Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

  • 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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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
\[dy/dx(x^2+2xy)=-y^2-2xy\]
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
|dw:1350260697079:dw|
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
Hello?
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?
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!

Not the answer you are looking for?

Search for more explanations.

Ask your own question