Differentiation: Definition and Fundamental Properties
~3 weeks · September · 5–10% · taught at double speed in BCThe limit machinery is built. Now it gets used for the thing it was built for, and the course starts paying.
This unit is the highest-leverage one in BC. Everything downstream — optimization, integration by substitution, differential equations, Taylor series — is either an application of these rules or a reversal of them. Fluency here is worth more than fluency anywhere else, and gaps here compound for the rest of the year.
The definition
The two words explain the shape of the definition, and they're Latin:
- Secant (secare, to cut) — a line that cuts through the curve at two points. Its slope is ordinary rise-over-run between those points: [f(a+h) − f(a)] / h. Nothing here is new; that's the slope formula from algebra.
- Tangent (tangere, to touch) — a line that touches at one point and matches the curve's direction there.
So the definition reads: take the slope of a secant, and slide the second point onto the first.
Computing a derivative from the definition
She'll be made to do this by hand for a couple of weeks before being allowed the shortcut rules. That's deliberate — the rules are meaningless if the definition isn't in the hands.
Find f′(x) for f(x) = x² − 3x.
| f(x+h) = (x+h)² − 3(x+h) | substitute x+h everywhere x appears |
| = x² + 2xh + h² − 3x − 3h | expand fully before subtracting anything |
| f(x+h) − f(x) = 2xh + h² − 3h | the x² and −3x cancel — they always do |
| ÷ h = 2x + h − 3 | every surviving term had an h, which is why the 0/0 resolves |
| f′(x) = 2x − 3 | let h→0 |
The structural point: every term without an h cancels, and every surviving term has an h to divide out. That's guaranteed, not lucky — it's the Factor Theorem from Unit 1 wearing a different hat.
Find f′(x) for f(x) = 1/x.
| [ 1/(x+h) − 1/x ] / h | set up the difference quotient |
| numerator = [x − (x+h)] / [x(x+h)] | common denominator first — the Unit 1 complex-fraction move |
| = −h / [x(x+h)] | |
| ÷ h = −1 / [x(x+h)] | cancel the h |
| f′(x) = −1/x² | let h→0, so x+h → x |
Find f′(x) for f(x) = √x.
| [ √(x+h) − √x ] / h | 0/0, and there's a root → rationalize |
| × (√(x+h) + √x)/(√(x+h) + √x) | conjugate, top and bottom |
| = (x + h − x) / [ h(√(x+h) + √x) ] | difference of squares kills the roots on top |
| = h / [ h(√(x+h) + √x) ] = 1/(√(x+h) + √x) | cancel |
| f′(x) = 1/(2√x) | h→0 makes the two roots identical |
Cross-check with the power rule below: √x = x1/2, so the derivative is ½x−1/2 = 1/(2√x). Agreement.
The second definition — and it's the one worth carrying
Two problems with it. First, it's nearly circular: you can't define the tangent line to a general curve without already having derivatives. (Circles are the exception — tangent means perpendicular to the radius — which is why the Greeks could do tangents to conics and nothing else for two thousand years.) Second, it doesn't generalize; "slope" stops meaning anything in higher dimensions.
Here's the alternative:
In words: the derivative is the multiplier in the best linear approximation to f near a. Zoom in far enough on a differentiable function and it becomes indistinguishable from a straight line. f′(a) is that line's slope, and E is the error you're making.
Why bother: this makes the error term explicit and nameable, and the error term is what the rest of the course is about. Linear approximation (Unit 4), Newton's method (Unit 4), Euler's method (Unit 7), and Taylor series (Unit 10) are all the same question — how good is this, and what's the next correction? Carrying this definition from September makes April easy.
Drag h toward zero. The secant through (x, f(x)) and (x+h, f(x+h)) turns into the tangent, and its slope settles on f′(x). Nothing here is a limit yet — it is what the limit is describing.
Notation
| Style | Written | Best for |
|---|---|---|
| Lagrange | f′(x), f″(x), f(4)(x) | Compact. "The derivative as a function." |
| Leibniz | dy/dx, d²y/dx², d/dx[…] | Shows the variables. Essential for chain rule, related rates, and all of integration. |
| Newton | ẋ, ẍ | Physics only, and only for time derivatives. |
Newton had calculus (he called derivatives "fluxions") by 1666, during the plague years, and didn't publish. Leibniz developed it independently and published in 1684. The resulting priority war was vicious and lasted decades.
Britain sided with Newton and kept his dot notation. The continent used Leibniz's dy/dx and ∫ — notation that suggests the correct manipulations. dy/dx looks like a fraction and behaves like one under the chain rule; ∫ f dx reads as "sum of height times width." British mathematics fell roughly a century behind as a result.
It's the strongest argument in the history of the subject that notation is not cosmetic. Interface design is leverage.
Differentiability
Differentiable ⟹ continuous. Not the reverse.
If f′(a) exists, the numerator [f(a+h) − f(a)] must be heading to 0 — otherwise, divided by a vanishing h, the quotient would blow up rather than converge. Numerator → 0 means f(a+h) → f(a), which is continuity.
The converse fails because a function can be perfectly connected and still have no well-defined direction at a point. |x| at 0 is the standard case: the secant slopes approach −1 from the left and +1 from the right. Both one-sided derivatives exist; they disagree; so the derivative doesn't exist.
The power rule, derived
(x + h)n expands into n+1 terms. The first two are all that matter here:
Check with n = 3: (x+h)³ = x³ + 3x²h + 3xh² + h³. First two terms fit the pattern.
