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

- anonymous

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- anonymous

... @ (0,1/2)

- anonymous

not that i tmatters anyway i'm just a little confused regarding the differentiation here

- anonymous

alright give me a second to remember lol

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## More answers

- anonymous

alright i'll upload a picture on what to do

- anonymous

and an example give me a sec

- anonymous

cool thanks

- anonymous

like i got as far as [4x+4y*y']/[4x+4y*y'-1] = 2x^2 + 2y^2 - x

- anonymous

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

- anonymous

by factor

- anonymous

right righ tok so and

- anonymous

so what do i do from what i have am i good so far

- anonymous

let me get a piece of paper lol

- anonymous

ama differentiate for u

- anonymous

thanks k cool coco

- anonymous

love it

- anonymous

so is coco checking if your differentiation is right?

- anonymous

one question. just verifying... for the secon part, everyting is raised to teh 2nd power?

- anonymous

i'm guessing

- anonymous

oh, ok

- anonymous

yes

- anonymous

ouch

- anonymous

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?

- anonymous

yeah i got completely something else than he did

- anonymous

the second is a chain and the first is just regular

- anonymous

i got 2x+2yy'=2(4x+4yy'-1)

- anonymous

=2x+2yy'= 8x+8yy'-2

- anonymous

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

- anonymous

yes

- anonymous

\[2yy'-8yy'=6x-2\]

- anonymous

ahh... i get it.. am so stupid! i was wondering where you got the chain rule from

- anonymous

yep

- anonymous

then you can pull out y'

- anonymous

i think it's\[\frac{-6x-2}{6y}\]

- anonymous

the - is outside parenthesis

- anonymous

\[-(6x-2)/6y\]

- anonymous

it's been a while though lol

- anonymous

blah wait i messed up lol

- anonymous

let me do this all over agani

- anonymous

i forgot something

- anonymous

it doesn't help that i'm also trying to help two people

- anonymous

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?

- anonymous

my chain rule was messed up lol it should be
\[2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)\]

- anonymous

yes
that's what i got

- anonymous

then what did you do

- anonymous

i'd say to combine the right side

- anonymous

it's going to be ugly hahah

- anonymous

dive 2xdx + 2ydy with 4xdx+4ydy...?

- anonymous

divide*

- anonymous

i think you'd have to combine the right side which is a pain

- anonymous

Hey, do you know is there anyway i can like mark this question so that i can come back and look @ it 2morrow?

- anonymous

no seriously?! ouch... calyne, your proff/teacher must be coldhearted to give you such a question.

- anonymous

um i think it should still be in the top left forner under notifications

- anonymous

it's not that bad tho

- anonymous

Oh, ok. thanks!

- anonymous

\[2x+2yy'=16x^3+16y^3y'+16x^2yy'+16y^2x-12x^2-4y^2-8xyy'+2x\]
\[2yy'=16x^3+16^3y'+16x^2yy'+16y^2x-12x^2-4y^2-8xyy'\]

- anonymous

...

- anonymous

did you get what me and coco got about my last when you used the chain rule

- anonymous

what

- anonymous

you use the chain rule yeah du^2/du * du/dx first thing overall ik

- anonymous

what you multiply that whole thing by dy/dx too

- anonymous

so whta did you get after you take dy/dx of both sides

- anonymous

what did you get damn

- anonymous

using the chain rule i got
2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)

- anonymous

sorry i got 2x+2yy' = 2(2x^2+2y^2-x) (4x + 4yy' - 1)

- anonymous

yeahhhh

- anonymous

alright so if you combine the terms on the right

- anonymous

is what i did ... what did you do after

- anonymous

idk that's gay

- anonymous

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.

- anonymous

not done yet haha

- anonymous

yeah i have no idea.. even wolfram alpha has omething like my end answer

- anonymous

unless your teacher is crazy and solved for like y lol

- anonymous

wait the answer to the differentiation is that or the tangent line

- anonymous

because if it's the tangent line my implicit differentiation may be correct

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