Question

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

Answer 1

with respect to x?

Answer 2

Yep!

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

oh ok,

what do you get for the left hand side?

Answer 8

I got 2x + 2yy'

Answer 9

good

Answer 10

now the righthand side

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

Answer 11

Would it be

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

Answer 12

looks good!

Answer 13

Now the simplification XD

Answer 14

hmmm, i don't think it simplifies very far,

implicit derivatives often don't simplify nice

Answer 15

What do you get?

Answer 16

please look @ my question after this @UnkleRhaukus

Answer 17

This is what I get at the end ...

\[\frac{ 2(4x^3 - 3x^2 +4xy^2 - y^2) }{ y(1-8x^2 - 8y^2 +4x )}\]

Answer 18

@UnkleRhaukus is this what you get as well?

Answer 19

what happened to the equals sign?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

Gave yourself a little extra work maybe though.. hmm

Answer 24

I got the same derivative* woops typo

Answer 25

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

Answer 26

...

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?

Answer 27

Well I only solved for y'

Answer 28

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.

Answer 29

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?

Answer 30

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.

Answer 31

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.

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

Please don't leave! :D

Answer 36

fine fine fine -_-

Answer 37

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

Answer 38

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

Answer 39

Here is the point (0,1/2)