derivatives 'defined' without limits?!?

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[nikitasstathatos]

Another issue is that it's a bit awkard to talk about tangents of linear functions. If we take f(x)=cx, then the tangent to the graph at any point coincides with the function, so there's infinite intersection points instead of a unique one, which means that we cannot calculate the most basic of derivatives! Seems pretty similar to the inflection point case, so perhaps it also has to be excluded.

[EricOfYork]

Minor arithmetic error at 9:10 the constant term of the quadratic should be m^2-2m+1, not just m^2+1
This leads to a simplified discriminant of 4m^2 -4m + 1, which factors to (2m-1)^2

[kappasphere]

9:25 The quadratic formula only works when m =/= 0, so you first have to check if m=0 is an option. And indeed, when you enter m=0, you get 1-x=0, which has exactly one solution.
I think one way to exclude these "bad" cases of m is to add that within a neighbourhood of x=a, the tangent line should either be all less/equal or all more/equal than the function in question.

[edwardlulofs]

Ohh. I think this is my favorite video of the last 12 months!
I love the math where you might think some procedure is correct and simpler than a more complicated definition - but it’s not correct.

It reminds me of a math lecture in 1970 where the professor showed that irrational numbers can’t be used as a base. That is based on 2 works, based on 10 works. But base pi does not work.

Excellent video! This made my day.

[blackyusylvean]

For the exponential case, it is doable using a non-elementary function called the Lambert W function, which is defined as the inverse of the function xe^x. Since said function is not bijective, the Lambert W function has infinitely many branches, however since we're looking to solve e^x = m(x-1)+1 in the *reals*, only the two real branches W[0] and W[-1] will work.
W[0](x) is defined for x in the interval (-1/e ; +inf) and W[-1] is defined for x in the interval [-1/e ; 0)
Therefore an equation involving the lambert W only has one solution if its argument is positive.

Solving e^x = m(x-1)+1 for x using the Lambert W:
exp(x) = m(x-1) + 1
=> exp(x) = mx - m + 1
=> 1 = (mx - m + 1)exp(-x)
=> -1/m = (-x + 1 - 1/m)exp(-x)
=> - 1/m exp(1 - 1/m) = (- x + 1 - 1/m)exp(-x + 1 - 1/m)

We have something of the form w = z exp(z), therefore solutions are all of the form z = lambertW[k](w). In our case, we have
- x + 1 - 1/m = lambertW(-1/m exp(1 - 1/m))
-> x = (1 - 1/m) - lambertW(-1/m exp(1 - 1/m))

As said before, in order to have only one possible solution, the argument of the lambertW must be positive so as to only be defined on the 0-branch.
Hence we must have -1/m exp(1 - 1/m) >= 0, which is only true if m is strictly negative. In fact, any negative m will do, so this does not work very well since we expected to get at least a positive slope.

The only m such that there is exactly one solution is m=0, in which case we have exp(x)=1 -> x=0, but that still isn't the value we were expecting (m=1)

[kantanlabs]

When I was young (in the sixties in France) the ideas of continuity and derivability were defined relying on the idea of bounded values for arbitrarily small intervals. Theses notions were tackled from a somehow topological point of view (compacity, open of closed sets…). I just remember that demonstrating that a function is derivable in a given interval or even a single point was much tedious than it is today with the modern formalism relying on limits and associated theorems.

[miraj]

At 7:41, probably easier to let y = sqrt(x). Then you have y = my^2 - m + 1 ==> my^2 - y + (1-m) = 0.
We want the discriminant = 0 so we need b^2 = 4ac ==> 1 = 4m(1-m) ==> 4m^2 - 4m + 1 = 0 ==> (2m-1)^2 = 0 ==> m = .5.

[knedl]

I'm not sure whether this definition generalises that well, since, for example, sin x = mx only has one solution for many different negative values of m. But I think this following definition could work:

First, define A := {all real values m for which m(x - a) + a - f(x) >= 0 on the interval (a - ε, a], for all small enough values of ε > 0}. This set should be of the form (-∞, c) or the form (-∞, c], for some real number c. This is because if some number is in the set, then all smaller numbers are in the set as well, since the value of m(x - a) increases when m decreases. Likewise, if a number is not in this set, all the bigger numbers also aren't a part of this set. Now you define the left derivative of the function f at a as the value c.

Similarly, we can define the right derivative. The set B := {all real values m for which m(x - a) + a - f(x) >= 0 on the interval [a, a + ε), for all small enough values of ε > 0} must be of the form (d, ∞) or [d, ∞), for some real number d. The argument is similar to the one above. Now we define the right derivative as the value d.

Finally, we define the derivative to be the values of the left and the right derivative, in the case that these two are equal. Otherwise, no derivative exists.

Your condition did not work in inflection points, because the function m(x - a) + a - f(x) changes sign for all values of m, as opposed to sign staying equal for some m (when m is the derivative) in other functions. But by separating derivatives into the left and the right derivative, we can just treat the two sides individually. We can then focus more on the sign of the function instead of the number of solutions, which brings us to this defintion.

