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So I do not understand the concept at all. Could somebody explain it to me, in very dumbed down terms, using the problem $x^2y + xy^2 = 6$?

All I have down is that it is equal to $4xy + x^2 + y^2 = 6$, and even that could be wrong. Help!

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So I do not understand the concept at all. Could somebody explain it to me, in very dumbed down terms, using the problem $x^2y + xy^2 = 6$?

All I have down is that it is equal to $4xy + x^2 + y^2 = 6$, and even that could be wrong. Help!

Yeap, it is. You're on the right track, applying the product rule, but forgot that you are differentiating with respect to $x$; so you need to apply the chain rule also. Also you must differentiate both sides

$\begin{align}x^2y + xy^2 & = 6 \\[1ex] \frac{\mathrm d x^2}{\mathrm d x} y + x^2 \frac{\mathrm dy}{\mathrm dx}+ \frac{\mathrm dx}{\mathrm dx}y^2 + x\frac{\mathrm dy^2}{\mathrm dx} & = \frac{\mathrm d 6}{\mathrm d x} \\[1ex] 2x\, y + x^2\, \frac{\mathrm dy}{\mathrm dx} + y^2 + 2x\,y\,\frac{\mathrm dy}{\mathrm dx} & = 0 \\[1ex] (x^2+2xy)\frac{\mathrm dy}{\mathrm dx} & = -(2xy+y^2) \\[1ex] \frac{\mathrm dy}{\mathrm dx} & = -\frac{2xy+y^2}{x^2+2xy} \end{align}$

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  • $\begingroup$ Thank you so much, after staring at it for ten minutes or so I understand it! Thanks! $\endgroup$ – Ethan Oct 16 '14 at 0:58
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It would actually depends on whether you are differentiating with respect to x or with respect to y ( d/dx or d/dy ). If you are differentiating with respect to x, then you would have 4xy (dy/dx) + 2x + 2y (dy/dx) = 0. Basically every time you see a y, take the derivative of if and put in dy/dx. Then solve for dy/dx to get the final answer.

Same goes with differentiating with respect to y. Every time you see a x, take the derivative and put dx/dy then solve for dx/dy. So the equation becomes: 4xy (dx/dy) + 2x (dx/dy) + 2y = 0.

Oh and one more thing, a derivative of a constant is zero so you would not have the 6 after you take the derivative.

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  • $\begingroup$ All my book says is "find dy/dx". What do you mean every time I see a y I take the derivative of it and put it in dy/dx? $\endgroup$ – Ethan Oct 16 '14 at 0:37
  • $\begingroup$ So every time you see a y, for example y^2, you would take the derivative of it, which is 2y, then you have to multiply it with dy/dx. Recall the chain rule, when you do D(y^2) it would equal to 2y D(y), which is 2y (dy/dx). $\endgroup$ – bodygued Oct 16 '14 at 0:40
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Hints:

$$2xy+x^2\cdot y'+y^2+2xy\cdot y'=0$$

Then express $y'$ by $x$ and $y$

May it helps :)

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