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i don't understand how you split it up, i'm confused.... that honestly made no sense

yay now my head hurts even worse

How Much More Questions Do You Have ?

Or Is This The Last Question ?

this is my first question...

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

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

Well, (x^2)y + x(y^2) = 6
d/dx[(x^2)y + x(y^2)] = d/dx (6)

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?

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?

\[x ^{2}y+xy^2=6\]

x^2(1) + y(2x) = 2xy + x^2

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

\[\frac{ dy }{ dx }[(x^2)y + x(y^2)] =\frac{ dy }{ dx }6\]

Now Use Product Rule

\[2xy + dy/dx(x^2) + y^2 + 2y(dy/dx)x =0\]

Now Take The Terms Including dy/dx On One Side

\[dy/dx(x^2)+2y(dy/dx)x = -y^2-2xy\]

Now Take Common dy/dx

\[dy/dx(x^2+2xy)=-y^2-2xy\]

then divide?

So,\[\frac{ dy }{ dx }=\frac{ -(y^2+2xy)}{x^2+2xy }\]

^ 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?

Well Its Your Own Decision Whether to Put U first or V First If We talk about the formula

|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

Depends Completely On You

so dy/dx2y = (x+1)(1) - (x-1)(1) / (x-1)^2

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

I am Back

\[y^2 = (x-1) /(x+1)\]

differentiate both sides with respect to x

oh okay i was lost there

i see where shes at now

\[2y dy/dx = [(x+1)-(x-1)] /(x+1)^2 \]

\[2y dy/dx = 2/(x+1)^2\]

\[dy/dx = y / (x+1)^2\]

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?

thank you so much!