Differentiability, and why it forces continuity while the reverse fails

0:00Here is the derivative, written as a limit. The change in the output over the change in the input, as the change in the input goes to nought. And attached to it is a caution that is easy to read past: this is said to be the derivative PROVIDED the limit exists. Everything in this video is what happens when it does not. So the first thing to do is to say precisely what that limit existing means, and it turns out to be two demands rather than one.

0:35Approach the input from below and you get one difference quotient. Approach it from above and you get another. The condition is that both of those settle on a number, and that the two numbers are the same. Finite, and equal. Read that as one sentence and you will describe half the failures wrongly. Read it as two tick boxes, ticked independently, and two famous examples stop being tricks. Everything that follows is built on keeping those two boxes apart.

1:11Differentiability on a stretch means the same thing at every input of it, with one adjustment at the two ends. At the lower end there is nothing below to approach from. Take a square and declare it only from one to two. At the input one the quotient from below asks about inputs the rule does not have, and the search returns nothing at all. The quotient from above returns two.

1:39At the input two it is the other way round: from below you get four, and from above, nothing. So at each end exactly one of the two is available, and it is the one that points INTO the stretch. At the lower end that is the derivative from the right; at the upper end it is the derivative from the left. Say those two names to yourself in that order, because they are easy to swap and the swap is invisible until an examination.

2:11And note what the refusal is about: the same square with no restriction has a derivative at both of those inputs. The rule is fine; the stretch is what removed one side. Four standard derivatives arrive from the previous class, and this video does not take them on trust. Each one is checked against the definition, at several inputs. The power rule first: the derivative of the n-th power is n times the power below.

2:45Twenty-four checks, four exponents against six inputs, and all twenty-four agree. That one entry alone covers every polynomial. Then the sine, the cosine and the tangent. These cannot be settled on a list of fractions, because the derivative of a sine at one is the cosine of one, and that is not a fraction at all. So a second test runs alongside the first: shrink the neighbourhood until the quotient is trapped in a band narrower than the tolerance, then ask whether that band sits where it should.

3:23The sine's quotient closes on the cosine at every tested input. The cosine's closes on the sine negated. The tangent's closes on one over the cosine squared, wherever the cosine is not nought. None of the three was quoted. The three rules for combining derivatives arrive the same way, from the previous class, unproved here. So they get the same treatment. Five simple rules, every pair of them, at four inputs, with every derivative in sight taken from the definition and none of them quoted.

4:03The sum rule: a hundred pairs, a hundred agreements. The product rule: a hundred, and a hundred. And notice what the product rule is not. It is not the product of the two derivatives, which is exactly why it is worth a name. The quotient rule: eighty-five pairs, and eighty-five agreements. Fifteen more pairs were refused outright, because the divisor was nought at the input -- the same proviso that hangs off the fourth closure result, arriving again for the same reason.

4:41Now the theorem. If a function has a derivative at an input, it is continuous there. One direction only. Before the proof, it is worth asking why an implication in this direction is worth stating at all, because that question has an answer you can measure. Take fourteen rules and five inputs: seventy pairs. Some of the rules are smooth, some have a corner, some jump, one runs away, one is pinned down only by a bound, and one climbs like a square root.

5:16Every derivative and every limit below is decided by the definition, against a stated list of candidates, with the point itself never sampled. Of the seventy pairs, fifty really do have a derivative. The proof is four lines and only one of them does anything. Write the change in the output as the difference quotient multiplied by the change in the input. That is the line. It is an identity, not an approximation -- checked here at over a thousand pairs and increments, with nought disagreements.

5:56Now take the limit of the product. The second factor goes to nought because it IS the change in the input. The first factor goes to the derivative, and the hypothesis is used exactly here and nowhere else: it tells you that number is finite. A finite number times nought is nought. So the change in the output goes to nought, which is precisely what continuity at the input says.

6:25Checked directly: the change in the output goes to nought at fifty-nine of the seventy pairs, and those fifty-nine are exactly the continuous ones. Look again at where the hypothesis was used. The proof never needed the two one-sided quotients to agree. It never needed them to settle on anything. It only needed the first factor not to run away. So the theorem is true with a weaker hypothesis than the one it is stated with, and that is measurable.

7:01Fifty pairs have a derivative. Fifty-eight have a difference quotient that merely stays bounded. Fifty-nine are continuous. Every pair on a lower rung is on the one above it, and both steps are real: eight pairs have a bounded quotient without having a derivative, and one is continuous without even a bounded quotient. That last one is the rule that climbs like a square root at nought. One more thing the proof tells you, and it is easy to miss.

