Differentiating twice, and what the second derivative is for

0:00You already know how to differentiate. Here is the whole of a new idea in one line. The derivative of a function is itself a function. So differentiate it. That is it. There is no new rule, no new table, nothing to learn that you do not already have. You take the derivative, you look at what came out, and you differentiate that. The result is called the second derivative, and everything after this line is either a name for it, a use for it, or a warning about when it is not there.

0:36Two of those three are usually rushed. This video slows down on all three, and it measures every claim it makes rather than asserting it. Take the input cubed. Differentiate: three times the square. Differentiate again: six times the input. That is the second derivative, and you have just computed one. So why does the topic feel harder than that? Because of three things nobody says out loud, and we are going to say all three.

1:11The first is cosmetic and it costs students more time than it should. The second derivative has five different spellings, and they all mean the same object. You will meet the doubled differential form, with a two on the top and a two on the bottom in different places. You will meet the function with two dashes after its name. You will meet the operator form, where a squared symbol acts on the function.

1:40You will meet the output letter with two dashes. And you will meet the subscript form, where a small two sits under the letter. Five spellings. One object. Every one of them names the same number at the same input, and if you can compute any one of them you can compute all five, because there is nothing to compute differently. Do not let the change of costume convince you that a new idea has arrived. Nothing has arrived except a second application of something you already do.

2:17The second thing is not cosmetic at all, and it is the only real mathematics in the definition. You may differentiate the derivative PROVIDED THE DERIVATIVE IS ITSELF DIFFERENTIABLE. That clause is easy to read past. It is where the whole topic lives. A function can have a first derivative everywhere and still have no second derivative somewhere. Not at a point where it does something dramatic. At a point where it looks perfectly ordinary.

2:51Here is the cleanest example there is. Take the input times its own modulus. On the right of nought that is the input squared. On the left it is minus the input squared. The two halves meet smoothly: the function is continuous, and its graph has no corner you can see. Differentiate it. On the right you get twice the input. On the left you get minus twice the input. Both of them go to nought at the join, so the first derivative exists everywhere, and at the join it is nought.

3:28So far so good. Now differentiate again. On the right the answer is two. On the left the answer is minus two. At the join there is no single answer, and there is no second derivative there at all. The first derivative of that function is twice the modulus. And twice the modulus has a corner at nought. The corner was not in the function. The corner was in its derivative. That is exactly what the conditional clause is warning you about.

4:03Now, this video does not want you to take that on trust, and it does not take it on trust itself. Behind it is a checker that measures every derivative from the function, as a limit of a difference quotient, and never quotes a rule. To measure a second derivative it samples the function at four spaced points ON ONE SIDE of the input, and combines them with weights chosen so that the value and the slope and the next term after that all cancel, and only the second derivative survives.

4:39Then it does the same on the other side. Then it demands that the two sides agree. That last demand is the entire point, and here is what happens without it. There is a cheaper reading you could take instead: one step either side of the input and the value in the middle. It is symmetric, it is quick, and it is a trap. On the input times its own modulus, at the join, that cheap reading answers NOUGHT. It answers nought at every step size, without hesitation, and it is confidently reporting a second derivative that does not exist.

5:20It does that because the errors from the two sides are equal and opposite and cancel each other out. The symmetric reading averages away exactly the disagreement you needed to see. Asked one side at a time, the same function answers two on the right and minus two on the left, the hull of those two is four wide, and the reading is refused. And lest you think the cheap reading is merely generous, on the plain modulus, which has no first derivative at the join and so cannot possibly have a second, that same cheap reading REFUSES. Its answer multiplies by ten every time the step is divided by ten.

6:07So it accepts the function with no second derivative and refuses the function with no first. Whatever it is measuring is not what we are asking about. Two sides, asked separately, and a first derivative demanded before a second is ever reported. That is the instrument, and everything below is measured with it. The corner is not the only way to lose the second derivative, and a ladder of examples makes the point better than a sentence.

6:42Take the modulus raised to three halves. That one is smooth to the eye and it does have a first derivative at the join, which is nought. But its second-derivative readings there do not settle. They GROW as the step shrinks instead of agreeing with each other, and the reading is refused. Away from the join it is granted at once. Now climb one rung. Take the cube of the modulus.

