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jamroz
Group Title
using implicit differentiation, find dy/dx of (x^2)y+x(y^2) = 6
 2 years ago
 2 years ago
jamroz Group Title
using implicit differentiation, find dy/dx of (x^2)y+x(y^2) = 6
 2 years ago
 2 years ago

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hba Group TitleBest ResponseYou've already chosen the best response.1
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).
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
i don't understand how you split it up, i'm confused.... that honestly made no sense
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
yay now my head hurts even worse
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
How Much More Questions Do You Have ?
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Or Is This The Last Question ?
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
this is my first question...
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
How Does Your Head Hurt ?
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
i've got about 16 more for this section, then 5 on a second section on it and a worksheet
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
i'm a junior in highschool taking calculus.. this isn't making sense
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Oh Wow Genius :D
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
obviously not if this isn't making sense. sorry it's just frustrating!
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
and it's not making a bit of sense
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Well, (x^2)y + x(y^2) = 6 d/dx[(x^2)y + x(y^2)] = d/dx (6)
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
i'm lost stop
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
how'd you get 2xy + dy/dx(x^2 + y^2 + 2y(dydx)x=0?
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
You dont know how to differentiate ?
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
I do, but it's confusing me
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
the d/dx of x^2y?
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
you use the multiplication rule right? udv + vdu?
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
\[x ^{2}y+xy^2=6\]
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
x^2(1) + y(2x) = 2xy + x^2
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Yes I am Using Product Rule
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
then how did you get 2xy then dy/dx x^2?
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
or do you get dy/dx and all that stuff in parenthasis because you're finding the d/dx of y
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ dy }{ dx }[(x^2)y + x(y^2)] =\frac{ dy }{ dx }6\]
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Now Use Product Rule
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
\[2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0\]
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Now Take The Terms Including dy/dx On One Side
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
\[dy/dx(x^2)+2y(dy/dx)x = y^22xy\]
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Now Take Common dy/dx
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
\[dy/dx(x^2+2xy)=y^22xy\]
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
then divide?
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
So,\[\frac{ dy }{ dx }=\frac{ (y^2+2xy)}{x^2+2xy }\]
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
^ Now Thats Your Answer Got It :)
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
question, when using the product rule
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
how do you know which one to put dy/dx with
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
like when we had x^2y and got dy/dx(x^2) + 2xy
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
why'd we put dy/dx with x^2 and not 2xy?
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Well Its Your Own Decision Whether to Put U first or V First If We talk about the formula
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
dw:1350260697079:dw
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
okay but then which one do i put the dy/dx with?
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
like the next problem is y2 = (x+1)/(x1)
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
so would i do 2y (dy/dx) = vdu  udv/v^2
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Depends Completely On You
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
so dy/dx2y = (x+1)(1)  (x1)(1) / (x1)^2
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
@Godsgirl Please Help Her Out I Gotta Leave Thanks :)
 2 years ago

Godsgirl Group TitleBest ResponseYou've already chosen the best response.0
Hey so wht do u need help with
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
I'm here. I have a question on where to put the dy/dx when solving
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
right now i'm working on y^2 = x1/x+1
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
so i take dy/dx y^2 = (x+1)(1)  (x1)(1) / (x+1)^2 right?
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
and the left side will equal 2y when i take dy/dx of it
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
now where do i keep dy/dx, when solving?
 2 years ago

Godsgirl Group TitleBest ResponseYou've already chosen the best response.0
here i already solved it
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
where do i put it? because i know i need to get it alone
 2 years ago

Godsgirl Group TitleBest ResponseYou've already chosen the best response.0
I sent it to u on ur messages
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
that's not the question i'm asking about..
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
the new one is
 2 years ago

Godsgirl Group TitleBest ResponseYou've already chosen the best response.0
which one are u asking about then
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
y^2 = x1/x+1
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
i'm not trying to come off as rude, i'm sorry!
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
just frustrated
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
\[y^2 = (x1) /(x+1)\]
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
differentiate both sides with respect to x
 2 years ago

Godsgirl Group TitleBest ResponseYou've already chosen the best response.0
oh okay i was lost there
 2 years ago

Godsgirl Group TitleBest ResponseYou've already chosen the best response.0
i see where shes at now
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
\[2y dy/dx = [(x+1)(x1)] /(x+1)^2 \]
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
\[2y dy/dx = 2/(x+1)^2\]
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
\[dy/dx = y / (x+1)^2\]
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
@jamroz Got It :)
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
so you keep dy/dx with the y on the left
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
but why? cause i had gotten that far, i just didn't know which side to leave dy/dx with?
 2 years ago

hba Group TitleBest ResponseYou've already chosen the best response.1
Now You Got Which Side You Had To Keep It :) Anyways I am Off To Sleep, Aww My Head Hurts Anyways Bye :)
 2 years ago

jamroz Group TitleBest ResponseYou've already chosen the best response.0
thank you so much!
 2 years ago
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