Technology Musings

Finding Derivatives without Limits or Differentials

JB

I always enjoy finding new ways of doing things.  I was reading Henle/Kleinberg's Infinitesimal Calculus, which is a great book.  It has, throughout the book, a main text and a side-text.  The side-text is infinitely more interesting than the main text.

Anyway, on pages 65-66, it gives a method for finding the slope of y = x^2 at a specific point, let's say its at (2,4).  The way you do it is kind of strange, but it seems to work.  You start by defining a line to intercept the equation:

y = mx + b

Now, we don't know what m and b are, but we *do* know what x and y are.  Therefore,

4 = 2m + b

We don't really care about b in the long run, so we want to find an equation for b to remove it from the mix:

b = 4 - 2m

Now our equation for the line becomes:

y = mx + 4 - 2m

Interestingly, we actually have a value for y - y = x^2, so this becomes:

x^2 = mx + 4 -2m

We can rearrange this into quadratic form with:

x^2 - mx + (2m - 4) = 0

Using the quadratic formula, we can do:

x = (m +- sqrt(m^2 - 4(2m - 4))) / 2

x = (m +- sqrt(m^2 - 8m + 16)) / 2

x = (m +- sqrt((m - 4)^2)) / 2

x = (m +- (m - 4)) / 2

Now, we want to take the "+" side of the "+-" since we are solving for "m".  Otherwise, we will lose "m".  Therefore, this becomes:

x = (2m - 4) / 2

x = m - 2

So, at the point we are looking at, what is the value of x?  At (2, 4), x is 2.  Therefore

2 = m - 2

m = 4

Therefore, the slope at (2, 4) is 4.  

Interestingly, we can also generalize this to get the full derivative for y = x^2.  To do this, we will introduce the variables p and q to be the specific x and y values at a specific point.  Therefore, the slope at (p, q) for y = x^2 is:

y = mx + b, therefore q = pm + b, therefore b = q - pm

q = p^2, therefore b = p^2 - pm

y = mx + p^2 - pm

y = x^2, therefore:

x^2 = mx + p^2 - pm

x^2 - mx + (pm - p^2) = 0

x = (m +- sqrt(m^2 - 4(pm - p^2)))/2

x = (m +- sqrt(m^2 - 4pm + 4p^2))/2

x = (m +- sqrt((m - 2p)^2))/2

x = (m +- (m - 2p))/2

x = (2m - 2p) / 2

x = m - p

m = x + p

Since p = x, this gives

m = x + x = 2x, which is the derivative of x^2

I don't know how many classes of functions you can do this successfully for, but I imagine it should cover quadratics at least.