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

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

with respect to x?
Yep!
I guess it would be chain rule on both sides but it gets a bit messy when I do it

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[x^2 + y^2 = 2x^2 + 2y^2 - x\\ \frac{d}{dx}\left(x^2 + y^2\right) = \frac{d}{dx}\left(2x^2 + 2y^2 - x\right)\\ \frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(2x^2)+\frac{d}{dx}(2y^2)-\frac{d}{dx}(x)\]
I just corrected the question, the right side is to the power of 2
Implicitly differentiate the terms with \(y\), like this: \[\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\]
oh ok, what do you get for the left hand side?
I got 2x + 2yy'
good
now the righthand side \[=\frac d{dx}(2x^2+2y^2-x)^2\]
Would it be 2(2x^2 + 2y^2 -x)(4x +4yy' -1) ?
looks good!
Now the simplification XD
hmmm, i don't think it simplifies very far, implicit derivatives often don't simplify nice
What do you get?
please look @ my question after this @UnkleRhaukus
This is what I get at the end ... \[\frac{ 2(4x^3 - 3x^2 +4xy^2 - y^2) }{ y(1-8x^2 - 8y^2 +4x )}\]
@UnkleRhaukus is this what you get as well?
what happened to the equals sign?
What I'm trying to solve is actually this question...
Oh lol forgot that... y' = to all that ^
Ooo I think the same derivative :) I think you did that correctly
Gave yourself a little extra work maybe though.. hmm
I got the same derivative* woops typo
Oh that's a relief! Haha at least i got that right XD
... 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?
Well I only solved for y'
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.
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?
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.
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.
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
\[\Large\rm y-\frac{1}{2}=m(x-0)\]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.
Alright so I'm gonna give that a try and tell you what I get :)
Please don't leave! :D
fine fine fine -_-
Lol so plugging in the x and y values I got -1... Tell me I'm not wrong :D
Let's look at the graph to sort of check our work easily.
|dw:1433147166496:dw|Here is the point (0,1/2)
Does the slope look like a -1? Hmmm not so much :[
Maybe its positive? haha
I don't know where i went wrong >.<
\[\Large\rm 2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)\]Plugging in our stuff gives us,\[\Large\rm 0+y'=2\left(0+\frac{1}{2}-0\right)(0+2y'-1)\]Ya? :d
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
positive 1? Ooo that seems to match the picture a lot better also! :) yay
Haha awesome! XD
so then my point-slope formula would be y - .5 = 1(x -0) which then simplifies to y = -x + 1/2 right??? :D
why -x? :o
Oh that was me going loco again :D haha sorry its y = x + 1/2
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
Look at this!!! Thanks soooo soo much!! I have a calculus midterm tomorrow, you have no idea how much I appreciate your help!!
yay team \c:/ np
Hahaha YaY :D

Not the answer you are looking for?

Search for more explanations.

Ask your own question