Taking logarithms first, when the base and the power both vary

0:00Here is a shape that no rule you have so far can touch. One expression raised to the power of another, where both of them depend on the input. Look at what you already have and watch each rule fail. The power rule differentiates the input to a fixed exponent. It wants the exponent to stand still, and here the exponent is moving. The exponential rule differentiates a fixed base to the power of the input. It wants the base to stand still, and here the base is moving too.

0:35So this is the one shape in the whole topic for which no earlier rule applies at all. And the fix is not a new rule. It is a change of clothes. Take logarithms of both sides first, and the shape you cannot differentiate turns into a shape you can. That is the whole technique, and it costs one condition, which is the part that gets marked. Call the whole thing y, and take the natural logarithm of both sides.

1:08On the right, the exponent comes down as a factor. That single move is the reason any of this works. A power has become a product. A product is a shape the rules in your hands handle perfectly well. And there is a second use of the same move, which we will come back to: a long product or quotient becomes a sum of logarithms, and a sum is easier still.

1:35So logarithmic differentiation is not a new differentiation rule. It is a rewriting, followed by the rules you already have. Nothing is being assumed here that the logarithm laws did not already give you. Notice what has to be true for that second line to exist at all. You have taken the logarithm of y, so y has to be positive. And you have written the logarithm of the base, so the base has to be positive too.

2:08Hold on to that. It arrives again later as a measurement. Now differentiate both sides, and look carefully at the left. The left side is the logarithm of an unknown function, so it needs the chain rule. The logarithm on the outside gives one over its input, and its input is y, so you get one over y. The chain rule then multiplies by the derivative of the inside. So the left-hand side is the derivative of y divided by y.

2:44It is not the derivative of y. That is the single commonest error in this whole technique, and it is worth naming out loud. You differentiate, you get something clean on the right, you write it down as your answer, and you have forgotten to multiply back by the function. The quantity on the left has a name worth having: the logarithmic derivative. It is the rate of change of a quantity as a fraction of the quantity itself.

3:15It is a perfectly good thing to want. It is just not what the question asked for. Everything from here on ends with the same last step: multiply by y again. Do that once in general and you have the formula for the whole shape. The right-hand side needs the product rule, because it is now the exponent times the logarithm of the base, and both of those move. Two terms come out of it.

3:46The first is the exponent, times the base's derivative, over the base. The second is the exponent's derivative, times the logarithm of the base. Multiply both by the function itself and you have the answer. Now read those two terms for what they are, because every later item is these same two terms with different contents. The first term is the contribution of the moving base. It is what you would get if the exponent stood still.

4:19The second term is the contribution of the moving exponent. It is what you would get if the base stood still. Which means the two rules you started with, the two rules that each failed on their own, are not rivals of this formula. They are its two halves. That is a strong claim, so it was measured rather than asserted. Five bases and five exponents were combined in every way, at four inputs, giving ninety cases where everything is defined.

4:53Every one of the ten pieces had its own derivative checked against its own difference quotient first, at every input it is used at: thirty-eight checks, all of them clean. That check is not a formality. The same sine offered with its derivative claimed to be the sine itself is refused at all four of its inputs. Then the power rule alone and the exponential rule alone were added together and compared with the two-term formula.

5:26They agree at every one of the ninety cases. The two halves really do add to the whole. So each half should be right on its own exactly where the other half contributes nothing, and that is exactly what the numbers say. The power rule alone is right at twenty-two of the ninety, and those twenty-two are precisely the cases where the moving exponent's term comes to nought. The exponential rule alone is right at four, and those four are precisely the cases where the moving base's term comes to nought.

6:03Both of those are exact set identities, with nothing on either side that the other does not have. Now put the formula itself on trial, because a rule you are told is correct is a rule you have not tested. Six ways of differentiating this shape were scored by identical machinery against each item's own difference quotient. The two-term formula, the power rule alone, the exponential rule alone, the two terms multiplied instead of added, the two derivatives written into each other's term, and the logarithmic derivative reported as the answer.

6:47Out of ninety: ninety, twenty-two, four, two, eighteen, and eight. The formula answers all of them. Nothing else comes close. And the test that accepts it is not accepting everything. The same answer a relative thousandth too large, or with the sign turned over, or a relative thousandth adrift, leaves exactly one case standing in each. That one case is the input raised to its own logarithm, at an input of one, where the base is one and the exponent is nought, so both contributions vanish at once and the curve is level.

7:29A relative twist of nought is still nought. An absolute thousandth is refused there. But here is the number that matters most, and it is about how you learn this rather than about the rule. At forty-one of the ninety cases, at least one of the five wrong rules is right as well. Only forty-nine of them rule out all five at once. So an example chosen because it is easy to differentiate is very often an example that cannot tell the formula from its own first half.

