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The equation itself is called "implicit" because it describes a set of points without making it obvious what that set of points looks like. For example, consid

The equation itself is called "implicit" because it describes a set of points without making it obvious what that set of points looks like. For example, consider an equation like `x^(2y) + y^(x^2) = -1`. This is a well-formed equation that describes a unique subset of the (real) plane, but it's not obvious what points are in it (or even, if any points are in it) at first glance. The "normal" rules of calculus apply to _functions_. But `x^(2y) + y^(x^2)` is not a function; neither syntactically (since it refers to both x and y) nor functionally (since the relation ignores the sign of x). Yet, despite not being a function, you can still compute an "implicit derivative" of the "equation". The reason this ends up working, hinted at by writing `y` as `y(x)`, is that the equation is equivalent to the graph of unknown "implicit function"(s). But even this is slightly different -- when you do `𝒟(left) = 𝒟(right)` -- the derivative is an operator that operates on _functions_, so this is actually an equation between two functions, something that doesn't normally come up in "normal" calculus. It turns out that all of the same algorithmic steps work out to be correct in this slightly different setting, but it does involve some things that are new, if you look at why the manipulations are happening and not just the algorithmic steps.

Tainnor 5y ago
> But `x^(2y) + y^(x^2)` is not a function; neither syntactically (since it refers to both x and y) nor functionally (since the relation ignores the sign of x). It is a function, if you interpret it as f(x) = x^(2y(x)) + y(x)^(x^2). Of course, we don't know what kind of function y is. And if y is given implicitly, there might be multiple possible ys. But that doesn't stop us from manipulating the composite function. Specifically, if there exist multiple ys, then the manipulated equations will be valid for all of these ys. So the only extra step is thinking about functions abstractly, instead of being given concrete functions. But otherwise it really is the "normal rules of calculus.

Tainnor 5y ago

> But `x^(2y) + y^(x^2)` is not a function; neither syntactically (since it refers to both x and y) nor functionally (since the relation ignores the sign of x). It is a function, if you interpret it as f(x) = x^(2y(x)) + y(x)^(x^2). Of course, we don't know what kind of function y is. And if y is given implicitly, there might be multiple possible ys. But that doesn't stop us from manipulating the composite function. Specifically, if there exist multiple ys, then the manipulated equations will be valid for all of these ys. So the only extra step is thinking about functions abstractly, instead of being given concrete functions. But otherwise it really is the "normal rules of calculus.