Implicit differentiation: x^3+y^3-6xy=0
Im a little confuse on the second part.
So, I took the derivative of all the terms, but I'm stuck at the product rule of (6xy):
3x^2+3y^2y'-(6xy'+6y)=0
-3x^2 -3x^2
+3y^2y'-(6xy'+6y)=-3x^2
Does that negative sign distrebute to (6x^2+6y), which make that plus sign negative? -
3y^2y'-6x^2-6y=-3x^2
Thus making me add it to both sides...

Question

Implicit differentiation: x^3+y^3-6xy=0 
Im a little confuse on the second part.
So, I took the derivative of all the terms, but I'm stuck at the product rule of (6xy):
3x^2+3y^2y'-(6xy'+6y)=0
-3x^2                               -3x^2
+3y^2y'-(6xy'+6y)=-3x^2
Does that negative sign distrebute to (6x^2+6y), which make that plus sign negative? ->
3y^2y'-6x^2-6y=-3x^2
Thus making me add it to both sides...

anonymous · Answer

Opps! -6xy'-6y not -6x^2-6y**

anonymous · Answer

distribute* lol

anonymous · Answer

Yes, you must share the negativity.

anonymous · Answer

You can think of factoring out the -6 then d/dx the (xy).

anonymous · Answer

Awesome! Thanks for the fast response!

anonymous · Answer

Thanks for the clear work that is easy to check!