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

- anonymous

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

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- UnkleRhaukus

with respect to x?

- anonymous

Yep!

- anonymous

I guess it would be chain rule on both sides but it gets a bit messy when I do it

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## More answers

- UnkleRhaukus

\[x^2 + y^2 = 2x^2 + 2y^2 - x\\
\frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx}\left(2x^2 + 2y^2 - x\right)\\
\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(2x^2)+\frac{d}{dx}(2y^2)-\frac{d}{dx}(x)\]

- anonymous

I just corrected the question, the right side is to the power of 2

- UnkleRhaukus

Implicitly differentiate the terms with \(y\), like this:
\[\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\]

- UnkleRhaukus

oh ok,
what do you get for the left hand side?

- anonymous

I got 2x + 2yy'

- UnkleRhaukus

good

- UnkleRhaukus

now the righthand side
\[=\frac d{dx}(2x^2+2y^2-x)^2\]

- anonymous

Would it be
2(2x^2 + 2y^2 -x)(4x +4yy' -1) ?

- UnkleRhaukus

looks good!

- anonymous

Now the simplification XD

- UnkleRhaukus

hmmm, i don't think it simplifies very far,
implicit derivatives often don't simplify nice

- UnkleRhaukus

What do you get?

- amilapsn

please look @ my question after this @UnkleRhaukus

- anonymous

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

- anonymous

@UnkleRhaukus is this what you get as well?

- UnkleRhaukus

what happened to the equals sign?

- anonymous

What I'm trying to solve is actually this question...

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- anonymous

Oh lol forgot that... y' = to all that ^

- zepdrix

Ooo I think the same derivative :)
I think you did that correctly

- zepdrix

Gave yourself a little extra work maybe though.. hmm

- zepdrix

I got the same derivative* woops typo

- anonymous

Oh that's a relief! Haha at least i got that right XD

- UnkleRhaukus

...
I got
2x + 2yy' = 2(2x^2 + 2y^2 -x)(4x +4yy' -1)
x + yy' = (2x^2 + 2y^2 -x)(4x +4yy' -1)
= +/- âˆš(x^2 + y^2)(4x +4yy' -1)
how do you simplify it?

- anonymous

Well I only solved for y'

- zepdrix

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.

- anonymous

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?

- zepdrix

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.

- zepdrix

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.

- anonymous

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

- zepdrix

\[\Large\rm y-\frac{1}{2}=m(x-0)\]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.

- anonymous

Alright so I'm gonna give that a try and tell you what I get :)

- anonymous

Please don't leave! :D

- zepdrix

fine fine fine -_-

- anonymous

Lol so plugging in the x and y values I got -1... Tell me I'm not wrong :D

- zepdrix

Let's look at the graph to sort of check our work easily.

- zepdrix

|dw:1433147166496:dw|Here is the point (0,1/2)

- zepdrix

Does the slope look like a -1?
Hmmm not so much :[

- anonymous

Maybe its positive? haha

- anonymous

I don't know where i went wrong >.<

- zepdrix

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

- anonymous

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

- zepdrix

positive 1? Ooo that seems to match the picture a lot better also! :) yay

- anonymous

Haha awesome! XD

- anonymous

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

- zepdrix

why -x? :o

- anonymous

Oh that was me going loco again :D haha sorry its
y = x + 1/2

- zepdrix

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

- anonymous

Look 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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- zepdrix

yay team \c:/
np

- anonymous

Hahaha YaY :D

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