8:06Now the condition, which is the most load-bearing sentence in the whole technique and the easiest to wave away. The base and the whole expression have to be positive. This is not a technicality bolted on at the end. It is the condition under which the second line of the derivation exists. You took the logarithm of the left side, so the left side had to be positive. You wrote the logarithm of the base on the right, so the base had to be positive.

8:40Take that away and there is nothing to differentiate, because there is nothing to write down. And it is worth seeing that the requirement is real rather than cautious. Minus one to the power of a half is not a number at all. There is no small adjustment that rescues it. The whole construction has gone. So the condition is doing real work. The question worth asking is whether it is doing exactly the right amount.

9:13Four readings of it were scored the same way, against whether the item has a slope at all, over a hundred and fifty-six cases built to include negative bases and a base that reaches nought. The reading as stated, that the base is positive, declares ninety-six of them and is wrong about none. Not one item it lets through fails to be a number. But it turns away twenty-six that have a slope all the same.

9:43Drop the condition entirely and you declare all hundred and fifty-six and are wrong about twenty-two. Ask only that the base is not nought and you declare a hundred and forty-eight and are wrong about eighteen. Demand a positive exponent as well and you buy nothing and cost yourself one more. So the condition is sufficient, and it is not necessary, and those are two different columns rather than one word.

10:14What it turns away is a negative base carrying a whole exponent. The input less five, cubed, has a perfectly good slope where the base is minus one, because a whole power is repeated multiplication and needs no logarithm at all. The same base to the power of a half has no value there. There is a subtler thing hiding in that population, and it is worth a minute. Of the hundred and fifty-six cases, a hundred and thirty-four have a value.

10:49Only a hundred and twenty-two have a slope. Twelve of them are numbers at a point and have no derivative there at all. And those twelve are exactly located, not approximately. They are the cases whose base is nought, and the cases whose exponent is a whole number at that one input and at neither input beside it. Take the input less five, raised to the input itself, at an input of four.

11:18The base is minus one and the exponent is four, so the value is one. That is a genuine number. But move a hair either way and the exponent is no longer a whole number, and a negative base to a fractional power is nothing at all. There is no neighbourhood for a difference quotient to live in. A value at a point is not a slope at that point, and that is why the condition asks about an interval rather than about a point.

11:51Which brings us to the thing worth doing for yourself on every one of these items. Supply the interval. A drill list of these will often give you eighteen expressions to differentiate and attach a domain to none of them. The logarithm of the input, raised to something, needs the input above one, because the logarithm has to be positive before it can be a base. The input raised to the sine of the input needs the input above nought.

12:24The sine raised to the sine needs an interval on which the sine itself is positive. A bracket like the input less three, raised to anything, needs the input above three. None of that is decoration. Without it the second line of your own working does not exist. Write the interval down first, before you differentiate anything. It is the part a marker is looking for. Now the second use, and it is the one that shows what the technique really is.

12:59Nothing here is raised to anything at all. Take a long product, or a quotient, or a root of a quotient, with several factors. Differentiating that directly means the product rule applied over and over, and a page of algebra. Take the logarithm first and the product becomes a sum of logarithms, the quotient becomes a difference, and a root becomes a factor of a half in front of everything.

13:29Then differentiate a sum, term by term, which is the easiest thing in calculus. Each term becomes that factor's derivative over that factor. And at the end, the same last step as always: multiply back by the original expression. This is the clearest demonstration that logarithmic differentiation is about representation and not about powers. Run one in full. A square root, of a quotient, with a linear factor and a quadratic on the top and another quadratic underneath.

14:09The root halves everything. The quotient turns into a difference. The product on top turns into a sum. So the logarithm of the whole thing is one half of the logarithm of the first factor, plus the logarithm of the second, minus the logarithm of the third. Differentiate that and you get one half of a bracket holding three reciprocal-shaped terms. One over the linear factor, twice the input over the first quadratic, and the derivative of the last quadratic over itself.

14:45Multiply back by the original expression, and you are done in four lines. That answer was put to the item's own difference quotient at four inputs and survives all four. And the same answer with the final multiplication left out survives none of the four. That is what the forgotten step costs, measured rather than scolded. Here is a case where logarithms are the shorter path and not the only one, which is worth knowing.

15:18A positive constant, raised to the power of the input. By logarithms: take the logarithm, the input comes down times the logarithm of the constant, differentiate, multiply back, and the answer is the function times the logarithm of the constant. By rewriting: a positive constant is the natural exponential of its own logarithm. So the whole thing is the natural exponential of the input times the logarithm of the constant, and the chain rule finishes it in one line.

