Intuition behind the implicit function theorem.

Question

Is there any intuition behind the implicit function theorem? Take $F(x,y)=0$ where $y=f(x)$. Then,

$$ \frac{dy}{dx}=-\frac{F_x}{F_y} $$

I see that the derivative is equal to the reciprocal of the partial derivatives, what is the intuition behind this?

Answer 1

The geometric intuition is that the unit tangent vector, say $T$, to the surface $F=0$ is perpendicular to the gradient of $F$ at each point. This is because the directional derivative of $F$ in the tangential direction is $\nabla F \cdot T$, which would be nonzero if the tangent weren't perpendicular to the gradient, which intuitively implies that $F$ would vary along the surface. Once you know that, this relation follows by just taking the ratio of the $y$ component of the tangent vector and the $x$ component of the tangent vector.

Another mnemonic for this relation is to formally write $dF=F_x dx + F_y dy = 0$, which intuitively means that a change in one variable must be matched by a change in the other such that the change in $F$ will vanish (a slightly less rigorous way of saying $\nabla F \cdot T=0$). Then formally solve for $\frac{dy}{dx}$.

Answer 2

The intuition for the implicit function theorem as I understand it is: the solutions to $F(x,y)=0$ are a union of graphs of functions.

In your case, it looks like you are talking about one function you've already picked, $y=f(x)$. Sometimes, you might need to have $x=g(y)$ to cover the whole solution set.

One of the typical examples given is $F(x,y)=x^2+y^2 -1$.

Since $dF/dy=2y$ is nonzero whenever you are not on the $x$-axis, you're guaranteed that the points outside of the $x$-axis can be represented as a union of graphs.

In this case, the graph of $f(x)=\sqrt{1-x^2}$ and $g(x)=-\sqrt{1-x^2}$ for $x\in [-1, 1]$ might be one way to express it.

However, some authors require that the domains of these functions be open sets. In that case, it would be impossible to pick up the points of the graph above using functions of $x$. But, thankfully, we could also use the equivalent functions of $y$ to cover those points:

$f'(y)=\sqrt{1-y^2}$ and $g'(y)=-\sqrt{1-y^2}$

They are able to cover everything where $dF/dx$ is nonzero (which is everything off the $y$-axis.)

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From your example it looks like you are rather looking for an explanation of why this applies to implicit differentiation, to which I would say: since you can "zoom in" to one of these functions no matter where you are in the solution set, you'll always be able to learn about the tangent (locally) using regular differentiation (since they are functions.)