implicit-differentiation
I understand the method used in implicit differentiation, it's just an application of the chain rule. But why can you define $y$ as a function of $x$?
In this equation for example:
$x^2 + y^2 = 1$
$2x + 2yy' = 0 $
Why isn't it just this?:
$2x + 2y = 0$
Thank you in advance.
When you write down an equation in $x$ and $y$ of the form $$F(x,y)=0,$$ then you can define a curve $\mathcal{C}$ in the plane by saying that $(x_0,y_0)\in\mathcal{C}$ if and only if $F(x_0,y_0)=0$... a point is on the curve if and only if it satisfies the equation of the curve.
Now for different $F(x,y)$ you get different curves. Some might be empty such as $x^2+y^2+1$ while others are the entire plane such as $\sin^2x+\cos^2x-1$.
Others are nicer such as $x^2+y^2-1$ which gives you a circle or the equation that gives a figure-of-eight. Now locally (i.e. near a specific point) these curves might not be that nice in that, no, they don't look the graphs of functions or differentiable functions.
For example, the point in the middle of the figure-of-eight does not look like the graph of a function. Similarly on the left- and right-hand sides of the circle there are vertical tangents --- this is not the behaviour of a differentiable function.
However away from such danger points --- such as anywhere else on the circle --- if you zoom in close enough to the point, locally, the circle looks like the graph of a function. So what we do is almost take each point $(x,y)$ on a case-by-case basis and say, O.K., near this point we could in theory define a function $y=y(x)$.
For the example of your circle. Suppose we are away from the vertical tangents and are thinking about a point $P=(x,y)$. The implicit function tells you that close to $P$ there is a function $y=y(x)$ whose graph is locally the same as the circle (in fact for the circle you have $y(x)=\pm\sqrt{1-x^2}$).
So we assume that, near $P$, we actually have (I actually get my students to write out $y=y(x)$ to help them see Chain Rules and not have $y'=1$) $$x^2+[y(x)]^2-1=0.$$
Now these are two functions (LHS and RHS) so have the same derivative: $$2x+2(y(x))\cdot\frac{dy}{dx}=0\Rightarrow x+y\frac{dy}{dx}=0...$$
With a bit of work, you can show that $R(x,y)$ is only defined for $y=x$ and $y=-x$ excluding the point $(0,0)$. Hence your function does not satisfy the conditions of the implicit function theorem, which states that your function should be at least continuously differentable to apply the method of implicit differentiation.
Note, you could extend your problem to complex numbers, in that case the domain of applicability of $R(x,y)$ would be extended. However, even in that case, it would still not be the entire plane because no definition of the square root or the inverse cosine is analytic on the entire complex plane. In particular, if you choose the most "natural" extension to the complex plane of the inverse cosine, it is not defined on the branch points $-1$ and $1$. Those correspond exactly to the cases $y=x$ and $y=-x$. In other words, naïve implicit differentiation does not work in that case either. You would therefore need analytic extensions of the square root and the inverse cosine such that they are valid in the case $y=x$. But, those extensions will not correspond to the definitions you adopted in the real case.
Best Answer
(1). $\{(x,y): x^2+y^2 =1\}$ is NOT the graph of a function.
(2). There IS a function $f(x)$ for $|x|<1$ such that $f(x)$ is differentiable and $x^2+f(x)^2=1.$ In fact there are 2 such functions. And we have $$0= \frac {d}{ dx} (1) =\frac {d}{ dx} (x^2+f(x)^2)=2x + \frac {d}{ dx }(f(x)^2)=2x+2f(x)f'(x).$$
Note: In differentiation, it matters what you differentiate BY. Just as in division, where it matters whether you divide by x or by y. The derivative of $y^2$ with respect to $y$ is $2y.$ The derivative of $y^2$ with respect to $x$ is $2y \frac {dy}{dx}.$