The goal here is to find the derivative of $f(x)$, where $f(x)$ is a non-linear continuous function.
Let $y_1=f(x)$ and $y_2=ax+b$, such that we will find $y_2$ to be the tangent line at $x=c$.
Set $y_1=y_2$ and solve for $x$
$$f(x)=ax+b$$
$$x=x_1,x_2,x_3,\dots,x_n$$
Find $x_p$ and $x_q$ such that $x_p$ and $x_q$ are the closest to $x=c$ where $x_p\le c\le x_q$.
Set $x_p=x_q(=c$ by squeezey theorem$)$, so that we have
$$c=x_{p,q}$$
where $x_p$ and $x_q$ are given in terms of $a,b$. Solve for $a$ to get
$$a=g(c)$$
which we will rewrite as
$$a(c)=g(c)$$
Finally, we have
$$\frac d{dx}f(x)=a(x)$$
Example,
$$y_1=x^2,\ y_2=ax+b$$
$$x^2=ax+b$$
$$x^2-ax-b=0$$
$$x_p=\frac{a-\sqrt{a^2+4b}}2=\ x_q=\frac{a+\sqrt{a^2+4b}}2$$
Set $c=x_p=x_q$
$$c=\frac{a-\sqrt{a^2+4b}}2=\frac{a+\sqrt{a^2+4b}}2$$
Solving the right side, we get
$$a^2+4b=0$$
So we have
$$c=\frac a2\implies a(c)=2c$$
So finally,
$$\frac d{dx}x^2=a(x)=2x$$
Another example,
$$y_1=\frac1x,\ y_2=ax+b$$
$$\frac1x=ax+b\implies ax^2+bx-1=0$$
$$x_p=\frac{-b+\sqrt{b^2+4a}}{2a},\ x_q=\frac{-b-\sqrt{b^2+4a}}{2a}$$
Set $c=x_p=x_q$
$$c=\frac{-b+\sqrt{b^2+4a}}{2a}=\frac{-b-\sqrt{b^2+4a}}{2a}$$
Solving the right side gives
$$b^2+4a=0\implies b^2=-4a$$
$$c=\frac{-b}{2a}$$
$$c^2=\frac{b^2}{4a^2}=\frac{-4a}{4a^2}=\frac{-1}a$$
$$a(c)=\frac{-1}{c^2}$$
$$\frac d{dx}\frac1x=a(x)=\frac{-1}{x^2}$$
And lastly, another example that highlights we do not actually have to solve for $x$,
$$y_1=x^3,\ y_2=ax+b$$
$$x^3=ax+b\implies x^3-ax-b=0$$
Two roots will be $c$, and one root will be $x_r$, so we have
$$\begin{align}x^3-ax-b&=(x-x_r)(x-c)^2\\&=x^3-(2c+x_r)x^2+(c^2+2x_rc)x-c^2r\end{align}$$
By equating parts, we have
$$2c+x_r=0,\ c^2+2x_rc=-a$$
$$x_r=-2c$$
$$c^2+2(-2c)c=c^2-4c^2=-3c^2=-a$$
$$a(c)=3c^2$$
$$\frac d{dx}x^3=a(x)=3x^2$$
I've done this for $\sqrt{x}$ and other polynomials, and in all cases, this method has worked out.
So, to what extent does this method work? Can it solve higher degree polynomial derivatives? Can someone derive power rule and other similar derivative rules? Can we take derivatives of transcendental functions using this?
As an example, can I use this to differentiate $e^x$? Indeed, it is possible to solve for $x$ in $e^x=ax+b$, but how do I separate the solutions? How can I make that work?
Has someone else found this? It seems pretty basic when you visualize what this all means, though apparently this wasn't very popular if someone discovered this a long time ago.