Implicit differentiation help please 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!
 A: 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.
A: 
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
$\begin{align}x^2y + xy^2 & = 6
\\[1ex]
 \frac{\mathrm d x^2}{\mathrm d x} y + x^2 \frac{\mathrm dy}{\mathrm dx}+ \frac{\mathrm dx}{\mathrm dx}y^2 + x\frac{\mathrm dy^2}{\mathrm dx} & = \frac{\mathrm d 6}{\mathrm d x} 
\\[1ex]
 2x\, y + x^2\, \frac{\mathrm dy}{\mathrm dx} + y^2 + 2x\,y\,\frac{\mathrm dy}{\mathrm dx} & = 0
\\[1ex]
(x^2+2xy)\frac{\mathrm dy}{\mathrm dx} & = -(2xy+y^2)
\\[1ex]
\frac{\mathrm dy}{\mathrm dx} & = -\frac{2xy+y^2}{x^2+2xy}
\end{align}$
A: Hints:
$$2xy+x^2\cdot y'+y^2+2xy\cdot y'=0$$
Then express $y'$ by $x$ and $y$ 
May it helps :)
