The power rule, and a polynomial's derivative assembled out of it and the sum rule

0:00By now the definition works, and it works on anything. Form the difference quotient, let the step shrink, read what it settles on. That is the whole machine. But it is slow. Doing x to the fortieth power that way means expanding x plus h to the fortieth power by hand. So here is the claim: for any whole number n, the derivative of x to the n is n, times x to the n minus one.

0:30Bring the exponent down in front, and knock one off it. You already know two cases. The derivative of x is one. The derivative of x squared is two x. Check them against the claim. n equals one gives one times x to the zero, which is one. n equals two gives two x. Both fit. Two cases is not a proof. What follows is two proofs - and the interesting thing about them is that they have almost nothing in common.

1:04First proof. Go back to the definition and just do the work. The quotient is x plus h, all to the n, minus x to the n, over h. That first piece expands. Every power of a sum does, and the coefficients come from a triangle you build by adding neighbours. For the sixth power the row reads one, six, fifteen, twenty, fifteen, six, one. Each of those multiplies a power of x and a power of h, with the powers of h climbing from nothing up to n as the powers of x fall away.

1:43So the expansion has seven terms when n is six. In general, n plus one. And the very first of them - the one where h has not appeared at all - is x to the n. Which is exactly what the definition then subtracts. That subtraction is the whole proof, so watch it carefully. The first term goes. Seven terms become six. And every one of the six that is left carries at least one factor of h, because h has appeared in all of them.

2:18So h comes out as a common factor - and then the definition divides by h, which cancels it. Now look at what that division did. Every term dropped one power of h. The term that carried exactly one h now carries none. Every other term still carries at least one. Six terms. One of them free of h, five of them still holding on to it. Send h to nothing and those five die. One term survives.

2:49And which one is it? The term that was multiplying the first power of h - which is the second entry in the row. For the sixth power, the second entry of the row is six. Not because anyone brought an exponent down. Because that is what you get when you build the triangle by adding neighbours. The multiplier in the answer is a number the triangle hands you, and the recipe is a description of what it hands you.

3:20That is worth saying plainly, because the recipe is usually taught as the reason, and it is not. It is the consequence. Ask the same question at every power up to the fourteenth and read off what the argument produces: not what it should produce - what it does. The first power gives a thing of degree zero with a one in front. The second, degree one with a two. The third, degree two with a three.

3:50All the way to the fourteenth: degree thirteen, with a fourteen in front. Nothing there was written down in advance. It was read off the answer. Second proof. It expands nothing at all. Start at the bottom. The derivative of x is one - and that is not assumed, it comes straight out of the definition. Now suppose you already know the answer for the power below. Write x to the n as x, times x to the n minus one.

4:23That is a product, and there is a rule for products: differentiate the first times the second, plus the first times the second differentiated. The first term gives one, times x to the n minus one. That is one copy of x to the n minus one. The second gives x, times what we already know - which is n minus one copies of x to the n minus two, and x times that is n minus one copies of x to the n minus one.

4:58So: one copy, plus n minus one copies. One plus n minus one is n. That arithmetic is the entire content of the step. Each rung reaches the next by the product rule and the rung below it. Nothing is ever expanded. Two arguments. Do they actually deliver? Take fourteen powers, read each at thirty-seven places - five hundred and eighteen readings. For every one of them, work out the derivative from the definition, with no rule of any kind involved.

5:35Then ask each proof what it predicts there. The induction: five hundred and eighteen agreements. Zero disagreements. The expansion: five hundred and eighteen agreements. Zero disagreements. Three routes, one answer, everywhere. And two of the three never mention the other's equipment. That last point is worth pressing, because the two proofs are usually presented as though they were the same argument told twice. They are not. Look at what each one consumes.

6:11To reach the fourteenth power, the induction takes thirteen product-rule steps and never expands anything. The expansion reads a row of fifteen entries and never uses the product rule. One needs a theorem about expanding powers of a sum. The other needs a theorem about differentiating products. Neither needs the other. They meet only at the answer - and they do meet: the two polynomials they produce are the same object.

6:43At the very bottom the induction takes no steps at all, because the definition has already answered. Now, agreeing five hundred and eighteen times only means something if disagreeing was possible. So here are three near-misses. First: bring the exponent down, but forget to knock one off it. So n times x to the n. Through the identical scoring: twenty-seven agreements, four hundred and ninety-one disagreements. Second: reduce the exponent correctly but use a multiplier one too small. Thirteen agreements, five hundred and five disagreements.

7:27Third: drop the exponent and use no multiplier at all - just x to the n minus one. Fifty agreements, four hundred and sixty-eight disagreements. All three are wrong in different places and different amounts. Which is what makes the two zeros above measurements rather than decoration. Everything so far needed n to be a whole positive number. The expansion needs it; you cannot expand a power of a sum otherwise.

8:01But the rule is normally used far beyond that - on one over x, on square roots, on anything with an exponent. That extension is usually stated and not proved. It is worth being honest that it is being taken on trust. We can at least measure part of it. Here is the honest way to do that. Take one over x to the n, differentiate it at six places by the definition, and then go looking for a single multiplier and a single exponent that reproduce all six readings.

8:37Not check a guess. Search - and refuse if nothing fits. One over x gives a multiplier of minus one and an exponent of minus two. One over x squared gives minus two, and minus three. Then minus three and minus four. Minus four and minus five. Minus five and minus six. Minus six and minus seven. Bring the exponent down, knock one off. The same pattern, on the other side of zero, found rather than assumed.

