The chain rule, and differentiating a relation without first solving it for y

0:00Here is a rule: take the input, double it, add one, and cube what you get. There is a way to differentiate this that needs nothing new. Multiply the bracket out. Twice the input plus one, all cubed, is eight cubes plus twelve squares plus six inputs plus one. That is an ordinary polynomial, and you already know how to differentiate one of those. Term by term: twenty-four squares, plus twenty-four inputs, plus six.

0:32Both of those identities were checked at twenty-five different inputs, exactly, with no exceptions. And now look at the answer, because it factorises. Twenty-four squares plus twenty-four inputs plus six is six times the square of the bracket you started with. Six, times the square of twice the input plus one. The bracket has come back. That factorisation is the whole of what happens next, so it is worth reading slowly.

1:07The rule was built in two stages. First, twice the input plus one. Then, cube it. Call the first stage the inner rule and the second the outer rule. The derivative of cubing is three times a square. The derivative of twice the input plus one is two. Three times the square of the bracket, times two, is six times the square of the bracket. Which is the answer the expansion gave.

1:36So the answer is the outer derivative, taken at what the inner rule produced, multiplied by the inner derivative. One factor for each stage. That is the rule, and this is the honest account of where it came from. A case was chosen that could already be done another way, the two answers agreed, and the general rule was then stated. The proof is normally skipped. This is not a complaint.

2:07It is a good way to meet a rule, and it is exactly the move that licenses the second half of this video. But a pattern that has been checked on one case is a pattern, not a theorem. So before using it anywhere, it is worth asking what a check like that can actually rule out. The way to find out is to write down the wrong rules as well, and score them all the same way.

2:36There are two derivatives in play and a small number of ways to combine them. The first is the rule itself: the outer derivative at the inner output, times the inner derivative. The second is the commonest mistake in this whole topic: the outer derivative taken at the input, rather than at what the inner rule made of it. The third exchanges the two rules altogether. The fourth keeps only the outer derivative and forgets the inner one.

3:08The fifth keeps only the inner one. And the sixth adds the two factors instead of multiplying them. Six candidates, one of them right. Now they get scored, and none of them is scored against the rule. They are scored against the composite's own derivative, taken from the definition, by narrowing a neighbourhood until the difference quotient is pinned finer than the tolerance. Seven outer rules, six inner rules, five inputs: two hundred and five composites where everything is defined.

3:46Not one derivative anywhere in this is quoted. Every outer and inner derivative used to build a candidate was itself checked against the definition, at every input it was used at -- a hundred and fifteen checks, all of them clean. So here is the table. The rule, as stated: right at two hundred and five, wrong at none. The outer derivative at the input instead: right at a hundred and nine.

4:16The two rules exchanged: a hundred and eight. The outer derivative alone: eighty-four. The inner derivative alone: forty-one. And adding rather than multiplying: seven. Look at that second row again. The commonest error in the topic gives the right answer at more than half the cases tested. That number is not an accident, and it is not noise. Every one of those rivals is right for exactly one reason, and the reason can be written down.

4:52Taking the outer derivative at the input is right precisely where the outer derivative happens to take the same value at the input as at the inner output. Those two sets of cases are the same set: there is not one case in two hundred and five where it is right for any other reason. Dropping the inner factor is right precisely where that factor is one -- or where the factor it kept is nought, which hides the loss just as well.

5:23Dropping the outer factor, the same way round. And adding is right precisely where the sum of the two factors happens to equal their product. So each mistake has a hiding place, and the hiding places are large. If your inner rule is the input itself, or one more than the input, or anything whose derivative is one, several wrong rules will agree with you all afternoon. Which brings back the question about checking a rule on one case.

5:57At a hundred and forty-six of the two hundred and five composites, at least one wrong rule gives the right answer as well. Only at fifty-nine does a single case rule out all five at once. The case that got chosen at the start of this video is one of the good ones: across its five inputs, not one wrong rule survives it. But that is a fact about the case and the inputs, not about the method.

6:27Take that same composite at the single input minus one, and two of the five wrong rules come through unscathed. So a rule verified on one example is a rule you have not yet tested. That is not a reason to distrust this one. It is a reason to know that the trusting was done on your behalf, and to be able to say which cases would have caught a wrong version.

6:55The rule extends, and it extends in the obvious way. Three stages: an innermost rule, a middle one, an outer one. The derivative is a product of three factors, one for each stage, each taken at whatever the stage below it produced. That was checked on a hundred and six three-stage composites, and it agreed at every one. And every one of the three factors is doing work. Drop the middle factor and only four of the hundred and six survive.

7:29Drop the innermost and forty-one survive -- more, because so many inner rules have derivative one, which is the hiding place again. The pattern is one factor per stage, and you can go as deep as you like. In practice the whole difficulty is naming the two rules before you differentiate anything. Take the sine of a square. The inner rule is squaring; the outer rule is the sine. The outer derivative is the cosine, taken at the square -- not at the input.

8:05The inner derivative is twice the input. So the answer is twice the input, times the cosine of the square. Checked at five inputs against the definition, and right at all five. One habit is worth building here. If you introduce a letter for the inner output, substitute it back at the end, so the answer is written in the letter you were asked about. Leaving the working letter in the answer is not a mathematical error, but it is an unfinished answer.

8:40The drill is always the same three steps: name the inner rule, name the outer rule, and take the outer derivative at the inner output. A sine of a square plus five. A cosine of a sine. A sine of a linear expression. A cosine of a square. Each of those is one application of the rule and nothing else. Some of them need a second rule as well -- a product, a quotient, or a third stage -- and it is worth knowing which before you start.

