Implicit differentiation: under what conditions can implicit differentiation not be used? is there a way too tell before solving? My calculus I book states "in the examples and exercises of this section it is always assumed that the given equation determines y implicitly as a differentiable function of x so that the method of implicit differentiation can be applied."

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*Is there another, simpler, way to say this?

What is the use of the word "determine" here mean? does it mean that "the explicit form of the equation in terms of y(x) will be differentiable"?
Is it the same thing to say: y' can be determined using implicit differentiation because we assume all functions y(x), in this sections, are differentiable?


*When is this assumption not true? is it ever not true? Can the implicit function of x be undifferentiable? I'm probably misunderstanding, or just not far enough along yet, but I thought primary benefit of using implicit differentiation was to differentiate equations that are difficult or impossible to express explicitly; how does one know that y(x) will be differentiable  before solving if they can't tell it's explicit form is differentiable?

 A: By the implicit function theorem, an equation of the form $f(x,y(x)) = 0$ defines a continuously differentiable function $y$ in a region of the plane provided that $f$ is continuously differentiable and $\partial_y f \ne 0$. So the problem is saying to assume that all of the equations are "nice enough" that you can make this assumption. For a simple example of an equation that would not satisfy this, $ f(x,y) = y- \lfloor x\rfloor$ is not continuously differentiable, so the implicit function theorem would fail for $y - \lfloor x \rfloor = 0$.
That said, they may give you problems where $\partial_y f = 0$ at some point anyways. In this case, $y(x)$ will be multivalued and thus not strictly speaking a function. For a simple example, $x^2 + y^2 - 1 = 0$ defines a circle, which has two points for every value of $x$. And sure enough it has points where $\partial_y f = 0$. However, if these points are isolated, the implicit function theorem can still be applied in the usual way piecewise between those points. In the circle example, you would get $y(x) = \sqrt{1-x^2}$ for one piece and $y(x) = -\sqrt{1-x^2}$ for the other piece.
