# Is there a 'lost calculus'?

Are there any 'lost' theorems of calculus that could be used to 'simplify' it? For example, are there ways to calculate derivatives without using limits, maybe by some forgotten methods in calculus?

• "There is no royal road to geometry." [Euclid] – Gerald Edgar Feb 14 '15 at 13:30
• I'm more of a visual learner even though I love math yet why is it a lot of math books and those used as textbooks use a lot of boring black and white printed pages. I have a somewhat hyperactive mind which makes it hard to focus even when I like to study, also a soporific presentation found in many math books ( where they don't really care about any visual or 'informal' presentations) makes in hard to focus. There may be no 'easy' way to learn math but there may be ways with more visual and artistic approaches. – 201044 Feb 15 '15 at 16:20
• @201044 Arnold quipped in a popular book "Bourbaki writes, somewhat derisively, that Barrow's book had 90 figures on twice as many pages. Bourbaki's own books have no figures on thousands of pages, and I don't know which is worse." books.google.com/books/about/… The irony is that the founding works of conventional calculus by Fermat, Newton and Leibniz are very visual and geometric, the analytization of teaching it happened exactly because that was considered the "royal road" to calculus. – Conifold Feb 16 '15 at 1:29
• According to historians, Archimedes and others of that era were shockingly close to discovering calculus thousands of years before Newton and Leibniz. Some have wondered about how much farther advanced we would be today if the ancients had discovered calculus way back then. – Inquisitive Feb 16 '15 at 19:23
• @201044 One of the best textbooks I have ever used was, "Calculus with Analytic Geometry; Second Edition" by Howard Anton. In my opinion, it was extremely well and painstakingly written. – Inquisitive Feb 16 '15 at 19:26

Descartes first introduced the idea in La Geometrie in 1637, and later simplified it to "the method of tangents" in 1638. Suppose we want to find the slope of the tangent to $y=x^2$ at $x=1$. The general equation of a line passing through $(1,1)$ is $y-1=m(x-1)$, and $m$ is the slope we are looking for. Since nearby lines intersect the graph at two points and the tangent only touches it at one algebraically the system
$$y=x^2,\ \ y-1=m(x-1)$$ must have a double root at $x=1$. Eliminating $y$ and factoring we get $$x^2-1-m(x-1)=(x-1)(x+1-m)=0,$$ so for $x=1$ to be a double root we must have $1-m=-1$ or $m=2$, which is the sought slope of the tangent.
This approach works for any polynomial $y=p(x)$, and more generally for rational and even algebraic functions given implicitly. For example, to find slope of the tangent at $x=a$ we write $y-p(a)=m(x-a)$ and look for the value of $m$ that makes $a$ a double root of $p(x)-p(a)=m(x-a)$. This only requires long division of $p(x)-p(a)$ by $x-a$, no limits or infinitesimals, and if the quotient is $q(x)$ then $m=q(a)$. To make the method computationally viable an algorithm more efficient than long division is needed for detecting double roots. Such method was provided by Jan Hudde, a talented Dutch mathematician who had to abandon mathematics for politics to save Netherlands from Spanish invasion, in two letters included into 1659 edition of Descartes' La Geometrie. It involves a clever modular reduction of polynomials that anticipates methods of modern algebraic geometry.
• I find this is a nice different way to come up with the formula, but in the end what you compute is $\displaystyle \lim_{x\to a} \frac{p(x)-p(a)}{x-a}$, just doing it "by hand" by performing long division of the polynomials and then substituting $x=a$, which after all is the standard method to compute such a limit taught early in a calculus course; the whole point of derivative rules is to avoid that cumbersome calculation. It's cool that people came up with fast algorithms for polynomial division, but for "real" applications it's obviously faster to compute these limits with the power rule. – Torsten Schoeneberg Feb 28 '19 at 22:05