jamroz
using implicit differentiation, find dy/dx of (x^2)y+x(y^2) = 6
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hba
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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).
jamroz
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i don't understand how you split it up, i'm confused.... that honestly made no sense
jamroz
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yay now my head hurts even worse
hba
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How Much More Questions Do You Have ?
hba
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Or Is This The Last Question ?
jamroz
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this is my first question...
hba
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How Does Your Head Hurt ?
jamroz
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i've got about 16 more for this section, then 5 on a second section on it
and a worksheet
jamroz
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i'm a junior in highschool taking calculus.. this isn't making sense
hba
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Oh Wow Genius :D
jamroz
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obviously not if this isn't making sense. sorry it's just frustrating!
jamroz
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and it's not making a bit of sense
hba
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Well, (x^2)y + x(y^2) = 6
d/dx[(x^2)y + x(y^2)] = d/dx (6)
hba
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2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0
jamroz
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i'm lost stop
jamroz
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how'd you get 2xy + dy/dx(x^2 + y^2 + 2y(dydx)x=0?
hba
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You dont know how to differentiate ?
jamroz
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I do, but it's confusing me
jamroz
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the d/dx of x^2y?
jamroz
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you use the multiplication rule right? udv + vdu?
hba
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\[x ^{2}y+xy^2=6\]
jamroz
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x^2(1) + y(2x) = 2xy + x^2
hba
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Yes I am Using Product Rule
jamroz
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then how did you get 2xy then dy/dx x^2?
jamroz
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or do you get dy/dx and all that stuff in parenthasis because you're finding the d/dx of y
hba
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\[\frac{ dy }{ dx }[(x^2)y + x(y^2)] =\frac{ dy }{ dx }6\]
hba
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Now Use Product Rule
hba
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\[2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0\]
hba
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Now Take The Terms Including dy/dx On One Side
hba
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\[dy/dx(x^2)+2y(dy/dx)x = -y^2-2xy\]
hba
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Now Take Common dy/dx
hba
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\[dy/dx(x^2+2xy)=-y^2-2xy\]
jamroz
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then divide?
hba
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So,\[\frac{ dy }{ dx }=\frac{ -(y^2+2xy)}{x^2+2xy }\]
hba
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^ Now Thats Your Answer Got It :)
jamroz
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question, when using the product rule
jamroz
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how do you know which one to put dy/dx with
jamroz
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like when we had x^2y and got dy/dx(x^2) + 2xy
jamroz
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why'd we put dy/dx with x^2 and not 2xy?
hba
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Well Its Your Own Decision Whether to Put U first or V First If We talk about the formula
hba
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|dw:1350260697079:dw|
jamroz
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okay but then which one do i put the dy/dx with?
jamroz
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like the next problem is y2 = (x+1)/(x-1)
jamroz
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so would i do 2y (dy/dx) = vdu - udv/v^2
hba
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Depends Completely On You
jamroz
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so dy/dx2y = (x+1)(1) - (x-1)(1) / (x-1)^2
hba
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@Godsgirl Please Help Her Out I Gotta Leave- Thanks :)
Godsgirl
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Hey so wht do u need help with
Godsgirl
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Hello?
jamroz
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I'm here. I have a question on where to put the dy/dx when solving
jamroz
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right now i'm working on y^2 = x-1/x+1
jamroz
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so i take dy/dx y^2 = (x+1)(1) - (x-1)(1) / (x+1)^2 right?
Godsgirl
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right
jamroz
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and the left side will equal 2y when i take dy/dx of it
jamroz
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now where do i keep dy/dx, when solving?
Godsgirl
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here i already solved it
jamroz
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where do i put it? because i know i need to get it alone
Godsgirl
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I sent it to u on ur messages
jamroz
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that's not the question i'm asking about..
jamroz
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the new one is
Godsgirl
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which one are u asking about then
jamroz
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y^2 = x-1/x+1
jamroz
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i'm not trying to come off as rude, i'm sorry!
jamroz
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just frustrated
hba
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I am Back
hba
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\[y^2 = (x-1) /(x+1)\]
hba
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differentiate both sides with respect to x
Godsgirl
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oh okay i was lost there
Godsgirl
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i see where shes at now
hba
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\[2y dy/dx = [(x+1)-(x-1)] /(x+1)^2 \]
hba
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\[2y dy/dx = 2/(x+1)^2\]
hba
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\[dy/dx = y / (x+1)^2\]
hba
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@jamroz Got It :)
jamroz
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so you keep dy/dx with the y on the left
jamroz
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but why? cause i had gotten that far, i just didn't know which side to leave dy/dx with?
hba
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Now You Got Which Side You Had To Keep It :) Anyways I am Off To Sleep, Aww My Head Hurts Anyways Bye :)
jamroz
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thank you so much!