Author Topic: implicit differentiation (Read 1580 times) Tweet Share

Pixon · « **Reply #1 on:** January 24, 2011, 10:14:37 pm »

omg I can't believe how frustrating it is to use latex...(now i have to check super duper hard to see if I've made a mistake...which is very likely in this mess)

$x^3-y^3+2x^2+2y^2-2x+2y=4+x^2y-xy^2$

$3x^2-3y^2* \frac{dy}{dx} +4x+4y* \frac{dy}{dx} -2+2* \frac{dy}{dx}=2xy+x^2* \frac{dy}{dx}-y^2-2yx* \frac{dy}{dx}$

$(-3y^2+4y+2-x^2+2yx)* \frac{dy}{dx} =2xy-y^2-3x^2-4x+2$

$\frac{dy}{dx}= \frac{2xy-y^2-3x^2-4x+2}{2xy-3y^2-x^2+4y+2}$

at x=-1

$\frac{dy}{dx}= \frac{(-1)2y-y^2-3(-1)^2-4(-1)+2}{2(-1)y-3y^2-(-1)^2+4y+2}$

$\frac{dy}{dx}= \frac{-2y-y^2-3+4+2}{-2y-3y^2-1+4y+2}$

$\frac{dy}{dx}= \frac{-2y-y^2+3}{2y-3y^2+1}$

at $x=-1, y=-1$

y=-1
$\frac{dy}{dx}=-1$

evaever · « **Reply #2 on:** January 24, 2011, 10:35:36 pm »

I believe you have made mistake(s). I noticed you did not copy the question correctly.

Pixon · « **Reply #3 on:** January 24, 2011, 10:43:22 pm »

lol of course I did that...wait a second

Alright I obviously didn't redo the whole question...if it's still wrong I might just crawl back into my hole and let someone more confident do it. lol

evaever · « **Reply #4 on:** January 24, 2011, 10:54:04 pm »

your dy/dx in terms of y is wrong

Pixon · « **Reply #5 on:** January 24, 2011, 10:57:42 pm »

sigh...lol i'll have one more look...you may not realise how hard it is to follow lines in latex...plus I've only just started using it
no matter...i'll do it properly by hand now then swap things around
Can I ask what the answer is so i don't post another wrong answer?

evaever · « **Reply #6 on:** January 24, 2011, 11:00:02 pm »

The numerical answer(s) are not really important.

Pixon · « **Reply #7 on:** January 24, 2011, 11:06:31 pm »

Well I don't see the mistake now. Show me my mistake...I'm curious.
Besides...I wanted the numerical answer to check my dy/dx.

Andiio · « **Reply #8 on:** January 24, 2011, 11:09:29 pm »

Quote from: Pixon on January 24, 2011, 11:06:31 pm

Well I don't see the mistake now. Show me my mistake...I'm curious.
Besides...I wanted the numerical answer to check my dy/dx.

Pixon I commend you for typing all of that up in latex LOL

evaever · « **Reply #9 on:** January 24, 2011, 11:11:10 pm »

Quote from: Pixon on January 24, 2011, 10:14:37 pm

omg I can't believe how frustrating it is to use latex...(now i have to check super duper hard to see if I've made a mistake...which is very likely in this mess)

$x^3-y^3+2x^2+2y^2-2x+2y=4+x^2y-xy^2$

$3x^2-3y^2* \frac{dy}{dx} +4x+4y* \frac{dy}{dx} -2+2* \frac{dy}{dx}=2xy+x^2* \frac{dy}{dx}-y^2-2yx* \frac{dy}{dx}$

$(-3y^2+4y+2-x^2+2yx)* \frac{dy}{dx} =2xy-y^2-3x^2-4x+2$

$\frac{dy}{dx}= \frac{2xy-y^2-3x^2-4x+2}{2xy-3y^2-x^2+4y+2}$

at x=-1

$\frac{dy}{dx}= \frac{(-1)2y-y^2-3(-1)^2-4(-1)+2}{2(-1)y-3y^2-(-1)^2+4y+2}$

$\frac{dy}{dx}= \frac{-2y-y^2-3+4+2}{-2y-3y^2-1+4y+2}$

$\frac{dy}{dx}= \frac{-2y-y^2+3}{4y-3y^2+1}$

at $x=-1, y=\pm 1\$

y=1
$Undefined$

y=-1
$\frac{dy}{dx}=\frac{-2}{3}$

$\frac{dy}{dx}= \frac{-2y-y^2+3}{2y-3y^2+1}$

Pixon · « **Reply #10 on:** January 24, 2011, 11:15:49 pm »

Ahh crud...that would have come from changing the signs in the first line when i copied it wrong. I don't really understand why you're posting a question you can do...but okay then.

Edit...fixed! lol

Romperait · « **Reply #11 on:** January 24, 2011, 11:18:07 pm »

Huge respect for attempting that with latex.

Edit: OP should be grateful, there aren't many people on this forum with the skill/will to do a question like that.

evaever · « **Reply #12 on:** January 24, 2011, 11:25:08 pm »

Quote from: Pixon on January 24, 2011, 11:15:49 pm

Ahh crud...that would have come from changing the signs in the first line when i copied it wrong. I don't really understand why you're posting a question you can do...but okay then.

Edit...fixed! lol

This is a question with an interesting feature that you missed.
Why did you say undefined at (-1,1)?

evaever · « **Reply #13 on:** January 24, 2011, 11:36:38 pm »

The graph of the relation is interesting too.

Pixon · « **Reply #14 on:** January 24, 2011, 11:46:04 pm »

I've seen a similar equation before...I know what it looks like too...I just missed it in all the latex.
I guess my mistake is in $\frac{dy}{dx}= \frac{2xy-y^2-3x^2-4x+2}{2xy-3y^2-x^2+4y+2}$
It simply shows that the point (-1,1) doesn't exist at all. I used the wrong term for 0/0 calling it undefined...and I know seeing as the limits would come to 0/0 anyways.

In further lines I would sub in y=1, it comes up with 0/0 and yeh...with limits it's almost always possible to take anything and make the limit 0/0.

BUT...I've probably missed whatever it is you're talking about anyways. So meh.

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Author Topic: implicit differentiation (Read 1580 times) Tweet Share

evaever

implicit differentiation

Pixon

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