[robert.ehrlich]

Your guess at 14:36 that if you are away from an inflection point is false. A good counter-example for this is the function f defined by f(x) =x²sin(1/x) for x not equal to 0 and f(0) = 0. This function has a derivative equal to 0 for x=0, the value of your "m" should be 0, but the equation f(x) = 0 has an infinity of solutions in any interval containing 0. It can be argued if you are really "away from an inflection point" at x=0 since there is an infinity of inflection points in any interval containing 0 (the sames as above) but at x=0 we don't have an inflection point.

[ianfowler]

Here's another nifty extension. Find the equation of the tangent to y = x^3 - 3x^2 - 3 at the point of inflection. Then intersect the tangent with the cubic. You will get an equation that has a triple root at the x-value of the inflection point. Works in general. Nice proof.

[ianfowler]

Another way to force the double root at the point of contact where x = a is to divide f(x) by (x - a)^2 and inspect the reminder. Which must be linear since we are dividing by a quadratic.

x^2 divided by (x - 3)^2 gives a quotient of 1 and a remainder if 6x - 9, the tangent equation.

A cubic example: Find the equation of the tangent to y = x^3 - 2x + 3 @(1, 2)
Divide x^3 - 2x + 3 by (x - 1)^2 = x^2 - 2x + 1
You will get a quatient of x + 2 and a remainder of x+1 so the tangent is : y = x + 1.

[jimbobago]

10:17 I get 8m^3, not 4m^3 but then 1/2 isn't a root.

[spicymickfool]

Descartes found derivatives via osculating circles. The radius of a tangent circle is perpendicular to the tangent line. Alternatively, find the intersection of $y=mx+b$ with a parabola. Select m and b so that the discriminant is 0. The result gives you the tangent line the slope of which is 2Ax+B as expected.

[pietergeerkens]

The factoring at 11:12 is incorrect.
(2m - 1) (2m^2 - 2m + 1) = 4m^3 - 6m^2 - 1
and not the required
4m^3 - 4m^2 + 4m - 1.

Likewise by substitution of 1/2, the value for the cubic at 1/2 is
1/2 - 1 + 2 - 1 = 1/2 <> 0.

This seems to follow from the earlier error at 9:10 for the constant term of the cubic.

[Miparwo]

Far more elegant and general is to use nilpotent units, for automatic differentiation.

[ivandebiasi]

I'm missing something or either in the case of f(x) = sqrt(x) and f(x) = sin(x) any value m<0 satisfy the condition "f(x) = m(x-a) + f(a) has only 1 solution"?

Why he completely ignored those cases?

[GabrielOliveira-qlcg]

I think there is something missing, no? For the exponential case, for exemple, at any point (a, e^a), e^x=m(x-a)+e^a will have a unique solution for any m<=0. Not only near but anywhere, actualy. Not only we need a better definition of "near", we also need stronger conditions on m. It seems to me that the particular slope m wich makes the right side coincide with the tangent line is special not only for making the solution unique at a neighborhood D about a, but also because it is in some sense "isolated" from other values that also makes the solution unique.

So, m should be a value that makes the solution for f(x)=m(x-a)+f(a) unique for some neighborhood D around x=a AND that for any neighborhood E around x=a contained in D, there will be a perturbation ε to the slope m s.t. f(x)=(m-ε)(x-a)+f(a) will have at least two solutions in E.

What makes inflection points different is that their "correct" value for m is isolated in this sense from one side only. That is, for inflection points with positive slope, ε should be strictly negative to break the uniqueness of the solution in the chosen neighborhood, while ε should be strictly positive for inflection points of negative slope. Also, I think in both cases we should be able to fit three solutions for the equation for any E with small enough ε. For non-inflection points, on the other hand, we should be able to pick both positive and negative values for ε and, for small enough E, the equation will have only two solutions in E even for arbitrarily small ε.

Of course, this is all geometric intuition and certainly should be made more rigorous. Maybe some conditions on the smoothness of f(x) are also needed? That could be tricky, since we would need a definition for smoothness prior to our definition of derivatives. Also, this approach is even more questionable from a "practical" point of view, but it could be useful for testing values of m found via the naive method, I guess.

[hbowman]

The most streamlined way to algebraically differentiate an algebraic variety at a point is to linearize it.

You express the variety as the intersection of the zeros of multivariable polynomials and change variables. For instance, if we have y²=x at (1, 1), we replace x with u+1 and y with v+1. Then we have v²+2v+1=u+1. So, -v²+u+2v is zero at (u, v)=0, and then we get the cotangent from the linear approximation u+2v=0, giving us that (1, 2) is in the cotangent space. Then it is a simple matter to find the vector perpendicular to the cotangent space. The slope of (1, 2) is 2, so the derivative is -½.

If the dimension of the cotangent space is "wrong", you're at a singular point on the variety.

[plislegalineu]

In Poland we use the uppercase letter delta (that triangle) for the discriminant and we use it a lot cuz when we learn the quadratic formula (which doesn't have a name here) we first calculate the discriminant and then plug it into under the square root

[mathunt]

I believe the string theorist Robert Diigraaf(sp) did talk where he started out with the well known derivative definition and presented all the generalisations. I believe that the one of the examples he uses is the one here

The square root example may have been easily done, by simply considering the equation as an equation in \sqrt{x}. The calculation is far easier.