Use implicit differentiation to find an equation of the tangent line to the curve at the curve at the given point: x^2 + y^2 = (2x^2 + 2y^2 - x)^2

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Use implicit differentiation to find an equation of the tangent line to the curve at the curve at the given point: x^2 + y^2 = (2x^2 + 2y^2 - x)^2

Mathematics
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... @ (0,1/2)
not that i tmatters anyway i'm just a little confused regarding the differentiation here
alright give me a second to remember lol

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alright i'll upload a picture on what to do
and an example give me a sec
cool thanks
like i got as far as [4x+4y*y']/[4x+4y*y'-1] = 2x^2 + 2y^2 - x
actualy i can't... i'll just type it out what my book says GUIDELINES FOR IMPLICIT DIFFERENTIATION 1. Differentiat both sides of the equatio with respects to x 2. collect al terms involve dy/dx on the left side of the equation and move all others to right 3.factor dy/dx out of the left side of the equation 4.solve for dy/dx by dividing both sides
by factor
right righ tok so and
so what do i do from what i have am i good so far
let me get a piece of paper lol
ama differentiate for u
thanks k cool coco
love it
so is coco checking if your differentiation is right?
one question. just verifying... for the secon part, everyting is raised to teh 2nd power?
i'm guessing
oh, ok
yes
ouch
umm. i suck @ algebra... and i think i am going to mess up if i do this. but, i am still going to go ahead and square all that. can someone then tell me if i was right?
yeah i got completely something else than he did
the second is a chain and the first is just regular
i got 2x+2yy'=2(4x+4yy'-1)
=2x+2yy'= 8x+8yy'-2
if you do implicit differentiation, then don't you do (for example, the first part): 2xdx+2ydy... then get all teh dx's and dy's on one side and divide so that dy/dx
yes
\[2yy'-8yy'=6x-2\]
ahh... i get it.. am so stupid! i was wondering where you got the chain rule from
yep
then you can pull out y'
i think it's\[\frac{-6x-2}{6y}\]
the - is outside parenthesis
\[-(6x-2)/6y\]
it's been a while though lol
blah wait i messed up lol
let me do this all over agani
i forgot something
it doesn't help that i'm also trying to help two people
ok, sorry guys but i am lost lol. but i wanna know how it's done. so outkast, once you did the chain rule, did you multiply or what?
my chain rule was messed up lol it should be \[2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)\]
yes that's what i got
then what did you do
i'd say to combine the right side
it's going to be ugly hahah
dive 2xdx + 2ydy with 4xdx+4ydy...?
divide*
i think you'd have to combine the right side which is a pain
Hey, do you know is there anyway i can like mark this question so that i can come back and look @ it 2morrow?
no seriously?! ouch... calyne, your proff/teacher must be coldhearted to give you such a question.
um i think it should still be in the top left forner under notifications
it's not that bad tho
Oh, ok. thanks!
\[2x+2yy'=16x^3+16y^3y'+16x^2yy'+16y^2x-12x^2-4y^2-8xyy'+2x\] \[2yy'=16x^3+16^3y'+16x^2yy'+16y^2x-12x^2-4y^2-8xyy'\]
...
did you get what me and coco got about my last when you used the chain rule
what
you use the chain rule yeah du^2/du * du/dx first thing overall ik
what you multiply that whole thing by dy/dx too
so whta did you get after you take dy/dx of both sides
what did you get damn
using the chain rule i got 2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)
sorry i got 2x+2yy' = 2(2x^2+2y^2-x) (4x + 4yy' - 1)
yeahhhh
alright so if you combine the terms on the right
is what i did ... what did you do after
idk that's gay
i tried a bunch of different stuff what was your final answer i'll tell you if it's correct nah flutter that the answer is x+1/2. is that what you got. if not we're wasting time.
not done yet haha
yeah i have no idea.. even wolfram alpha has omething like my end answer
unless your teacher is crazy and solved for like y lol
wait the answer to the differentiation is that or the tangent line
because if it's the tangent line my implicit differentiation may be correct

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