# Not following the derivation of $\frac{dy}{dx}=-\frac{F_x}{F_y}$

I've seen similar questions to mine asked on the forum, but I haven't seen answers that address the part I'm confused about.

My calculus textbook (Thomas from Pearson) derives the following formula to "take some of the algebra out of implicit differentiation":

Suppose the function $$F(x,y)$$ is differentiable and the equation $$F(x,y)=0$$ defines $$y$$ implicitly as a differentiable function of $$x$$. Then at any point where $$F_y\neq 0$$, we have $$\frac{dy}{dx}=-\frac{F_x}{F_y}$$.

(The formula itself is pretty intuitive to me, except for the negative sign.) I feel like I am misinterpreting the derivation given, as it seems to be using $$F(x,y)$$ to denote two different functions and treating them as if they are the same. The derivation goes like this:

Suppose that (1) the function $$F(x,y)$$ is differentiable and that (2) the equation $$F(x,y)=0$$ defines $$y$$ implicitly as a differentiable function of $$x$$. Since $$w=F(x,y)=0$$, the derivative $$\frac{dw}{dx}$$ must be zero.

As I understand this, they are defining a new function $$w:\{(x,y):F(x,y)=0\}\rightarrow\{0\}$$, a level curve of the original $$F(x,y)$$, which is zero everywhere on its domain, and we're to suppose that its domain defines $$y$$ implicitly in terms of $$x$$. But then they continue:

... Computing the derivative [of the equation $$w=F(x,y)=0$$] from the chain rule, we find $$0=\frac{dw}{dx}=F_x\frac{dx}{dx}+F_y\frac{dy}{dx}=F_x+F_y\frac{dy}{dx}.$$ Therefore, we have $$\frac{dy}{dx}=-\frac{F_x}{F_y}.$$

This is where I get confused. In the example questions, it is clear that $$F_x$$ and $$F_y$$ denote the partial derivatives of the original function $$F(x,y)$$ of which $$w$$ is a level curve. But this use of the chain rule seems to assume that those are also the partials of w (which is a constant function, and should have zero derivatives, no?). I'm interpreting this as a special case of $$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}$$ where $$t=x$$, and where $$\frac{\partial w}{\partial x}$$ and $$\frac{\partial w}{\partial y}$$ are written as $$F_x$$ and $$F_y$$. But I'm not seeing how the former and the latter partials are equivalent. Why can we assume both that $$\frac{dw}{dx}=0$$ and that $$F_x=\frac{\partial w}{\partial x}$$, when $$F_x$$ is not zero in general? Or is that assumption not actually being made by using the chain rule this way? What am I missing or getting wrong here? I'd really appreciate if someone would set me on the right track so that I can get some intuition for why this theorem works. Thanks!

• There is no need to use $w$ in the derivation. You are starting with $F(x,y) = 0$ a constant function. So, its derivative $dF/dx = 0$.
– Doug
Commented Jun 20, 2022 at 18:28

Yes, there are several abuses of notation here. What is happening is you're first given a smooth function $$F:\Bbb{R}^2\to\Bbb{R}$$; for simplicity assume that at every point $$p\in\Bbb{R}^2$$, we have $$\frac{\partial F}{\partial y}(p)\neq 0$$. The implicit function theorem tells us that if you fix such a point $$p=(a,b)$$, then you can find some smooth function $$\eta:I\subset\Bbb{R}\to\Bbb{R}$$ such that $$\eta(a)=b$$ and for all $$t\in I$$, we have $$F(t,\eta(t))=0$$. So, the function $$w:I\to\Bbb{R}$$ defined as $$w(t)=F(t,\eta(t))$$ is smooth and is zero at every point; i.e is the constant zero function. So, we obviously have that $$w'=0$$. But now what does the chain rule tell us (note that $$w$$ is the composition of $$F$$ with the function $$t\mapsto (t,\eta(t))$$, so chain rule is indeed the way to go)? It tells us for each $$t\in I$$, \begin{align} 0&=w'(t)=\frac{\partial F}{\partial x}\bigg|_{(t,\eta(t))} \cdot 1+\frac{\partial F}{\partial y}\bigg|_{(t,\eta(t))}\cdot \eta'(t) \end{align} Rearranging this equation, we get \begin{align} \eta'(t)&=-\frac{\frac{\partial F}{\partial x}\bigg|_{(t,\eta(t))}}{\frac{\partial F}{\partial y}\bigg|_{(t,\eta(t))}}. \end{align} Hopefully with the different notation, it's clear what the different functions are, and how the chain rule is being applied, and where everything is evaluated.

