What intuition stands behind implicit differentiation

Question

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

1. If i want to know how the equation changes as x changes, i'm taking derivative with respect to x
2. $\frac{d}{dx}y^2+\frac{d}{dx}x^2=\frac{d}{dx}1$
3. We can consider $y$ as a function, $y = f(x)$
4. Therefore: $\frac{d}{dx}(f(x))^2+\frac{d}{dx}x^2=\frac{d}{dx}1$
5. We can calculate how (f(x))^2 changes as f(x) changes, using chain rule.
6. $\frac{df(x)}{dx}\frac{d}{df(x)}(f(x))^2+\frac{d}{dx}x^2=\frac{d}{dx}1$
7. This is equal: $2f(x)\frac{df(x)}{dx}f(x)+\frac{d}{dx}x^2=\frac{d}{dx}1$
8. As $x$ changes, $x^2$ changes as $2x$, therefore $2f(x)\frac{df(x)}{dx}f(x)+2x=\frac{d}{dx}1$
9. As x changes, 1 doesn't changes, therefore it is 0. $2f(x)\frac{df(x)}{dx}f(x)+2x=0$
10. We don't know derivative of $f(x)$ but we can solve it
11. If we solve the derivative, we get $f'(x) = -\frac xy$

3. Questions

1. My way of thinking is right?
2. What does mean the final answer? It looks strange, it doesn't tell me nothing comparing to norma, explicit derivative of a function.
3. There is a difference between $\frac{dy}{dx}$ and $\frac{d}{dx}y$ ?
4. 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.

Answer 1

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 with respect to $t$

$$2x(t)x'(t) + 2y(t)y'(t) = 0$$

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))$

Answer 2

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:

$$ qy^{q-1} \frac{dy}{dx} = px^{p-1}$$

Answer 3

The way I think about it reflects a marginally different perspective on the derivative itself. Essentially, I view the fundamental operation as the "differential", and the derivative is merely the ratio of two differentials. For a general overview of how I teach this, you can see my arXiv paper. For a more theoretical version of the justification for this, see Fite's paper.

Imagine that every variable in your equation is dependent on some "invisible" variable, let's call it $q$. Therefore, $x$ is actually $x(q)$ and $y$ is actually $y(q)$. The differential operator $d()$ takes its parameter and evaluates it at $q$ and $q + \epsilon$ (where $\epsilon$ is a fixed hyperreal infinitesimal value). So to expand standard notation, $dx = d(x)$ which is actually $d(x(q)) = x(q + \epsilon) - x(q)$.

Since we are just doing differentials and not derivatives, you can take the differential of anything. You don't have a variable that you are differentiating "with respect to", and therefore there is no denominator. So, $d(x^2) = 2x\,dx$; $d(xy) = x\,dy + y\,dx$; $d(xyz) = xy\,dz + xz\,dy + yz\,dx$; and so forth. There is no limit to the number of variables or where they are in the equation. Just use standard rules, but don't divide by $dx$.

Then, if you wish, you can solve for any derivative (i.e., ratio of differentials) that you want.

In this methodology, there is no difference between explicit differentiation, implicit differentiation, and multivariable differentiation. They all follow the exact same pattern.

So, for $x^2 + y^2 = 1$, you can simply differentiate both sides:

$$d(x^2 + y^2) = d(1) \\ d(x^2) + d(y^2) = d(1) \\ 2x\,dx + 2y\, dy = 0$$

Now, we can algebraically solve for any derivative we wish. If we want to find $\frac{dy}{dx}$ we just manipulate for it:

$$ 2x\,dx + 2y\, dy = 0 \\ 2y\,dy = -2x\,dx \\ \frac{dy}{dx} = \frac{-2x}{2y} \\ \frac{dy}{dx} = -\frac{x}{y} $$

For me, this combines both the intuitive and the operational approach. They are one and the same. It just requires an adjustment on what you view the differential and the derivative to be. If the differential is just a subtraction between two infinitesimally-close points of a presumed independent variable, then all of the basic rules just fall out, as well as the ability to do algebraic manipulations.