Question

use implicit differentiation to find dx/dy.
x^4y^2 -x^3y +2xy^3=0

Answer 1

dx/dy, not dy/dx?

Answer 2

yes

Answer 3

Alrighty then. Well, I'm going to take the derivative of everything like normal, using product rule or whatever I would normally need to do. But in order to do this properly, I need to keep track of when I actually took the derivative of y and when I took the derivative of x. Now some peopel think the dy/dx or d/dx notation is confusing, so I'm just going to mark each time I take the derivative of x and go from there.

Answer 4

\[x ^{4}y ^{2}-x ^{3}y+2xy ^{3}= 0\]
So product rule with the first term now.
\[(4x ^{3}y ^{2})+2x ^{4}\] After the first product rule I have this. I marked when I took the derivative of x by putting it in parenthesis. Now for the second term:
\[-(3x ^{2}y)-x ^{3}\] Again, I mark the term that contains my x-derivative. Last term:
\[(2y ^{3})+6xy ^{2}\]
Putting everything together I have:
\[(4x ^{3}y ^{2})+2x ^{4}-(3x ^{2}y)-x ^{3}+(2y ^{3})+6xy ^{2}=0\]
So now what I need to do is separate my x-derivatives (the parenthesis terms) and my y-derivatives by having them on the opposite side of the equal sign. Now since we want dx/dy, after we move everything to opposite sides, we will divide both sides by all of the x-derivatives. So basically, dx/dy will equal the y-derivatives divided by the x-derivatives. Now if we had to solve for dy/dx, we would do the same thing but divide both sides by all the y-derivatives, having dy/dx equal the x-derivatives divided by the y-derivatives.
Okay, I know this is the great wall of china of explanations, but hopefully it makes sense. So now I leave it to you to separate the x-derivatives from the y-derivatives and then divide both sides by all the x-derivatives. Please let me know if anything is kind of confusing about what I did or said x_x

Answer 5

Parenthesis is the way you marked where your dx/dy terms are?
That's .... weird. :\