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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
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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UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2with respect to x?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess it would be chain rule on both sides but it gets a bit messy when I do it

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2\[x^2 + y^2 = 2x^2 + 2y^2  x\\ \frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx}\left(2x^2 + 2y^2  x\right)\\ \frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(2x^2)+\frac{d}{dx}(2y^2)\frac{d}{dx}(x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just corrected the question, the right side is to the power of 2

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2Implicitly differentiate the terms with \(y\), like this: \[\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2oh ok, what do you get for the left hand side?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2now the righthand side \[=\frac d{dx}(2x^2+2y^2x)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would it be 2(2x^2 + 2y^2 x)(4x +4yy' 1) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now the simplification XD

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2hmmm, i don't think it simplifies very far, implicit derivatives often don't simplify nice

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0please look @ my question after this @UnkleRhaukus

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is what I get at the end ... \[\frac{ 2(4x^3  3x^2 +4xy^2  y^2) }{ y(18x^2  8y^2 +4x )}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@UnkleRhaukus is this what you get as well?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2what happened to the equals sign?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What I'm trying to solve is actually this question...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh lol forgot that... y' = to all that ^

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ooo I think the same derivative :) I think you did that correctly

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Gave yourself a little extra work maybe though.. hmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I got the same derivative* woops typo

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh that's a relief! Haha at least i got that right XD

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.2... 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well I only solved for y'

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh right!!! XD But then Im gonna be left with y in the equation as well, how would I plug in the slope in the pointslope formula at the end for finding the equation of the tangent line?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Left 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1y' = 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay  so we plug in x=0 and y=1/2 and then solve for y'? Then plug that number in for 'm' in the pointslope formula, correct? :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large\rm y\frac{1}{2}=m(x0)\]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 pointslope form.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright so I'm gonna give that a try and tell you what I get :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Please don't leave! :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol so plugging in the x and y values I got 1... Tell me I'm not wrong :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Let's look at the graph to sort of check our work easily.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433147166496:dwHere is the point (0,1/2)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Does the slope look like a 1? Hmmm not so much :[

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe its positive? haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know where i went wrong >.<

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large\rm 2x+2yy'=2(2x^2+2y^2x)(4x+4yy'1)\]Plugging in our stuff gives us,\[\Large\rm 0+y'=2\left(0+\frac{1}{2}0\right)(0+2y'1)\]Ya? :d

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wait no I was right!! Just went a little kookoo at the end! I get .5/.5 and that's 1 not 1 XD

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1positive 1? Ooo that seems to match the picture a lot better also! :) yay

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so then my pointslope formula would be y  .5 = 1(x 0) which then simplifies to y = x + 1/2 right??? :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh that was me going loco again :D haha sorry its y = x + 1/2

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Mmmm 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Look at this!!! Thanks soooo soo much!! I have a calculus midterm tomorrow, you have no idea how much I appreciate your help!!
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