Assuming y' = dy/dx, then 0 = 2 y y' + 2 x -> -x/y = y'
I've honestly never understood what exactly implicit differentiation means or what is implicit about it, it seems to just be the normal rules of calculus?
Why would you expect dy/dx to be 2y?
e: ah, didn't see you were referencing a problem from the page
Because he's for some reason differentiating with respect to y.
You're right, "implicit differentiation" is nothing more than just differentiating both sides of an equation. Pedagogically, it is a bit of a step up from "here's a single expression that denotes a function, differentiate it". You have to recognize that differentiation is an operation on functions, and then you have to realize that something like "9=y^2+x^2" really means "9=y(x)^2 + x^2 for all values of x" (or equivalently as an equation of functions, the constant function 9 = y^2 + id^2) and that it's valid to apply the differentiation operator to both sides.
Many calculus courses probably gloss over subtleties like this or expect students to be able to "intuitively" grasp them (and many very well may), but I can see it leading to confusion.
An "explicit function" is something like y = √(x³ + 1), where you have a formula like √(x³ + 1) that tells you how to compute a value of the function for any given value of the independent variable. An "implicit function" is the function implicitly defined by an equation like y² - x³ = 1; you don't have a formula. In this case, it actually defines a relation, because for any value of x there are two values of y. You could sort of reasonably claim that by sticking "y =" before a formula, you are defining an implicit function, just a boring degenerate one, but we conventionally reserve the term "implicit function" for things that we can't handle in the explicit-function fashion.
It's called "implicit differentiation" because we're differentiating the implicit-function representation—instead of just taking the derivative of one formula to find the answer, we take the derivatives of the formulas on both sides of the equation. (We know this is safe: if a = b, then we can substitute them for each other in any context, for example da/dx = db/dx.) Or, equivalently, we subtract them from each other, take the derivative of the resulting expression, and set it to 0.
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.
> 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.
I've honestly never understood what exactly implicit differentiation means or what is implicit about it, it seems to just be the normal rules of calculus?
Why would you expect dy/dx to be 2y?
e: ah, didn't see you were referencing a problem from the page