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anonymous

  • one year ago

How can I implicitly differentiate the following... x^2 + y^2 = (2x^2 + 2y^2 - x)^2

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  1. UnkleRhaukus
    • one year ago
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    with respect to x?

  2. anonymous
    • one year ago
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    Yep!

  3. anonymous
    • one year ago
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    I guess it would be chain rule on both sides but it gets a bit messy when I do it

  4. UnkleRhaukus
    • one year ago
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    \[x^2 + y^2 = 2x^2 + 2y^2 - x\\ \frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx}\left(2x^2 + 2y^2 - x\right)\\ \frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(2x^2)+\frac{d}{dx}(2y^2)-\frac{d}{dx}(x)\]

  5. anonymous
    • one year ago
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    I just corrected the question, the right side is to the power of 2

  6. UnkleRhaukus
    • one year ago
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    Implicitly differentiate the terms with \(y\), like this: \[\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\]

  7. UnkleRhaukus
    • one year ago
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    oh ok, what do you get for the left hand side?

  8. anonymous
    • one year ago
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    I got 2x + 2yy'

  9. UnkleRhaukus
    • one year ago
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    good

  10. UnkleRhaukus
    • one year ago
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    now the righthand side \[=\frac d{dx}(2x^2+2y^2-x)^2\]

  11. anonymous
    • one year ago
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    Would it be 2(2x^2 + 2y^2 -x)(4x +4yy' -1) ?

  12. UnkleRhaukus
    • one year ago
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    looks good!

  13. anonymous
    • one year ago
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    Now the simplification XD

  14. UnkleRhaukus
    • one year ago
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    hmmm, i don't think it simplifies very far, implicit derivatives often don't simplify nice

  15. UnkleRhaukus
    • one year ago
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    What do you get?

  16. amilapsn
    • one year ago
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    please look @ my question after this @UnkleRhaukus

  17. anonymous
    • one year ago
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    This is what I get at the end ... \[\frac{ 2(4x^3 - 3x^2 +4xy^2 - y^2) }{ y(1-8x^2 - 8y^2 +4x )}\]

  18. anonymous
    • one year ago
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    @UnkleRhaukus is this what you get as well?

  19. UnkleRhaukus
    • one year ago
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    what happened to the equals sign?

  20. anonymous
    • one year ago
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    What I'm trying to solve is actually this question...

  21. anonymous
    • one year ago
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    Oh lol forgot that... y' = to all that ^

  22. zepdrix
    • one year ago
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    Ooo I think the same derivative :) I think you did that correctly

  23. zepdrix
    • one year ago
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    Gave yourself a little extra work maybe though.. hmm

  24. zepdrix
    • one year ago
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    I got the same derivative* woops typo

  25. anonymous
    • one year ago
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    Oh that's a relief! Haha at least i got that right XD

  26. UnkleRhaukus
    • one year ago
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    ... I got 2x + 2yy' = 2(2x^2 + 2y^2 -x)(4x +4yy' -1) x + yy' = (2x^2 + 2y^2 -x)(4x +4yy' -1) = +/- √(x^2 + y^2)(4x +4yy' -1) how do you simplify it?

  27. anonymous
    • one year ago
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    Well I only solved for y'

  28. zepdrix
    • one year ago
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    It probably would have been easier to simply plug in your coordinate from this point,\[\Large\rm 2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)\]And simply add all the stuff up before solving for y'. Especially since you're getting a bunch of 0's in there.

  29. anonymous
    • one year ago
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    Ohhh right!!! XD But then Im gonna be left with y in the equation as well, how would I plug in the slope in the point-slope formula at the end for finding the equation of the tangent line?

  30. zepdrix
    • one year ago
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    Left with y in the equation? No no, you're plugging 0s in for the Xs, and 1/2s in for the Ys, ya? :) Only #s and `Y'`s left over.

  31. zepdrix
    • one year ago
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    y' = m <-- this i the slope you're looking for. We would STILL have to do the work to solve for y' to find our slope, but im simply saying that if you plug in the coordinate BEFORE solving for y', it's a ton easier, because you're plugging in a bunch of 0s.

  32. anonymous
    • one year ago
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    Okay - so we plug in x=0 and y=1/2 and then solve for y'? Then plug that number in for 'm' in the point-slope formula, correct? :D

  33. zepdrix
    • one year ago
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    \[\Large\rm y-\frac{1}{2}=m(x-0)\]Yes good :) Figure out that m! And since our line is tangent to the curve, it "touches" right at this coordinate point, so both the curve and this tangent line share the point, we can use it for our point-slope form.

  34. anonymous
    • one year ago
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    Alright so I'm gonna give that a try and tell you what I get :)

  35. anonymous
    • one year ago
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    Please don't leave! :D

  36. zepdrix
    • one year ago
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    fine fine fine -_-

  37. anonymous
    • one year ago
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    Lol so plugging in the x and y values I got -1... Tell me I'm not wrong :D

  38. zepdrix
    • one year ago
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    Let's look at the graph to sort of check our work easily.

  39. zepdrix
    • one year ago
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    |dw:1433147166496:dw|Here is the point (0,1/2)

  40. zepdrix
    • one year ago
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    Does the slope look like a -1? Hmmm not so much :[

  41. anonymous
    • one year ago
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    Maybe its positive? haha

  42. anonymous
    • one year ago
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    I don't know where i went wrong >.<

  43. zepdrix
    • one year ago
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    \[\Large\rm 2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)\]Plugging in our stuff gives us,\[\Large\rm 0+y'=2\left(0+\frac{1}{2}-0\right)(0+2y'-1)\]Ya? :d

  44. anonymous
    • one year ago
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    Oh wait no I was right!! Just went a little kookoo at the end! I get -.5/-.5 and that's 1 not -1 XD

  45. zepdrix
    • one year ago
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    positive 1? Ooo that seems to match the picture a lot better also! :) yay

  46. anonymous
    • one year ago
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    Haha awesome! XD

  47. anonymous
    • one year ago
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    so then my point-slope formula would be y - .5 = 1(x -0) which then simplifies to y = -x + 1/2 right??? :D

  48. zepdrix
    • one year ago
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    why -x? :o

  49. anonymous
    • one year ago
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    Oh that was me going loco again :D haha sorry its y = x + 1/2

  50. zepdrix
    • one year ago
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    Mmmm nice! :) Seems to have worked out correctly! Here is the graph of the function and tangent line just in case you wanna see what it looks like: https://www.desmos.com/calculator/eibccmgpxy

  51. anonymous
    • one year ago
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    Look at this!!! Thanks soooo soo much!! I have a calculus midterm tomorrow, you have no idea how much I appreciate your help!!

  52. zepdrix
    • one year ago
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    yay team \c:/ np

  53. anonymous
    • one year ago
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    Hahaha YaY :D

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