7:11This one has a second derivative everywhere, including at the join, and the answer is six times the modulus. Notice: six times the MODULUS, not six times the input. Six at an input of one, and six at minus one too. Twelve at two, and twelve at minus two. Measured at every one of those. So the cube of the modulus survives two differentiations. Does it survive a third? Feed the measured second derivative straight back into the same instrument. At the join the reading from the right runs to six and the reading from the left runs to minus six.

7:55A hull twelve wide. Refused. That refusal is a genuine disagreement, not an instrument that failed, and you can see it in the two numbers themselves: six on one side, minus six on the other, each of them settled at once. One step to the right of the join, the same nest of measurements answers six without complaint. On the plain cube, where nothing is joined, it answers six everywhere.

8:25So here is a function with a first derivative everywhere, a second derivative everywhere, and no third derivative at one point. The conditional clause does not stop at the second floor. It applies at every floor, and something can run out on any of them. With the warnings out of the way, the routine work is genuinely routine, and it is worth seeing a batch of it. Ten functions. For each one, differentiate, then differentiate what you got.

8:59A square with a slope and a shift gives two. The twentieth power gives three hundred and eighty times the eighteenth. The input times its cosine gives minus twice the sine, minus the input times the cosine. The logarithm gives minus the reciprocal of the square. A cube times the logarithm gives the input times six logarithms plus five. The exponential times the sine of five times the input gives the exponential times ten cosines less twenty-four sines.

9:36The exponential of six times the input, times the cosine of three times it, gives the same exponential times twenty-seven cosines less thirty-six sines. The angle whose tangent is the input gives minus twice the input over the square of one plus the square. The logarithm of the logarithm gives minus one plus the logarithm, over the square of the input times its logarithm. And the sine of the logarithm gives minus the sine plus the cosine of the logarithm, over the square of the input.

10:16Ten items, three inputs each, thirty separate readings. Every single first derivative and every single second derivative in that list was confirmed against a measurement taken off the function alone. And the same thirty answers, put a relative thousandth out, survive not one reading between them. The agreement is not the tolerance being kind. There is a pattern hiding in that list which is worth pulling out, because it tells you where the work is.

10:51The first differentiation is usually easy. It is the SECOND one that catches people, and it catches them in a very specific way. Look at what the first derivative turned out to be. If it is a single lump, differentiating it again is one more move. But if the first derivative came out as a PRODUCT of two things that both move, then the second step needs the product rule, and that is where the slips happen.

11:22Of the ten, exactly four are like that. Six are not. The input times its cosine is one: its derivative carries the input times a sine. The cube times the logarithm is another: its derivative carries a square times a logarithm. And the two exponential-and-wave items are the other two. Those two need the product rule TWICE, because their first derivatives carry two product terms each. That count is not an impression. Each first derivative was split into the terms it is written as, the split was checked to add back up to the measured derivative at all thirty readings, and then both halves of every claimed product were MEASURED to move.

12:11A constant is not half of a product rule. Twenty times the nineteenth power is not a product in this sense, because the twenty does not move, and the test says so. Four of ten. If you are budgeting your attention across a set of these, that is where to spend it. Now the third thing, and it is the biggest, because it is a change of task that usually arrives with no announcement.

12:41Everything so far has asked you to COMPUTE something. Here is a question of a completely different shape. Take two constants, one on a sine and one on a cosine. Show that the second derivative plus the function itself comes to nought. Read that again. You are not being asked what the second derivative is. You are being asked to confirm that a stated relation among the function and its derivatives holds.

13:11Nothing is unknown. Nothing is being solved for. The answer is already on the page and your job is to demonstrate it. That is a different task with a different write-up, and if you have been drilled on computing derivatives it can be genuinely disorienting the first time. The method, once you see it, is completely fixed. Differentiate as far as the relation needs. Substitute. Collect. Watch it cancel. For the sine and cosine pair: differentiate twice and you get the original function back with a minus sign in front. Add the function and you have nought. Done.

13:55That is the whole technique. What makes it feel hard is not the mathematics; it is that nobody told you the question had changed. Here is the layout that turns that technique into something you can do half asleep. Take three times an exponential at one rate, plus two times an exponential at a faster rate. Show that the second derivative, less five first derivatives, plus six of the function, is nought.

