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What does dy/dx really mean in the context of implicit differentiation

The problem is that I can't relate it back to the limit definition. If you can't write y explicitly as a function of x, often y approaches multiple values as x approach some value, or at some x, y has multiple values hence the limit does not exist and therefore the derivative will not exist

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  • $\begingroup$ Before getting into implicit differentiation, are you clear on the question of what it means for an equation to implicitly define a function? ${}\qquad{}$ $\endgroup$ Feb 13 '15 at 0:40
  • $\begingroup$ No, functions the way I learned it is something that only has one output. When y is written implicitly in x, this one-to-one relationship breaks down. How can you still call it a function? $\endgroup$ Feb 14 '15 at 8:20
  • $\begingroup$ The way I was taught the concept of a function also implies that a function has only one output for each input. The equation $x^2+y^2=1$ is satisfied by two different $y$ values for each of all but two $x$ values in $[-1,1]$. But there is only one differentiable function on a small neighborhood of $(4/5,3/5)$ that satisfies $x^2+f(x)^2=1$. ${}\qquad{}$ $\endgroup$ Feb 14 '15 at 14:19
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For implicit differentiation you will have an equation that -- at least locally -- is satisfied by the points on a curve through the plane.

Imagine that we find a parameterization of the curve such that the set in question is $$ \{ (f_x(t),f_y(t)) \mid t\in [0,1] \} $$ and otherwise "sufficiently well-behaved". Then at the point $(f_x(t_0),f_y(t_0))$ the implicit derivative ought to be the limit

$$ \lim_{h\to 0}\frac{f_y(t_0+h)-f_y(t_0)}{f_x(t_0+h)-f_x(t_0)} $$

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