using implicit differentiation, find dy/dx of (x^2)y+x(y^2) = 6

- anonymous

using implicit differentiation, find dy/dx of (x^2)y+x(y^2) = 6

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- hba

The easiest way to do this is to use implicit differentiation on it as it stands.
You differentiate using the normal rules but you treat y as an unknown function of x. This means f(y) differentates to f '(y)*(dy/dx) by the chain rule.
xy + x + y^2 = 6
First implicit differentiation gives
[1*y + x*1*(dy/dx)] + 1 + 2y*(dy/dx) = 0
.....[from xy].....................[from y^2]
You can rearrange this to give dy/dx = (function of x and y)
Second implicit differentiation gives
dy/dx + [1*(dy/dx) + x*(d2y/dx2)] + [2*(dy/dx)*(dy/dx) + 2y*(d2y/dx2)] = 0
...................[from x*(dy/dx)]......................[from 2y*(dy/dx)]
Now you replace each dy/dx by its function of x and y found above.
Then rearrange it to give d2y/dx2 = (function of x and y).

- anonymous

i don't understand how you split it up, i'm confused.... that honestly made no sense

- anonymous

yay now my head hurts even worse

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## More answers

- hba

How Much More Questions Do You Have ?

- hba

Or Is This The Last Question ?

- anonymous

this is my first question...

- hba

How Does Your Head Hurt ?

- anonymous

i've got about 16 more for this section, then 5 on a second section on it
and a worksheet

- anonymous

i'm a junior in highschool taking calculus.. this isn't making sense

- hba

Oh Wow Genius :D

- anonymous

obviously not if this isn't making sense. sorry it's just frustrating!

- anonymous

and it's not making a bit of sense

- hba

Well, (x^2)y + x(y^2) = 6
d/dx[(x^2)y + x(y^2)] = d/dx (6)

- hba

2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0

- anonymous

i'm lost stop

- anonymous

how'd you get 2xy + dy/dx(x^2 + y^2 + 2y(dydx)x=0?

- hba

You dont know how to differentiate ?

- anonymous

I do, but it's confusing me

- anonymous

the d/dx of x^2y?

- anonymous

you use the multiplication rule right? udv + vdu?

- hba

\[x ^{2}y+xy^2=6\]

- anonymous

x^2(1) + y(2x) = 2xy + x^2

- hba

Yes I am Using Product Rule

- anonymous

then how did you get 2xy then dy/dx x^2?

- anonymous

or do you get dy/dx and all that stuff in parenthasis because you're finding the d/dx of y

- hba

\[\frac{ dy }{ dx }[(x^2)y + x(y^2)] =\frac{ dy }{ dx }6\]

- hba

Now Use Product Rule

- hba

\[2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0\]

- hba

Now Take The Terms Including dy/dx On One Side

- hba

\[dy/dx(x^2)+2y(dy/dx)x = -y^2-2xy\]

- hba

Now Take Common dy/dx

- hba

\[dy/dx(x^2+2xy)=-y^2-2xy\]

- anonymous

then divide?

- hba

So,\[\frac{ dy }{ dx }=\frac{ -(y^2+2xy)}{x^2+2xy }\]

- hba

^ Now Thats Your Answer Got It :)

- anonymous

question, when using the product rule

- anonymous

how do you know which one to put dy/dx with

- anonymous

like when we had x^2y and got dy/dx(x^2) + 2xy

- anonymous

why'd we put dy/dx with x^2 and not 2xy?

- hba

Well Its Your Own Decision Whether to Put U first or V First If We talk about the formula

- hba

|dw:1350260697079:dw|

- anonymous

okay but then which one do i put the dy/dx with?

- anonymous

like the next problem is y2 = (x+1)/(x-1)

- anonymous

so would i do 2y (dy/dx) = vdu - udv/v^2

- hba

Depends Completely On You

- anonymous

so dy/dx2y = (x+1)(1) - (x-1)(1) / (x-1)^2

- hba

@Godsgirl Please Help Her Out I Gotta Leave- Thanks :)

- anonymous

Hey so wht do u need help with

- anonymous

Hello?

- anonymous

I'm here. I have a question on where to put the dy/dx when solving

- anonymous

right now i'm working on y^2 = x-1/x+1

- anonymous

so i take dy/dx y^2 = (x+1)(1) - (x-1)(1) / (x+1)^2 right?

- anonymous

right

- anonymous

and the left side will equal 2y when i take dy/dx of it

- anonymous

now where do i keep dy/dx, when solving?

- anonymous

here i already solved it

- anonymous

where do i put it? because i know i need to get it alone

- anonymous

I sent it to u on ur messages

- anonymous

that's not the question i'm asking about..

- anonymous

the new one is

- anonymous

which one are u asking about then

- anonymous

y^2 = x-1/x+1

- anonymous

i'm not trying to come off as rude, i'm sorry!

- anonymous

just frustrated

- hba

I am Back

- hba

\[y^2 = (x-1) /(x+1)\]

- hba

differentiate both sides with respect to x

- anonymous

oh okay i was lost there

- anonymous

i see where shes at now

- hba

\[2y dy/dx = [(x+1)-(x-1)] /(x+1)^2 \]

- hba

\[2y dy/dx = 2/(x+1)^2\]

- hba

\[dy/dx = y / (x+1)^2\]

- hba

@jamroz Got It :)

- anonymous

so you keep dy/dx with the y on the left

- anonymous

but why? cause i had gotten that far, i just didn't know which side to leave dy/dx with?

- hba

Now You Got Which Side You Had To Keep It :) Anyways I am Off To Sleep, Aww My Head Hurts Anyways Bye :)

- anonymous

thank you so much!

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