14:26Differentiate once. Both exponentials come back multiplied by their own rates. Differentiate again. Both come back multiplied by the squares of their rates. Now do not substitute in a rush. Set it out as a column, one column per exponential. Under the first exponential: twelve from the second derivative, minus thirty from the first, plus eighteen from the function. Twelve, minus thirty, plus eighteen. Nought. Under the second exponential: eighteen, minus thirty, plus twelve. Nought again.

15:07Both columns vanish, so the whole thing vanishes. That layout is the thing to keep. Once you have written one of these as a column of coefficients that must sum to nought, every other one of them is the same picture with different numbers. Some of these relations arrive with a square root in the denominator, and there is a technique for those which is the real lesson of the hard ones.

15:36Take the angle whose sine is the input. Its derivative is the reciprocal of the square root of one less the square. You could differentiate that again head-on. You can, and it is unpleasant, and it is where most of the arithmetic errors live. Do not. Clear the radical first. Multiply both sides by the square root of one less the square. Now the square root of one less the square, times the first derivative, equals one.

16:09The radical has moved from a denominator to a factor, and a factor is something the product rule is happy with. Differentiate that line. The product rule gives you the square root times the second derivative, plus the first derivative times the derivative of the square root. That derivative of the square root supplies minus the input over the same square root, and multiplying the whole line through by the square root once more clears it for good.

16:43What you are left with is: one less the square, times the second derivative, less the input times the first derivative, equals nought. Clean, radical-free, and it holds at every input where the function is defined. That was measured at three of them. There is a shorter route to the same place, and it is worth having both. Go back to the cleared line: the square root of one less the square, times the first derivative, equals one.

17:16Instead of differentiating that, SQUARE it first. One less the square, times the square of the first derivative, equals one. No radical survives at all. Now differentiate. The left side is a product of two things, so the product rule gives minus twice the input times the square of the first derivative, plus one less the square times twice the first derivative times the second. The right side is a constant, so all of that is nought.

17:50Every term now carries a factor of twice the first derivative except the first one, which carries the first derivative squared. Divide the whole line by twice the first derivative. And out drops the same relation: one less the square, times the second derivative, less the input times the first, is nought. Shorter, and it is where you will most often meet the subscript notation, because writing the first and second derivatives with small numbers underneath keeps a line like that readable.

18:27The technique to take away is not either route in particular. It is this: CLEAR THE RADICAL BEFORE THE SECOND DIFFERENTIATION. Every hard item of this kind does it, one way or the other. So how much of this topic is actually that second kind of task? More than you would guess, and it is worth being precise. The checker behind this video holds ten stated relations of this shape. Three of them are the ones worked through above.

19:00The other seven are set as questions: six of the sort you meet in a numbered exercise, and one from the harder mixed set at the end. Each of the ten was tested at three inputs. Thirty readings. And here is what matters: for every reading, the function's own value, its MEASURED first derivative and its MEASURED second derivative were substituted into the stated relation and nothing else was. All thirty hold.

19:32Then the same thirty were run again with the largest term in each relation put a relative thousandth out. Not one survives. So the relations are being confirmed, not waved through. The functions themselves cover the ground: sines and cosines of a logarithm, exponentials at two rates, exponentials at equal and opposite rates, a relation that rearranges into a logarithm, the square of an inverse tangent, and an exponential of an inverse cosine.

20:09That last one takes exactly the clearing-the-radical route we just walked, which is why it is worth having both routes in hand. Now a word about what those questions actually are, because nobody usually says it and it changes how they feel. A relation among a function and its derivatives has a name. It is called a differential equation. Solving one is a large topic you will meet later. But CONFIRMING that a given function satisfies one is not large at all, and you have just done it eight times.

20:49Counting them up: six of them come from the numbered exercise, and two more from the mixed set at the end, including the circle we are coming to. Eight questions in all, of the same shape. So when you meet differential equations properly, you will have already done the verification half of the subject, without anyone telling you that was what you were doing. That is worth knowing for two reasons. It makes these questions feel purposeful rather than arbitrary. And it means that when the phrase turns up later, you will recognise the shape instead of meeting it cold.

