Question

[high school calculus] implicit differentiation, finding value for dy/dx=0

Answer 1

You know that dy/dx=0 when y=-2x, ie, at points on a certain line. You also know that you're looking for points which are actually *on* the curve in question, ie, points where x^(2)+xy-y^(3)=0. So really, you're looking for intersection points between the line y=-2x and the original curve.

The reason you need to plug back into the original equation is that something like x^(2)+xy-y^(3)=377 would have the same implicit derivative, but would be a totally different curve, so the solution points will not be the same. So you somehow have to use the fact that the RHS was 0 instead of something else to find the solution.

Answer 2

Remember, when you type a fraction on one line, and the numerator or denominator has more than one term, wrap it in parentheses.

With that said, implicit differentiation results in this:

* 2x+y+xy′−3y^(2)y′=0, from which
 * y′=(2x+y)/(3y^(2)−x).

The point on the graph of x^(2)+xy−y^(3)=0 where dy/dx=0 is indeed where y=−2x (and also x≠3y^(2), be sure to check this later); this means that you're looking for a point that is

1. on the graph of x^(2)+xy−y^(3)=0 and
2. on the graph of y=−2x, and also
 * not on the graph of x=3y^(2).

If you're finding points on the intersection between two curves given as graphs of equations, and one equation is already solved for one of the variables, you can substitute in, as you've done since introductory algebra.

Answer 3

Your first equation says, "iff (x,y) is a point on the graph, then x² + xy  - y³ = 0 is true"

Your second equation says, "iff (x,y) is a point on the graph with rate of change dy/dx, then dy/dx = (2x+y)/(3y²-x) is true"

Your third equation says, "iff (x,y) is a point on the graph with rate of change 0, then y = -2x is true"

So you have a system of **two equations** concerning x and y. Plugging this into the first equation is the method of **substitution**, you're eliminating y from the equation so you can solve for x