AA Topic 5.12 · teacher page · HL

Running first principles

The same limit as 5.1, done with algebra, and the cancellation that makes it legal.

The one thing to do with the figure

Put the 5.1 animation back on the board first.

This is deliberately the same picture, so say so. At 5.1 the limit was estimated from a table because analytic methods were not required; now they are.

The two readouts are the quotient and 2x + h, and they are identical for every h. That identity IS the algebra, shown numerically before it is proved.

Cancel first, take the limit last

Setting h = 0 at the start gives 0/0. Dividing by h is legal for every h that is not zero, and once the h is gone the limit is safe.

Students who understand that one sentence stop finding the method mysterious.

The answers

x²Quotient = 2x + h, limit 2x.
x³Quotient = 3x² + 3xh + h², limit 3x².
Higher derivatives of x⁴At x = 2: f′ = 32, f″ = 48, f‴ = 48, f(4) = 24, f(5) = 0.
1. Quotient at x = 3, h = 0.16.1.
2. Why not h = 0B, it gives 0/0.
3. f‴(2) for x⁴48.

Where the marks go

1 markExpanding correctly and showing the leading terms cancel.

1 markDividing by h before taking the limit.

1 markKeeping the limit notation until the limit is taken.

What each wrong answer tells you

They giveWhat it means
6 for Q1Gave the limit rather than the quotient at h = 0.1. Worth separating: one is exact at that h, the other is what it tends to.
Dropping limWriting the quotient equal to the derivative before h has gone. It is not, and examiners mark it.
Using the power ruleIf the question says from first principles, this scores nothing however right.
48 for f(4)Off by one. The fourth derivative is the constant 24.

Other things they will say

"Why does the x² cancel?" Because f(x+h) and f(x) share every term that has no h in it. If they did not, the quotient would blow up as h shrinks.

"What about |x| at zero?" A good question. The quotient is −1 from the left and +1 from the right, so the limit does not exist. That is what non-differentiable means, and it is why the definition needs a limit.

"Do I need this if I have the power rule?" For the exam, yes, when asked. More usefully it is the only honest answer to where the power rule came from.

A possible order

 What is happening
1Replay 5.1. Remind them the answer was estimated, not proved.
2The definition, and the x² case in full on the board.
3The animation, with both readouts agreeing at every h.
4The cubic, and the pattern towards the power rule.
5Higher derivatives and the notation.
6Questions, and the |x| aside if there is time.

Two things not to say

Do not drop the limit symbol while working on the board. Students copy what they see, and the missing lim is a real mark.

Do not present this as a formality before the real rules. It is the only derivation they will meet, and it is where the power rule actually comes from.

Questions to set

Three tiers, ramping the way practice should: the method on its own, then the method inside something real, then a challenge. Set the tier the class in front of you needs rather than one undifferentiated sheet. Answers are given so these can go straight onto a board.

1Fluency

The method on its own, with friendly numbers. Set these first and move on quickly once they are secure.

  1. Differentiate x squaredUse first principles on f(x) = x².
    ((x + h)² − x²)/h = (2xh + h²)/h = 2x + h, which tends to 2x.
  2. Differentiate x cubedDo the same for f(x) = x³.
    (3x²h + 3xh² + h³)/h = 3x² + 3xh + h², which tends to 3x². At x = 2 that is 12.

2In context

The same skill inside a real situation, where the first job is working out what is being asked.

  1. A reciprocalUse first principles on f(x) = 1/x.
    (1/(x + h) − 1/x)/h = −1/(x(x + h)), which tends to −1/x². At x = 2 that is −0.25.
  2. Why the h cancelsExplain why every first-principles calculation must cancel an h before the limit is taken.
    Before cancelling, the expression is 0/0 at h = 0 and says nothing. Cancelling produces an expression that is defined at h = 0, and its value there is the limit.

3Challenge

Reasoning, working backwards, or spotting an error. These are where the top grades are decided.

  1. State the definitionWrite the definition of the derivative and say which part students usually omit.
    f′(x) = lim as h tends to 0 of (f(x + h) − f(x))/h. The omitted part is "lim as h tends to 0": without it the expression is a chord gradient, not a derivative, and the words carry a mark.
  2. Where it failsExplain why f(x) = |x| has no derivative at x = 0 using first principles.
    The quotient is h/h = 1 for h > 0 and −h/h = −1 for h < 0. The two one-sided limits differ, so no single limit exists. A function can be continuous there and still have no derivative.

Practicalities

Works on a phone. Nothing is loaded from any other site, so it runs behind a school firewall, and nothing a student does is saved or sent anywhere.