21:31One question in this family is the odd one out, and it asks for something that catches people because it looks like a mistake. Take the angle whose cosine is the input. Find its second derivative, and give the answer IN TERMS OF THE OUTPUT. Not in terms of the input. In terms of the answer the function itself hands back. Start normally. The first derivative is minus the reciprocal of the square root of one less the square. The second derivative is minus the input over the three-halves power of one less the square.

22:12That is a perfectly good answer, in the input. Now convert. Call the output the angle. Then the input is the cosine of that angle, by the definition of the function. And the square root of one less the square of the cosine is the SINE of the angle. Substitute both. Minus the cosine of the output over the cube of the sine of the output. Which is to say: minus the cotangent of the output, times the square of its cosecant.

22:46Both forms were measured against the same function at five inputs and both agree at all five. Each of them a relative thousandth out agrees nowhere. But now the part that is easy to lose, and it is the reason this question exists. That substitution was only legitimate because of WHERE the output lives. The square root of one less the square is never negative. Writing it as the sine of the output is only right if the sine of the output is not negative either, and that is true precisely because the inverse cosine's outputs run from nought to half a turn.

23:29Read the root as MINUS the sine of the output instead, and you get an answer that is right at none of those five inputs. The branch is not decoration. It is load-bearing. Now the hardest thing in this topic, and the one place where a wrong rule is genuinely tempting. Suppose a curve is given through a parameter. The across is one rule in the parameter, the height is another rule in the same parameter.

24:01You already know the first derivative for that case. It is the height's parameter derivative over the across's parameter derivative. A ratio of the two. So what is the SECOND derivative? Here is the reflex, and it is wrong. The reflex says: the first derivative was the two first parameter derivatives divided. So the second derivative must be the two SECOND parameter derivatives divided. It is a beautiful piece of pattern-matching and it is simply not true.

24:36Here is why. The first derivative you found is a rule in the PARAMETER, not in the across. To differentiate it with respect to the across, you have to do the same thing you did the first time. Differentiate that ratio with respect to the parameter. Then divide by the across's parameter derivative AGAIN. Say it as a slogan and it will stay: DIVIDE TWICE, DO NOT DIFFERENTIATE TWICE. The second division is the chain rule doing its job, and it is the step that everybody drops.

25:15That is a claim about which rule is right, so let us not assert it. Let us measure it. Here is the difficulty in measuring anything about a parametric curve: if you check a formula using the very ratio the formula is built from, you are marking your own work. So the checker does something else. It makes the CURVE the truth. At each point it takes the across's rule and INVERTS it. Given a value of the across, it hunts the parameter that produces it by halving, on a window that has first been certified strictly one-to-one by walking a ladder of thirty-three values across it.

25:58Nothing in that certification comes from a derivative. It is the values themselves, each strictly beyond the last. What comes back is the height as a function of the across, with no parameter left anywhere in it. And then the same four-point reading used everywhere else in this video is taken on THAT. Before any of it, both claimed parameter derivatives of every curve were put to their own coordinate's difference quotient. Twenty-four checks, all passed. A control pair whose two coordinates claim each other's derivatives is refused at every one of six.

26:39Then six candidate rules were scored, over twelve points on five curves. Dividing twice takes all twelve. Dividing the two second parameter derivatives one by the other, the reflex, takes NONE. The half-measure that divides the second parameter derivative by the first one: none. Differentiating the ratio and forgetting the second division: none. The one-line form with the across's parameter derivative squared instead of cubed: none. And the one-line form done correctly, with it cubed, takes all twelve, because it IS the same rule written out flat.

27:22Both of the rules that take all twelve take none of them once their answer is put a relative thousandth out. Nothing here is being scored on a difference finer than the tolerance. There is one more thing that scoring turned up, and it is the sharpest illustration of the conditional clause in the whole video. The parametric rule divides by the across's parameter derivative. Twice. So what happens where that derivative is nought?

27:56Take a parabola given through a squared parameter, at the parameter value nought. That is the nose of the curve. There, the across stops moving. It reaches a minimum and turns round. So on every window round that point, the across takes the same value twice, and there is no way to write the height as a function of the across at all. The inversion refuses. There is no curve to differentiate, so there is nothing for any candidate to be right about.

