anonymous
  • anonymous
How can I implicitly differentiate the following... x^2 + y^2 = (2x^2 + 2y^2 - x)^2
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
UnkleRhaukus
  • UnkleRhaukus
with respect to x?
anonymous
  • anonymous
Yep!
anonymous
  • anonymous
I guess it would be chain rule on both sides but it gets a bit messy when I do it

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

UnkleRhaukus
  • 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
  • anonymous
I just corrected the question, the right side is to the power of 2
UnkleRhaukus
  • UnkleRhaukus
Implicitly differentiate the terms with \(y\), like this: \[\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\]
UnkleRhaukus
  • UnkleRhaukus
oh ok, what do you get for the left hand side?
anonymous
  • anonymous
I got 2x + 2yy'
UnkleRhaukus
  • UnkleRhaukus
good
UnkleRhaukus
  • UnkleRhaukus
now the righthand side \[=\frac d{dx}(2x^2+2y^2-x)^2\]
anonymous
  • anonymous
Would it be 2(2x^2 + 2y^2 -x)(4x +4yy' -1) ?
UnkleRhaukus
  • UnkleRhaukus
looks good!
anonymous
  • anonymous
Now the simplification XD
UnkleRhaukus
  • UnkleRhaukus
hmmm, i don't think it simplifies very far, implicit derivatives often don't simplify nice
UnkleRhaukus
  • UnkleRhaukus
What do you get?
amilapsn
  • amilapsn
please look @ my question after this @UnkleRhaukus
anonymous
  • 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
  • anonymous
@UnkleRhaukus is this what you get as well?
UnkleRhaukus
  • UnkleRhaukus
what happened to the equals sign?
anonymous
  • anonymous
What I'm trying to solve is actually this question...
anonymous
  • anonymous
Oh lol forgot that... y' = to all that ^
zepdrix
  • zepdrix
Ooo I think the same derivative :) I think you did that correctly
zepdrix
  • zepdrix
Gave yourself a little extra work maybe though.. hmm
zepdrix
  • zepdrix
I got the same derivative* woops typo
anonymous
  • anonymous
Oh that's a relief! Haha at least i got that right XD
UnkleRhaukus
  • UnkleRhaukus
... 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
  • anonymous
Well I only solved for y'
zepdrix
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
Alright so I'm gonna give that a try and tell you what I get :)
anonymous
  • anonymous
Please don't leave! :D
zepdrix
  • zepdrix
fine fine fine -_-
anonymous
  • anonymous
Lol so plugging in the x and y values I got -1... Tell me I'm not wrong :D
zepdrix
  • zepdrix
Let's look at the graph to sort of check our work easily.
zepdrix
  • zepdrix
|dw:1433147166496:dw|Here is the point (0,1/2)
zepdrix
  • zepdrix
Does the slope look like a -1? Hmmm not so much :[
anonymous
  • anonymous
Maybe its positive? haha
anonymous
  • anonymous
I don't know where i went wrong >.<
zepdrix
  • 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
  • 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
  • zepdrix
positive 1? Ooo that seems to match the picture a lot better also! :) yay
anonymous
  • anonymous
Haha awesome! XD
anonymous
  • 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
  • zepdrix
why -x? :o
anonymous
  • anonymous
Oh that was me going loco again :D haha sorry its y = x + 1/2
zepdrix
  • 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
  • anonymous
Look at this!!! Thanks soooo soo much!! I have a calculus midterm tomorrow, you have no idea how much I appreciate your help!!
zepdrix
  • zepdrix
yay team \c:/ np
anonymous
  • anonymous
Hahaha YaY :D

Looking for something else?

Not the answer you are looking for? Search for more explanations.