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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).
i don't understand how you split it up, i'm confused.... that honestly made no sense
yay now my head hurts even worse

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Other answers:

  • hba
How Much More Questions Do You Have ?
  • hba
Or Is This The Last Question ?
this is my first question...
  • hba
How Does Your Head Hurt ?
i've got about 16 more for this section, then 5 on a second section on it and a worksheet
i'm a junior in highschool taking calculus.. this isn't making sense
  • hba
Oh Wow Genius :D
obviously not if this isn't making sense. sorry it's just frustrating!
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
i'm lost stop
how'd you get 2xy + dy/dx(x^2 + y^2 + 2y(dydx)x=0?
  • hba
You dont know how to differentiate ?
I do, but it's confusing me
the d/dx of x^2y?
you use the multiplication rule right? udv + vdu?
  • hba
\[x ^{2}y+xy^2=6\]
x^2(1) + y(2x) = 2xy + x^2
  • hba
Yes I am Using Product Rule
then how did you get 2xy then dy/dx x^2?
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
then divide?
  • hba
So,\[\frac{ dy }{ dx }=\frac{ -(y^2+2xy)}{x^2+2xy }\]
  • hba
^ Now Thats Your Answer Got It :)
question, when using the product rule
how do you know which one to put dy/dx with
like when we had x^2y and got dy/dx(x^2) + 2xy
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
okay but then which one do i put the dy/dx with?
like the next problem is y2 = (x+1)/(x-1)
so would i do 2y (dy/dx) = vdu - udv/v^2
  • hba
Depends Completely On You
so dy/dx2y = (x+1)(1) - (x-1)(1) / (x-1)^2
  • hba
@Godsgirl Please Help Her Out I Gotta Leave- Thanks :)
Hey so wht do u need help with
I'm here. I have a question on where to put the dy/dx when solving
right now i'm working on y^2 = x-1/x+1
so i take dy/dx y^2 = (x+1)(1) - (x-1)(1) / (x+1)^2 right?
and the left side will equal 2y when i take dy/dx of it
now where do i keep dy/dx, when solving?
here i already solved it
where do i put it? because i know i need to get it alone
I sent it to u on ur messages
that's not the question i'm asking about..
the new one is
which one are u asking about then
y^2 = x-1/x+1
i'm not trying to come off as rude, i'm sorry!
just frustrated
  • hba
I am Back
  • hba
\[y^2 = (x-1) /(x+1)\]
  • hba
differentiate both sides with respect to x
oh okay i was lost there
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 :)
so you keep dy/dx with the y on the left
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 :)
thank you so much!

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