28:29Same story on a circle at the side, where the across is at its extreme and turns round. The rule under test refuses too, and it refuses honestly: it cannot even be written down there, because it divides by nought. But now look at what the WRONG rule does at the circle. The two second parameter derivatives divided one by the other: at that point that is nought over minus four. It hands back NOUGHT, without hesitation.

29:02So the wrong rule is not merely wrong about the number. It is confident where there is no number at all. That is the difference between a formula you have memorised and a formula you understand. The proviso is not small print attached to the answer. It is part of the answer. Now something rather beautiful, which is usually set as a manipulation exercise and almost never explained. Take a circle. Not through a parameter this time; just a circle with a centre and a radius.

29:40Form this combination: one plus the square of the slope, raised to the three-halves power, divided by the second derivative. The question asks you to show that this is a constant, and that it does not depend on where the centre is. You can do that with algebra. Differentiate the circle's equation implicitly once to get the slope in terms of the two offsets. Differentiate again to express the vertical offset through the second derivative. Substitute both back into the circle's own equation, and both offsets disappear.

30:19What is left is the radius. This video did it the other way. Five circles, three places on each, fifteen readings, every slope and every second derivative measured off the arc itself. All fifteen come to minus the radius of their own circle. So it does not depend on where the centre is, which is what the question asks. And it does not depend on where you STAND on the circle either, which the question does not say and which is the more interesting half.

30:53Two of the five circles have the same radius about different centres, and they give the same number. Circles of different sizes give different numbers, and each reading names exactly one of the four sizes on offer. Here is what that combination actually is. It is the reciprocal of the curvature. Curvature measures how sharply a curve bends, and for a circle the answer had better be the same everywhere and had better depend only on the radius. It is, and it does.

31:30That is the one genuinely geometric use of a second derivative in this whole topic, and it usually goes past unnamed. Now you have its name. Which brings us to the question this topic never quite answers. What IS the second derivative for? You have been shown how to compute one and you have been shown several relations it satisfies. You have not been shown a meaning. So here are three, briefly, and one warning.

32:02First: it is a rate of a rate. If the function is a position and the first derivative is a speed, the second derivative is an acceleration. That is where the idea came from and it is still the best picture of it. Second: it is a bend. The circle just showed you that, and it generalises. Where the second derivative is large the curve turns sharply; where it is nought the curve is momentarily straight.

32:33Third: it is a relation. Almost every law of physics you will meet is a statement that some function and its first two derivatives stand in a fixed relation to each other. That is what all those verification questions were rehearsing. And the warning. There is a fourth use, and it is famous: the second derivative tells you whether a turning point is a maximum or a minimum. That is true, and it is coming, and it is NOT part of this topic. Do not import the test into these questions. Nothing here asks for it and nothing here gives you the conditions under which it works.

33:17There is a real risk of learning this topic as pure notation and never being told any of that. If you take one meaning away, take the first one: a rate of a rate. Three things to carry away. One. There is no new rule here. Differentiate, look at what came out, differentiate again. The five spellings are five spellings of one object. Two. The second derivative exists only where the first one is itself differentiable, and that is a real restriction, not a formality. The input times its own modulus has a first derivative everywhere and no second at the join. The cube of the modulus has a second everywhere and no third.

34:08Three. For a curve given through a parameter, divide twice. Do not differentiate twice. And underneath all of it, one habit: when a question hands you a relation and asks you to confirm it, that is a different task from computing a derivative. Differentiate as far as the relation needs, substitute, and set the coefficients out in a column. Everything asserted in this video was measured before it was said.

34:39A checker written before the narration takes every value from a definition rather than from a formula: every exponential and logarithm is a bracket from a truncated series with its tail bounded, every angle is found by halving, and every derivative, first or second, is the limit of a difference quotient taken from both sides separately. It records forty-eight claims, none wrong, thirteen liveness controls and four of four deliberately planted faults caught.

35:15A separate harness plants ninety-seven deliberate defects in that checker one at a time, a weight changed, a guard removed, a rule swapped for its commonest misreading, and reruns it against each. All ninety-seven are caught. Alongside them run twenty-seven rewrites that change nothing this topic can see, and none of those is wrongly caught. Differentiate the derivative. Check that you were allowed to. And when the parameter is in the way, divide again.