What intuition stands behind implicit differentiation I'm trying to undestand implicit differentation
Let's take as a an example equation y^2 + x^2 = 1
1. How i think about how the equation works
I think the function as : if x changes then the y term have to hold value of "y^2 + x^2" equal 1. Therefore the equation defines some set of numbers at x cordinates and y cordinates.
2. How i think about how differentate the equation


*

*If i want to know how the equation changes as x changes, i'm taking
derivative with respect to x

*$\frac{d}{dx}y^2+\frac{d}{dx}x^2=\frac{d}{dx}1$

*We can consider $y$ as a function, $y = f(x)$

*Therefore: $\frac{d}{dx}(f(x))^2+\frac{d}{dx}x^2=\frac{d}{dx}1$

*We can calculate how (f(x))^2 changes as f(x) changes, using chain rule.

*$\frac{df(x)}{dx}\frac{d}{df(x)}(f(x))^2+\frac{d}{dx}x^2=\frac{d}{dx}1$

*This is equal: $2f(x)\frac{df(x)}{dx}f(x)+\frac{d}{dx}x^2=\frac{d}{dx}1$

*As $x$ changes, $x^2$ changes as $2x$, therefore $2f(x)\frac{df(x)}{dx}f(x)+2x=\frac{d}{dx}1$

*As x changes, 1 doesn't changes, therefore it is 0.  $2f(x)\frac{df(x)}{dx}f(x)+2x=0$

*We don't know derivative of $f(x)$ but we can solve it

*If we solve the derivative, we get $f'(x) = -\frac xy$


3. Questions


*

*My way of thinking is right?

*What does mean the final answer? It looks strange, it doesn't tell me nothing comparing to norma, explicit derivative of a function.

*There is a difference between $\frac{dy}{dx}$ and $\frac{d}{dx}y$ ?

*Why i want to know ? Because i want to know how to interpretate steps and solution, not only algorithmically solve some book's problems.


PS. I'm barely after highschool - Therefore i don't know yet set theorem and other high level math things. I'm learning calculus on my own.
 A: As a first thought, consider a parametrization of some part of the solution set as $t \mapsto (x(t),y(t))$ for $t \in [a,b]$. Suppose both $x(t), y(t)$ are differentiable with respect to $t$. Then we have the equation:
$$x(t)^2 + y(t)^2 = 1$$
Which becomes, differentiating wrt $t$ 
$$2x(t)x'(t) + 2y(t)y'(t) = 0$$
Assuming $x' \neq 0, y \neq 0$, we have
$$-\frac{x(t)}{y(t)} = \frac{y'(t)}{x'(t)}$$
So we've related the coordinates of the parametrization and the derivatives of the coordinates. A particularly nice case would be if $x(t) = t$, in which case we can thing of it as $y(x)$ and the right side is $\frac{dy}{dx}$.
What's really going on here is the study of the level set of some function $F(x,y)$. One way of seeing this is that if a smooth path $t \mapsto(x(t),y(t))$ traverses a level set then the composition $F(x(t),y(t)) = k$ is constant, so it's derivative is zero. The multivariable chain rule gives use relationships between $x,y,x',y'$ in terms of $F$. The implicit function theorem tells us that under some simple conditions on $\frac{\partial F}{\partial x},  \frac{\partial F}{\partial y}$we can locally solve in such a way as to parametrize with $x(t) = x$, so basically $y(t) =y(x)$. Then $x' = 1$ and the relation gets simpler. 
I meant to add this, but the "condition" is simply that $
y'(t) \perp \Delta F(\gamma(t))$
A: You can think of it as nudge both side of the equation and do the derivative.
Example:
For function,
$$y^q = x^p$$
If you increase $x \to x + \Delta x $ then $y \to y+\Delta y $
Or,
$$ (y+ \Delta   y)^q = ( x+ \Delta x )^p$$
Now substract this from original eqn
$$ (y+ \Delta y  )^q - y^q = 
(x+ \Delta x)^p - x^p$$
divide both side by $\Delta x $
$$ \frac{ (y+ \Delta y )^q - y^q }{ \Delta x}=\frac{ (x+ \Delta x)^p - x^p}{\Delta x}$$
Now on left side,
$$ \frac{ (y+ \Delta x  )^q - y^q }{ \Delta x} =  \frac{ (y+ \Delta x  )^q - y^q }{ \Delta y} \frac{\Delta y}{\Delta x}$$
Now taking limit on both side of equation as the deltas drop down to zero. You get derivatives!
i.e:
$$ qy^{q-1} \frac{dy}{dx} = px^{p-1}$$
