AA Topic 5.12 · Analysis and Approaches HL

Cancel the h, then let it go

The limit definition done with algebra rather than a table, and what higher derivatives are for.

Higher Level

At 5.1 you estimated this limit from a table, because that was all that was asked. Now you do it exactly. Shrink h and watch the chord, then read the algebra underneath that explains why the answer is exactly 2x.

h = 1.20
–the quotient
–2x + h
4.00the limit, 2x

Both readouts are the same number, which is the whole point: the quotient is 2x + h, exactly, for every h that is not zero.

The definition

f′(x) = limh→0 f(x + h) − f(x)h

Letting h be zero straight away gives 0⁄0, which means nothing. So you do the algebra first, cancel the h, and only then let it go.

f(x+h) − f(x)= (x+h)² − x² Substitute. Nothing clever yet.
expand= 2xh + h² The x² terms cancel, which is what makes this work.
divide by h= 2x + h Every term had an h in it, so the division is exact. This is the step that removes the 0/0 problem.
let h → 0= 2x Now it is safe: there is no h in the denominator any more.
so f(x) = x² has f′(x) = 2x
Cancel first. Take the limit last.

The same for a cubic

(x+h)³ − x³= 3x²h + 3xh² + h³ ÷ h= 3x² + 3xh + h² h → 0= 3x² which is the power rule, now proved rather than stated

Write "lim" on every line until you take it. Dropping the limit symbol and bringing it back at the end is a standard way to lose a mark in a "from first principles" question, because the expression genuinely is not equal to the derivative until h goes to zero.

Higher derivatives

Differentiate again, and again. The notation is dnydxn or f(n)(x), with the bracket distinguishing the fourth derivative f(4) from a fourth power.

f(x) = x⁴f′f″f‴f(4)f(5)
at x = 24x³ = 3212x² = 4824x = 48240

A polynomial of degree n has an (n+1)th derivative of zero, every time. That is worth recognising, and it is why Maclaurin series of polynomials terminate.

Continuity and differentiability, informally. A function is continuous if you can draw it without lifting the pen, and differentiable if it also has no corners. |x| is continuous at 0 but not differentiable there: the chord gradient is −1 from the left and +1 from the right, so the limit does not exist. You will not be asked to test for this formally, but the idea explains why the definition needs a limit at all.

Your turn

1. For f(x) = x², simplify f(x+h) − f(x)h at x = 3 with h = 0.1.

2. Why can you not just put h = 0 at the start?

3. For f(x) = x⁴, find f‴(2), the third derivative at x = 2.

Where the marks go

Expanding correctly, and showing the x² terms cancelling. If the leading terms do not cancel, the h will not divide out and nothing works.

Dividing by h before taking the limit, and saying so.

If a question says "from first principles", using the power rule scores nothing at all, however right the answer.

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