7:36The hypothesis is about ONE input. Not a neighbourhood of it. Here is a rule trapped between a square and the negative of a square, and pinned nowhere else. At nought it has a derivative, and the derivative is nought. At every other input tested it is not even continuous, let alone differentiable. And the theorem still delivers at the one input where its hypothesis holds: that rule is continuous at nought.

8:07A theorem about a point is a theorem about a point. Now run the implication the other way. Of the twenty pairs with no derivative, nine are continuous anyway and eleven are not. Mixed. That mixture is what makes the converse FALSE rather than merely unproved. If every non-differentiable pair had turned out discontinuous, the two ideas would have been the same idea and neither theorem would have had anything to say.

8:41Nine of them are the counterexamples, and the simplest one is the one everybody reaches for. The modulus. Continuous everywhere, and the argument for that came earlier. Now its difference quotient at nought, twice. For a small negative increment, the modulus of that increment is the increment negated, so the quotient is minus one. For a small positive increment, the modulus is the increment itself, so the quotient is one.

9:16Both are finite. They differ. First box ticked, second box not, and the derivative at nought therefore does not exist. Notice the shape of that refutation. Nothing about the function is badly behaved. One equality fails, at one input. Shift the modulus anywhere you like and the same two numbers appear at the shift -- checked at six different shifts -- while one step either side of it the derivative is minus one below and one above.

9:50The failure is at a single input, and it is not a property of writing a modulus down. Here is a rule with a modulus in it that has a derivative everywhere, the origin included: the input times its own modulus. So far the two-part condition has been taken on its word. Let us score it. The truth this is measured against is not the condition itself; it is the derivative as it is first stated: a single two-sided limit of the difference quotient.

10:27Six ways of deciding whether a function has a derivative here run side by side over the same seventy pairs. Each is scored three ways: how many it declares, how many of those declarations are wrong, and how many real derivatives it never sees. The two-part condition itself: fifty declared, nought wrong, nought missed. Perfect agreement, and that is a measurement rather than a definition restated. Now the rivals. Continuity, read as if it were the definition: fifty-nine declared, nine of them wrong, and nought missed.

11:10Look at that last number. Missing none IS the theorem -- every function with a derivative came out continuous -- and here it arrives as a count rather than as a quotation. The two slopes disagree: sixty declared, ten wrong. That reading lets through anything whose slopes do not visibly disagree, and a quotient that has run away does not visibly disagree with anything. Each half of the condition on its own: fifty-seven and seven, both times.

11:44And the hypothesis the proof actually eats: fifty-eight and eight. Every rival declares MORE than the real thing, and not one of them misses a single derivative. The two-part condition is the tightest of the six, and now you know that rather than assuming it. The second famous failure, and it fails the other half. The staircase rule, on the stretch from nought to three. At the input one the function takes the value one.

12:17Come from below with a small negative increment and the function has dropped to nought, so the change in output is minus one and the quotient is minus one divided by the increment. As the increment shrinks that grows past every bound. It is not finite. Come from above and the function is still one, so the change in output is nought and the quotient is nought. One side gives a number, the other gives nothing at all.

12:48The same thing happens at the input two. Put the two failures side by side, because they are not the same failure. The modulus passes the finiteness check and fails the equality check. The staircase fails the finiteness check -- and once one of your two quotients does not exist, there is nothing left for the equality check to compare, so it fails that one too. The modulus hands you two numbers that differ.

13:20The staircase hands you one number and a refusal. A student who has only met the first will write down two numbers at the staircase, one of which does not exist. That is the whole reason for keeping the two tick boxes apart. And there is a second difference: the staircase is not continuous at those inputs either, so it fails the theorem's conclusion as well as its hypothesis, while the modulus fails only the hypothesis.

13:54One last question, and it is a standard one. Does a function exist that is continuous everywhere and fails to be differentiable at exactly two inputs? Add two shifted moduli. Each summand is continuous everywhere, so the sum is. Between and beyond the two shifts the sum is a straight line, and a straight line has a derivative everywhere. Across a wide sweep of inputs the sum is continuous at every one of them, and it fails to have a derivative at exactly two: the two shifts.

14:32The derivative takes exactly three values across that sweep, one for each of the three straight stretches: minus two, nought, and two. And at each shift the two one-sided quotients differ by exactly two, which is the single modulus's gap of two, arriving twice. Exactly two is the part to check, and the three straight stretches are what check it. So here is the shape of it. One definition with two halves, and they fail separately.

15:07A stretch has one one-sided derivative available at each end, and it is the one pointing inwards. Four standard derivatives and three rules for combining them, all checkable against the definition rather than memorised. One theorem, one direction, and a proof whose only real step is a multiplication -- with the hypothesis used exactly once, to know a number is finite. A converse that is false, and false in a measurable way: nine counterexamples in seventy pairs, not nought.

15:46And two of them that fail for different reasons, which is worth saying out loud every time you use the words.