Question

find dy/dx.
x^2y+xy^2=6

Answer 1

Really, these are just like any other differentiation problem you may have done before, but now we have to actually organize. You're supposed to take the derivative of everything just like you normally would, but everytime you take the derivative of a y variable you want to mark it. Now some mark it by putting y' next to it, dy/dx next to it, either way, as long as you mark it and know where it is. I'll show ya what I mean, derivative now

So x^2y is a product rule and becomes
2xy + x^2!!

The exclamation points are there because that is a term in which I took the derivative of a y-variable. All I'm doing is marking it so I can keep track of which terms had y-derivatives and which terms did not. So next is xy^2 this becomes

y^2 + 2xy!!

Again, I'm marking the tern where I took a y-derivative. And of course that constant you have goes away in differentiation, giving us:

2xy +x^2!! + y^2 + 2xy!! = 0

Now what I want to do is have every term I marked on one side of the equation and everything else on the other. So that's what I'll do.

x^2!! + 2xy!! = -2xy -y^2

Once you get to this point youre basically done. ALl the terms with y-derivatives on one side and everythign else on the other. The final step is divide both sides by all the y-derivative terms. Technically, those exclamation points reprsent the dy/dx youre solving for, so if you divide both sides by the two terms, you'll isolate dy/dx. SO the final answer is

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

Now of course I just do the exclamation points to avoid clutter. If you like it better visually, we can do the whole problem but with dy/dx as a marker instead (just real quick)

x^2y + xy^2 = 6

\[2xy+x^{2}*\frac{ dy }{ dx }+y^{2} + 2xy*\frac{ dy }{ dx }=0\]
\[x^{2}*\frac{ dy }{ dx }+2xy*\frac{ dy }{ dx }= - (2xy+y^{2})\]
\[\frac{ dy }{ dx }(x^{2} + 2xy) = -(2xy+y^{2})\]
\[\frac{ dy }{ dx }=\frac{ -(2xy+y^{2}) }{ x^{2}+2xy }\]

Same exact thing, but I put dy/dx everytime. Its up to you if you want to write it everytime or maybe just checkmark each y-derivative. The process is the same no matter what, find the terms with y-derivatives and have them on opposite sides of the ewuation as everything else. THen at the end divide both sides by sll the y-derivatives.

Answer 2

ok thanks...but how would i do it if they gave me x=tany

Answer 3

x = tany, derivative is
1 = sec^2(y)!!

Well, now you already have y-derivatives on the other side from everything else, so now just divide by all the y-derivatives

\[\frac{ dy }{ dx }= \frac{ 1 }{ \sec^{2}y }\implies \frac{ dy }{ dx }= \cos^{2}y\]

Answer 4

ohhhh ok....thanks i think i understand now

Answer 5

Alright, cool :3