anonymous
  • anonymous
using implicit differentiation, find dy/dx of (x^2)y+x(y^2) = 6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
hba
  • 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
  • anonymous
i don't understand how you split it up, i'm confused.... that honestly made no sense
anonymous
  • anonymous
yay now my head hurts even worse

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hba
  • hba
How Much More Questions Do You Have ?
hba
  • hba
Or Is This The Last Question ?
anonymous
  • anonymous
this is my first question...
hba
  • hba
How Does Your Head Hurt ?
anonymous
  • anonymous
i've got about 16 more for this section, then 5 on a second section on it and a worksheet
anonymous
  • anonymous
i'm a junior in highschool taking calculus.. this isn't making sense
hba
  • hba
Oh Wow Genius :D
anonymous
  • anonymous
obviously not if this isn't making sense. sorry it's just frustrating!
anonymous
  • anonymous
and it's not making a bit of sense
hba
  • hba
Well, (x^2)y + x(y^2) = 6 d/dx[(x^2)y + x(y^2)] = d/dx (6)
hba
  • hba
2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0
anonymous
  • anonymous
i'm lost stop
anonymous
  • anonymous
how'd you get 2xy + dy/dx(x^2 + y^2 + 2y(dydx)x=0?
hba
  • hba
You dont know how to differentiate ?
anonymous
  • anonymous
I do, but it's confusing me
anonymous
  • anonymous
the d/dx of x^2y?
anonymous
  • anonymous
you use the multiplication rule right? udv + vdu?
hba
  • hba
\[x ^{2}y+xy^2=6\]
anonymous
  • anonymous
x^2(1) + y(2x) = 2xy + x^2
hba
  • hba
Yes I am Using Product Rule
anonymous
  • anonymous
then how did you get 2xy then dy/dx x^2?
anonymous
  • anonymous
or do you get dy/dx and all that stuff in parenthasis because you're finding the d/dx of y
hba
  • hba
\[\frac{ dy }{ dx }[(x^2)y + x(y^2)] =\frac{ dy }{ dx }6\]
hba
  • hba
Now Use Product Rule
hba
  • hba
\[2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0\]
hba
  • hba
Now Take The Terms Including dy/dx On One Side
hba
  • hba
\[dy/dx(x^2)+2y(dy/dx)x = -y^2-2xy\]
hba
  • hba
Now Take Common dy/dx
hba
  • hba
\[dy/dx(x^2+2xy)=-y^2-2xy\]
anonymous
  • anonymous
then divide?
hba
  • hba
So,\[\frac{ dy }{ dx }=\frac{ -(y^2+2xy)}{x^2+2xy }\]
hba
  • hba
^ Now Thats Your Answer Got It :)
anonymous
  • anonymous
question, when using the product rule
anonymous
  • anonymous
how do you know which one to put dy/dx with
anonymous
  • anonymous
like when we had x^2y and got dy/dx(x^2) + 2xy
anonymous
  • anonymous
why'd we put dy/dx with x^2 and not 2xy?
hba
  • hba
Well Its Your Own Decision Whether to Put U first or V First If We talk about the formula
hba
  • hba
|dw:1350260697079:dw|
anonymous
  • anonymous
okay but then which one do i put the dy/dx with?
anonymous
  • anonymous
like the next problem is y2 = (x+1)/(x-1)
anonymous
  • anonymous
so would i do 2y (dy/dx) = vdu - udv/v^2
hba
  • hba
Depends Completely On You
anonymous
  • anonymous
so dy/dx2y = (x+1)(1) - (x-1)(1) / (x-1)^2
hba
  • hba
@Godsgirl Please Help Her Out I Gotta Leave- Thanks :)
anonymous
  • anonymous
Hey so wht do u need help with
anonymous
  • anonymous
Hello?
anonymous
  • anonymous
I'm here. I have a question on where to put the dy/dx when solving
anonymous
  • anonymous
right now i'm working on y^2 = x-1/x+1
anonymous
  • anonymous
so i take dy/dx y^2 = (x+1)(1) - (x-1)(1) / (x+1)^2 right?
anonymous
  • anonymous
right
anonymous
  • anonymous
and the left side will equal 2y when i take dy/dx of it
anonymous
  • anonymous
now where do i keep dy/dx, when solving?
anonymous
  • anonymous
here i already solved it
anonymous
  • anonymous
where do i put it? because i know i need to get it alone
anonymous
  • anonymous
I sent it to u on ur messages
anonymous
  • anonymous
that's not the question i'm asking about..
anonymous
  • anonymous
the new one is
anonymous
  • anonymous
which one are u asking about then
anonymous
  • anonymous
y^2 = x-1/x+1
anonymous
  • anonymous
i'm not trying to come off as rude, i'm sorry!
anonymous
  • anonymous
just frustrated
hba
  • hba
I am Back
hba
  • hba
\[y^2 = (x-1) /(x+1)\]
hba
  • hba
differentiate both sides with respect to x
anonymous
  • anonymous
oh okay i was lost there
anonymous
  • anonymous
i see where shes at now
hba
  • hba
\[2y dy/dx = [(x+1)-(x-1)] /(x+1)^2 \]
hba
  • hba
\[2y dy/dx = 2/(x+1)^2\]
hba
  • hba
\[dy/dx = y / (x+1)^2\]
hba
  • hba
@jamroz Got It :)
anonymous
  • anonymous
so you keep dy/dx with the y on the left
anonymous
  • anonymous
but why? cause i had gotten that far, i just didn't know which side to leave dy/dx with?
hba
  • hba
Now You Got Which Side You Had To Keep It :) Anyways I am Off To Sleep, Aww My Head Hurts Anyways Bye :)
anonymous
  • anonymous
thank you so much!

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