If the $$x,y$$ are confusing (and I believe they are), you can write the chain rule computation as follows: for each $$t\in I$$, \begin{align} 0&=w'(t)=(\partial_1F)_{(t,\eta(t))}\cdot 1+(\partial_2F)_{(t,\eta(t))}\cdot \eta'(t), \end{align} and hence \begin{align} \eta'(t)&=-\frac{(\partial_1F)_{(t,\eta(t))}}{(\partial_2F)_{(t,\eta(t))}} \end{align}

It is an abuse of notation to use $$y$$ to refer to both the coordinate, and also the name of the implicitly defined function, and to use $$F$$ as both the original function, and the new composed function $$w$$, but unfortunately, it is standard practice.

• Here is a similar answer where I deal with a similar situation. Commented Jun 20, 2022 at 18:32
• Thanks! That was extremely helpful. Good to know that I should expect to see that notation practise elsewhere. Commented Jun 23, 2022 at 18:37

Let's give the new function a name: $$Y$$. Thus we are assuming there is a point $$(x_0, y_0)$$ where $$F(x_0, y_0) = 0$$ and $$F_y(x_0, y_0) \ne 0$$, and a differentiable function $$Y$$ defined in a neighbourhood of $$x_0$$ such that $$F(x, Y(x)) = 0$$ in this neighbourhood, with $$Y(x_0) = y_0$$. Then we can differentiate both sides of the equation $$F(x, Y(x)) = 0$$ using the chain rule, obtaining $$F_x(x, Y(x)) + F_y(x, Y(x)) Y'(x) = 0$$ so that $$Y'(x) = - \frac{F_x(x, Y(x))}{F_y(x, Y(x))}$$

$$\frac{dw}{dt}=\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}$$ where $$t=x$$, and where $$\frac{\partial w}{\partial x}$$ and $$\frac{\partial w}{\partial y}$$ are written as $$F_x$$ and $$F_y$$.

Then $$\dfrac{\mathrm d w}{\mathrm d t}=0$$ , $$\dfrac{\mathrm d x}{\mathrm d t}=1$$ , $$\dfrac{\mathrm d y}{\mathrm d t}=\dfrac{\mathrm dy}{\mathrm d x}$$ , and so so:$$0=F_x\cdot 1+F_y\cdot\dfrac{\mathrm d y}{\mathrm d x}$$

Why can we assume both that $$\mathrm dw/\mathrm dx=0$$ and that $$F_x=∂w/∂x$$, when $$F_x$$ is not zero in general?

We do not assume $$\mathrm dw/\mathrm dx=0$$, rather it is true because $$w$$ is established as a constant function ($$w=0$$).

However, $$w$$ is also a function in two variables, one which is $$x$$ and the other $$y$$.   $$w=F(x,y)$$.   Therefore, when the partial derivative with respect to the first argument is not zero, then $$y$$ must be related to $$x$$ in some manner to make make $$w$$ a constant function, and that relation makes $$\dfrac{\mathrm d y}{\mathrm d x}=-~\dfrac{F_x(x,y)}{F_y(x,y)}$$

Alternatively: Let $$\partial_n F(\cdots)$$ be the partial derivative with respect to the $$n^{\rm th}$$ argument of the function $$F$$.

Let $$F(x,h(x))=0$$ , so by the chain rule:

\begin{align}\dfrac{\mathrm d ~~}{\mathrm d x}F(x,h(x))&= \dfrac{\mathrm d x}{\mathrm dx}~\partial_1F(x,h(x))+\dfrac{\mathrm d h(x)}{\mathrm d x}~\partial_2F(x,h(x))\\[2ex]\therefore\qquad\dfrac{\mathrm d h(x)}{\mathrm d x}&=-\,\dfrac{\partial_1 F(x,h(x))}{\partial_2 F(x,h(x))}\end{align}

So let $$y=h(x)$$ ...