9:13Twenty of these answers were compared against each one's own derivative, taken from the definition, and all twenty agreed. The same test, handed each answer with a fifth added to it, refused every single one -- so it is a test and not a rubber stamp. Now the second half, and it starts with two equations that look alike and are not. The first: the first letter minus the second letter equals a constant.

9:45The second: the second letter, plus the sine of the second letter, equals the cosine of the first. The first can be rearranged in one step. The second cannot be rearranged at all. There is no formula, in any of the functions you have met, that gives the second letter in terms of the first. And yet the second equation clearly says something about how the two letters move together.

10:13So the question is whether you can get a derivative out of an equation you cannot solve. Before hunting a derivative, check there is something there to differentiate. Fix the first letter and ask how many values of the second answer the equation. For that second relation, across a window either side of nought, the two sides cross exactly once -- at every one of thirteen different inputs tested. Exactly one value answers it, every time.

10:46So the relation does determine the second letter. It just refuses to say so with a formula. That is worth saying out loud, because it is the difference between a curve you cannot write down and no curve at all. And it is why the derivative is a fair thing to ask for. Here is the whole of the new method, and it is one permission. The second letter is allowed to be a function of the first, even though you cannot write it down.

11:19So when you differentiate a term containing it, the chain rule applies to it like anything else. The derivative of the second letter is the thing you are hunting. The derivative of the square of the second letter is twice the second letter, times the thing you are hunting. The derivative of the sine of the second letter is the cosine of the second letter, times the thing you are hunting.

11:46That factor is the chain rule, and forgetting it is the only mistake this method really has. Everything after that is collecting terms. Take the easy relation first, precisely because the answer is already known. The first letter minus the second equals a constant. Rearranged, the second letter is the first letter minus that constant, so the derivative is one. Now do it the other way, without rearranging. Differentiate both sides with respect to the first letter.

12:23The first letter gives one; minus the second letter gives minus the hunted derivative. And the constant on the right gives nought. Not the constant -- nought. So one minus the derivative is nought, and the derivative is one. Both routes, same answer, and the second route never needed the rearrangement. Now the one that cannot be rearranged. The second letter, plus its sine, equals the cosine of the first.

12:55Differentiate everything with respect to the first letter. The second letter gives the hunted derivative. Its sine gives the cosine of the second letter, times that same derivative. The cosine of the first gives minus the sine of the first. So the derivative appears twice on the left and nowhere on the right. Factor it out: the derivative, times one plus the cosine of the second letter, equals minus the sine of the first.

13:28Divide. The derivative is minus the sine of the first letter, over one plus the cosine of the second. That answer comes with a condition, and the condition is not handed down from anywhere. You read it off the denominator. One plus the cosine of the second letter is nought exactly where that cosine reaches minus one. Sweep the second letter across eight hundred and one inputs, a fiftieth apart, from minus eight to eight.

14:02The denominator comes within a hundredth of nought at twenty-eight of them, and every one of those twenty-eight sits beside a half turn or its negative. Nowhere else. The half turn itself is never named here -- it is pinned by the cosine crossing minus one inside a bracket a hundred million millionths wide. So the condition is: the second letter avoids the odd multiples of a half turn. Attach a condition when a denominator earns one, and not by reflex.

14:38The easy relation above has no denominator that can vanish, and so needs no condition at all. And now the same treatment as the first half, because this method deserves it too. Eight relations, at twenty-four points, and at every point the second letter was found from the relation itself by bisection -- squeezed into a bracket narrower than a ten-thousand millionth. That bracket is the truth here. The derivative of the curve is taken from those brackets, by the definition, and never from the algebra that produced the answer.

15:19The method scores twenty-four out of twenty-four. Forgetting the chain rule on the second letter scores six right and ten wrong -- and at eight more points it gives no answer at all, because the derivative you were hunting drops out of the equation entirely. That is the luckier failure: you notice it. And the six it gets right are exactly the two relations where the second letter never sits inside anything, so there was no chain to forget.

15:50Reading the derivative of a constant as the constant scores nought right and twelve wrong. One more thing about that answer, because it looks unfinished and is not. Minus the sine of the first letter, over one plus the cosine of the second. Both letters are in it. For a relation you cannot solve, that is normally the only available form, and it is a complete answer. To evaluate it at a point on the curve you need both coordinates of that point, which you have, because a point on the curve is what you were given.

16:30An answer in both letters is not a failure to finish. It is the shape the answer has. The hardest version of this adds one more move. The cosine of the second letter equals the first letter, times the cosine of a fixed angle added to the second letter. Differentiate both sides, and the hunted derivative appears on both. Substitute for the first letter from the original relation, collect over a common denominator, and the numerator that appears is the sine of a difference of two angles.

17:08That difference collapses, by an identity, to the sine of the fixed angle alone. The derivative is the square of the cosine of the fixed angle added to the second letter, over the sine of the fixed angle. Checked at three points against the curve's own derivative, found by bisection: right at all three. It is the same method throughout -- differentiate, collect, solve -- with a trigonometric identity doing the tidying at the end.

17:42So: name the inner rule and the outer rule before you differentiate anything. Take the outer derivative at the inner output, and multiply by the inner derivative. One factor per stage, however deep it goes. Substitute back, so the answer is in the letter you were asked about. For a relation, the one new permission is that the second letter may be differentiated, and doing so leaves the derivative you want sitting in the equation.

18:14Collect it, solve for it, and read any condition off the denominator that appears. An answer with both letters in it is finished. And when someone shows you a rule checked on one example, you now know the question to ask: which wrong rules would that example also have let through?