# How could I build a “derivative-like” object defined w/ circles (instead of secant/tangent lines) that reduces to the usual derivative?

This is, maybe, a stupid question—but it's stupid in a fun way (at least for me).

We know the derivative of a function, $$f(x)$$, at some point, $$(x_0, f(x_0))$$ is: $$\lim_{h \rightarrow 0}\frac{f(x_0 + h) - f(x_0)}{h}$$ ... which is just the slope of the line tangent to that point.

What if we look at the circle whose center is the point $$(x_0, f(x_0))$$ and whose radius, $$r$$, is small enough such that the circle intersects $$f(x)$$ at only two points. Consider the angle between these two points, $$\theta$$, as $$r \rightarrow 0$$. It's just an intuition, but I feel like $$\theta$$ should correspond w/ the slope of the tangent line at that point, and that it should also be $$0$$ for functions like $$|x|$$ when $$x = 0$$.

$$(x - x_0)^2 + (y - f(x_0))^2 - r^2 = 0 \tag{1}$$ The roots of this equation (when $$r$$ is small enough, there should only be two) would give us the points of intersection, let's call those points $$(x_1, f(x_1))$$ and $$(x_2, f(x_2))$$. Then, we'd look at something like:

$$\theta' = \tan^{-1}\frac{f(x_1) - f(x_0)}{x_1 - x_0} + \tan^{-1}\frac{f(x_2) - f(x_0)}{x_2 - x_0}$$

(Parts of this are looking mighty familiar!)

Ok, so here's the actual question: How can I define this in such a way that we're looking at the roots of the equation as $$r \rightarrow 0$$? I don't know any notation that refers to roots, specifically, in such a way that I can say something like:

$$\lim_{r \rightarrow 0}~~~\text{roots of left-side of eqn. in (1)}$$

Is there such a notation? Also, separately, is there anything obviously stupidly wrong with this approach such that it's inviable mathematically?

• For your definition of $\theta'$, why are you adding the angles that both radii make with the horizontal? Shouldn't the angle for each radii separately converge to the angle the tangent makes with the horizontal? I don't understand the point of adding both angles. – Andrew Paul Jul 25 '19 at 19:30
• @AndrewPaul Oops! You're right. I should subtract the points direclty. I'm going to leave it as is, though, so that people reading John's answer below don't get confused. – AmagicalFishy Jul 25 '19 at 20:38

$$(x - x_0)^2 + (y - f(x_0))^2 - r^2 = 0 \tag{1}\label{eq1}$$

note that, for any fixed $$x_0,y_0$$ and $$r$$, the only variables are $$x$$ and $$y$$. Also, let $$y_0 = f(x_0)$$ for simplicity. You can expand the LHS of \eqref{eq1}, gather the terms of multiple of powers of $$y$$ to get a quadratic equation in $$y$$ which you can solve, as long as the discriminant is positive, to get $$2$$ values of $$y$$ in terms of $$x$$. In particular, you get

\begin{align} 0 & = (x - x_0)^2 + (y - y_0)^2 - r^2 \\ & = x^2 - 2x_0 x + x_{0}^{2} + y^2 - 2y_0y + y_{0}^{2} - r^2 \\ & = y^2 - 2y_0 y + (x^2 - 2x_0 x + x_{0}^{2} + y_{0}^{2} - r^2) \tag{2}\label{eq2} \end{align}

Letting $$c = x^2 - 2x_0 x + x_{0}^{2} + y_{0}^{2} - r^2$$ for simplicity, you can then use the quadratic formula you get

$$y = \frac{2y_0 \pm \sqrt{4y_{0}^{2} - 4c}}{2} = y_0 \pm \sqrt{y_{0}^{2} - c} \tag{3}\label{eq3}$$

Let these $$2$$ values be $$y_1,y_2$$. To find the corresponding values of $$x_1,x_2$$ requires solving for $$y_i = f(x_i)$$ for $$i=1,2$$ in \eqref{eq3}. Note depending on how large your $$r$$ is and what the curve is, there may be more than one value of $$x_i$$ for a given $$y_i$$, but for most functions as long as $$r$$ is small enough, there should only be $$1$$ solution, as you stated.

Next, you state

$$\theta' = \tan^{-1}\frac{f(x_1) - f(x_0)}{x_1 - x_0} + \tan^{-1}\frac{f(x_2) - f(x_0)}{x_2 - x_0} \tag{4}\label{eq4}$$

However, as Andrew Paul's comment indicates, you're adding basically the same angle twice if $$r$$ is very small so $$x_1$$ and $$x_2$$ are very close to $$x_0$$. One way to adjust your equation to make it work is to take the average of the $$2$$ angles, i.e., divide by $$2$$ on the RHS. However, as you indicated earlier, it seems you're more interested in the angle between $$x_1$$ and $$x_2$$ as these points approach $$x_0$$, i.e., you want

$$\theta' = \tan^{-1}\frac{f(x_2) - f(x_1)}{x_2 - x_1} \tag{5}\label{eq5}$$

$$\lim_{r \rightarrow 0}~~~\text{roots of left-side of eqn. in (1)}$$

Note using $$r = 0$$ in \eqref{eq1} gives

$$(x - x_0)^2 + (y - f(x_0))^2 = 0 \tag{6}\label{eq6}$$

Since you have two values squared adding to $$0$$, each must be $$0$$, i.e., $$x = x_0$$ and $$y = f(x_0)$$. Thus, using limits in this manner doesn't seem to help much. Instead, if you use something like \eqref{eq3} to get your $$(x_1,f(x_1))$$ and $$(x_2,f(x_2))$$, you can use them in \eqref{eq5}. Note this will become a function of $$x_0,y_0,r$$ so, for any fixed $$x_0,y_0$$, you can take the limit of $$r \to 0$$. Based on the definition of a derivative, in most cases for well-behaved functions, you'll then get

$$\theta = \lim_{r \to 0}\left(\tan^{-1}\frac{f(x_2) - f(x_1)}{x_2 - x_1}\right) = \tan^{-1}(f'(x_0)) \tag{7}\label{eq7}$$

i.e., the slope of the tangent, which you've already noted.

However, in some other cases such as $$f(x) = |x|$$ that you mention, you'll also get that $$\theta = 0$$ at $$x = 0$$, even though $$f(x)$$ is not differentiable there. In general, you can use your procedure with any continuous function, even at points where the function may not be differentiable.

The way I would do it is to consider the circle tangent to the curve and let its radius grow to infinity. In the limit, the circle at the point would approach the regular tangent.

You could also consider the circle that passes through $$(x, f(x)), (x-h, f(x-h)), (x+h, f(x+h))$$ and let $$h \to 0$$.