9:12Now polynomials, and this is the part where a long formula turns out to be no new fact at all. A polynomial is a sum of constant multiples of powers. You already have a rule for a sum: differentiate the pieces and add. And you already have a rule for a power. Put them together and the derivative of a polynomial is: each coefficient multiplied by its exponent, each exponent knocked down by one.

9:43That is not a theorem to memorise. It is two things you have, used one after the other. Scored on four polynomials at six places each - twenty-four readings - term-by-term against the definition: twenty-four agreements, zero disagreements. One thing in that sentence is easy to skip past. The constant term disappears. It has no x in it, so it does not change, so its rate of change is zero. The definition gives zero for a constant at every place you ask.

10:21Watch what happens if you leave it in anyway - carry the constant term through instead of dropping it. Six agreements, eighteen disagreements. And the six are not luck: they are the one polynomial of the four whose constant term was already zero. So the constant is not being dropped by convention. It is being dropped because it is nothing. A worked one. Six x to the hundredth, minus x to the fifty-fifth, plus x.

10:54Nobody is expanding x plus h to the hundredth power. Take it term at a time. Six times a hundred is six hundred, exponent drops to ninety-nine. Minus one times fifty-five is minus fifty-five, exponent drops to fifty-four. And x becomes one. So the answer has degree ninety-nine, and exactly three of its coefficients are not zero: six hundred, minus fifty-five, and one. One line of work for a hundredth power. That is what the rule bought.

11:33Another. One plus x plus x squared plus x cubed, all the way up to x to the fiftieth. Differentiate term by term: one, plus two x, plus three x squared, up to fifty x to the forty-ninth. Fifty terms, the highest exponent forty-nine. The constant vanished, which is why there are fifty and not fifty-one. Now evaluate it at x equals one. Every power of one is one, so every term collapses to its own coefficient, and the answer is one plus two plus three, all the way to fifty.

12:13That sum is one thousand two hundred and seventy-five. You can get it by adding the fifty numbers up, or by the shortcut - fifty times fifty-one, halved. Both give one thousand two hundred and seventy-five, and so does evaluating the polynomial. The bookkeeping is the question. That is the point of it. Here is one that looks harder than it is. Take x to the k, divided by k, for k running from a hundred down to two. Then add x, and add one.

12:49Every term is a power divided by its own exponent. So when you differentiate, the exponent comes down and cancels the division exactly. Every term becomes a plain power with nothing in front of it. The derivative is x to the ninety-ninth, plus x to the ninety-eighth, all the way down to x, plus one. A hundred terms, and the number of them with anything other than a one in front is zero.

13:20At x equals one, every term is one, so the total is a hundred. At x equals zero, every term with an x in it vanishes, and only the final one survives. The total is one. So the derivative at one is exactly a hundred times the derivative at zero, and you never had to write out the hundred terms to see it. Two more shapes worth recognising. First: a polynomial whose coefficients are themselves powers of some fixed number.

13:55Take x to the fifth, plus three x to the fourth, plus nine x cubed, plus twenty-seven x squared, plus eighty-one x, plus two hundred and forty-three. The threes are constants. They are not the variable, so nothing happens to them - they ride through untouched. The derivative comes out five x to the fourth, plus twelve x cubed, plus twenty-seven x squared, plus fifty-four x, plus eighty-one - and the two hundred and forty-three is gone.

14:29Second: something that does not look like a polynomial at all. x to the fifth minus thirty-two, over x minus two. Divide it out. It goes exactly - remainder nothing - leaving x to the fourth, plus two x cubed, plus four x squared, plus eight x, plus sixteen. That is a polynomial, so the rule applies. Differentiating the quotient and differentiating the original ratio give the same readings: two, twenty-six and one hundred and ninety-four.

15:07And the shapes with negative exponents, which is where the extension earns its keep. Something like one over x cubed, times five plus three x. You could reach for the product rule. But multiply it out first and it is just a sum of two powers - five over x cubed, plus three over x squared. Now it is a polynomial in disguise and every term goes through the rule separately.

15:35Three such expressions, each differentiated whole and again as separate powers added together, at every place: a hundred and eight agreements, zero disagreements. Three readings refused - one for each expression, all at x equals zero, because that input is not in the domain. Which is not something differentiating did. Each of those expressions already had nothing at zero. Expanding first is not a trick. It is the shorter route, and it lands in the same place.

16:13Step back and look at what actually carried the weight here. One proof needed a theorem about expanding a power of a sum. The other needed a theorem about differentiating a product. Neither one needed the answer written down first, and neither one is a restatement of the other. Everything after that was assembly. The polynomial rule is the power rule plus the sum rule, in that order, with nothing new added.

16:44The exercises are the same assembly with the pieces arranged differently: divide first, or multiply out first, or notice that a coefficient cancels an exponent. And the one genuine gap - exponents that are not whole numbers - is a place where the rule is used far beyond what either proof establishes. So: the derivative of x to the n is n x to the n minus one, and you now have two independent reasons to believe it.

17:18The expansion argument says the multiplier is the second entry of a row of the triangle, and that entry is n. The induction argument says one copy plus n minus one copies is n copies. Both were scored against the definition over five hundred and eighteen readings without a single disagreement, and three near-miss versions of the rule were scored on the identical line and broke. The polynomial theorem needed no separate proof, and the constant term goes because it is nothing, not because someone said so.

17:58The extension below zero was searched for rather than assumed, and found. What is worth keeping is not the recipe. It is that bringing the exponent down is something that happens to be true, for a reason, twice over.