15:57Three constants at three inputs, nine pairs. Both routes answer all nine, and the two answers bracket each other everywhere. The second route is worth the time, because that rewriting is the standard way to handle any constant base, and because it shows the logarithms here were a convenience rather than a necessity. Now the shape we started with, with something in it. The input, raised to the sine of the input, for positive inputs.

16:31The general formula gives the function, times a bracket holding the sine over the input plus the cosine times the logarithm. That is a correct answer and you can stop there. But multiply the bracket out and something becomes visible. The first term is the input to the power one less than the sine, times the sine. The second is the original function, times the cosine, times the logarithm. There they are again: the moving base's contribution and the moving exponent's, standing apart.

17:09Both forms were measured separately at four inputs. Each survives the item's own difference quotient every time, and the two bracket each other every time. The expanded form is worth writing out, because it is the only place those two contributions stand where you can see them. And notice that the answer still contains the original function as a factor. For a variable power it usually will, and that is not a sign you have gone wrong.

17:43Here is the item that makes the case for the technique better than any argument could. One plus the input, times one plus its square, times one plus its fourth power, times one plus its eighth power. Find the slope at an input of one. Multiply that out directly and you have a polynomial of degree fifteen. By logarithms it is four terms: for each factor, that factor's derivative over that factor.

18:15At an input of one those four terms are a half, one, two and four. They sum to fifteen halves. The product itself is sixteen there. Sixteen times fifteen halves is a hundred and twenty. That number was then checked two more ways that share no line of working with the first. The item's own difference quotient settles on a hundred and twenty. And those four factors multiply out to the sum of the first sixteen powers of the input, checked at four inputs.

18:49The slope of that sum at one is one plus two plus three, all the way up to fifteen. Which is a hundred and twenty. Four lines, against a page. One thing you cannot do, and it is worth being blunt about. You cannot usefully take the logarithm of a sum. There is no law that turns the logarithm of a sum into anything shorter. Over twenty pairs of positive fractions, the logarithm of the sum agrees with the sum of the logarithms not once.

19:24Over the same twenty pairs, the logarithm of the product agrees with the sum of the logarithms every single time. That is the difference between the two, as a count. So when you meet three variable powers added together and set equal to a constant, you cannot start by taking logarithms. You name the three summands first. You differentiate the sum, which gives you three derivatives adding to nought. Then you go after each summand separately, by logarithms, and put the three results back.

20:01The splitting is not tidiness. It is the only legal order of operations. Sometimes the derivative you are looking for turns up on both sides of your own working, and then there is one more algebraic step: collect it. Take one letter raised to the power of the other, set equal to the second raised to the power of the first. Take logarithms of both sides and the exponents come down: the second times the logarithm of the first equals the first times the logarithm of the second.

20:40Differentiate, and the unknown derivative appears in two places, once on each side. Gather it onto one side, divide, and you have it. Now, that answer was not checked against the algebra that produced it, because that would only be checking arithmetic. Instead, at three inputs, the second letter was hunted from the relation itself, by halving, down to a bracket narrower than a ten-thousand million million millionth. The branch found is not the obvious one where the two letters are equal. It really does satisfy the relation, at all three.

21:21The derivative of that hunted branch was then taken as a limit of difference quotients, and compared with the collected formula. They agree at all three. So the collecting step is right, and it is right against a truth that never touched the algebra. One last thing, and it is a small lesson in reading rather than in calculus. Two letters written side by side, and two letters with one raised to the power of the other, look very similar on a page and mean entirely different things.

21:58Both were worked through. Read as a power, taking logarithms turns the relation into a linear equation, the second letter comes out in closed form as the first over one plus its logarithm, and the slope is the logarithm over the square of that same bracket. Read as a product, nothing solves for the second letter, so it has to be hunted, and the slope is a different expression entirely.

22:29Each survives its own difference quotient at all three inputs. Each satisfies its own relation at all three. And the two answers agree at none of the three. So the reading is a choice, not a detail, and it is worth a second look before you start differentiating. One more small thing while we are here. In a sum of terms, watch for the one that is a constant raised to a constant power.

22:58Its difference quotient settles on nought at every input, while the same test refuses a thousandth. It is there to be recognised and dropped, not differentiated. Three things. First, this is not a new rule. It is a rewriting, followed by the rules you already have, and it ends with a multiplication you must not forget. Second, the two terms of the formula are the two rules you started with, and each of them is right on its own exactly where the other contributes nothing.

23:36Twenty-two cases out of ninety for one, four for the other, and at forty-one of the ninety some wrong rule looks right too. One worked example will not tell you which rule you have. Third, the condition is the mathematics, not the small print. It declares ninety-six items of a hundred and fifty-six and is wrong about none of them, and it turns away twenty-six that were fine. Sufficient, and not necessary, and you should be able to say which of those you are relying on.

24:11Write down the interval before you write down the derivative, and the hardest mark on the page is already yours.