The coefficients are the rows of Pascal's triangle (1, 1 / 1, 2, 1 / 1, 3, 3, 1 / …), each entry the sum of the two above it. You don't need the whole row — just that the h¹ coefficient is n.
| [ (x+h)n − xn ] / h | the definition |
| = [ xn + n xn−1h + (h² terms) − xn ] / h | binomial expansion |
| = [ n xn−1h + (h² terms) ] / h | the xn cancels |
| = n xn−1 + (terms still carrying an h) | divide through |
| → n xn−1 | every leftover term dies as h→0 |
The rule extends to all real n — negative, fractional, irrational — but proving that needs logarithmic differentiation (later this unit). For now: it works for everything, and you saw it verified above for n = ½ and n = −1 by direct computation.
The power rule only applies once something is in xn form. Most power-rule errors are actually algebra errors committed before the calculus started.
- √x = x1/2 → derivative ½x−1/2 = 1/(2√x)
- ∛(x²) = x2/3 → derivative (2/3)x−1/3
- 1/x³ = x−3 → derivative −3x−4 = −3/x⁴
- x/√x = x1−1/2 = x1/2
- 5/x = 5x−1 → derivative −5x−2
The rules themselves: xa·xb = xa+b · xa/xb = xa−b · (xa)b = xab · x−a = 1/xa · x0 = 1 · x1/n = ⁿ√x
Rewrite first. Differentiate second.
Set n = 1. The rule gives 1·x⁰ = 1, and y = x is a line of slope 1 ✓. Set n = 0: 0·x⁻¹ = 0, and y = 1 is flat ✓.
Any version of the power rule you write down under pressure has to survive both. If it does not, you have the exponent in the wrong place.
Linearity: constants and sums
The constant rule is immediate from the definition: [c − c]/h = 0 for every h. Geometrically, a horizontal line has slope zero.
(fg)′ ≠ f′g′ and (f/g)′ ≠ f′/g′. Ever. Quick check: f = g = x. Then (x·x)′ = (x²)′ = 2x, but f′g′ = 1·1 = 1. Not equal. The rectangle picture below shows exactly what the naive version leaves out.
Product rule
| [ f(x+h)g(x+h) − f(x)g(x) ] / h | the definition applied to the product |
| insert − f(x+h)g(x) + f(x+h)g(x) | adding and subtracting the same thing changes nothing but creates two groupable pairs |
| = f(x+h)·[g(x+h) − g(x)]/h + g(x)·[f(x+h) − f(x)]/h | group and factor |
| → f(x)g′(x) + g(x)f′(x) | each bracket is a difference quotient; f(x+h) → f(x) by continuity |
The inserted term is the algebraic version of the corner in the picture. Adding and subtracting a middle term is a standard move worth recognizing — it reappears in the proof of the chain rule and in error analysis generally.
Quotient rule
| let Q = f/g, so f = Q·g | rewrite the quotient as a product |
| f′ = Q′g + Qg′ | product rule |
| Q′ = (f′ − Qg′)/g | solve for Q′ |
| = (f′ − (f/g)g′)/g | substitute back what Q is |
| = (f′g − fg′)/g² | multiply top and bottom by g |
Worth doing once, because it means you only ever have to remember the product rule. If the quotient rule's sign order deserts you mid-test, this reconstructs it in four lines.
Unlike the product rule, this one is not symmetric. f′g − fg′, not the reverse. The mnemonic most people carry: "low d-high minus high d-low, over low squared." Bottom times derivative of top, minus top times derivative of bottom, all over bottom squared.
Reconstruction check if you're unsure: try f = x, g = 1. Then f/g = x, so the answer must be 1. Plugging in: (1·1 − x·0)/1 = 1. ✓. Reversing the order would give −1. ✗.
Prove that differentiable ⟹ continuous. BC asserts it. The proof is three lines and it clarifies what the implication actually depends on.
| Assume f′(a) exists. | |
| f(a+h) − f(a) = h · [f(a+h) − f(a)]/h | multiply and divide by h — legal since h ≠ 0 in a limit |
| take h → 0: the bracket → f′(a), a finite number; h → 0 | |
| so f(a+h) − f(a) → 0·f′(a) = 0, i.e. f(a+h) → f(a). ∎ | which is continuity |
The step that carries the weight is finite. If the difference quotient blew up — a vertical tangent — the product 0·∞ would be indeterminate and the argument would collapse. That is precisely the case where a function is continuous but not differentiable.
The general Leibniz rule — the product rule for the n-th derivative. Rarely taught in the US, standard in the Russian sequence, and it's a genuinely satisfying structural fact.
Those are the binomial coefficients — the same numbers as in (a+b)n, the same Pascal's triangle used to derive the power rule. Check n = 1: C(1,0)f g′ + C(1,1)f′g = fg′ + f′g. ✓ Check n = 2: fg″ + 2f′g′ + f″g.
Why it's true, informally: each differentiation must land on either f or g. Over n rounds, the number of ways to hit f exactly k times is C(n,k). Differentiation and binomial expansion have the same combinatorial skeleton, which is not a coincidence — it's the same reason factorials appear in Taylor series.
A problem in the tradition: find the 10th derivative of x²·ex at 0. By Leibniz, only three terms survive, because x² dies after two differentiations: C(10,0)x²ex + C(10,1)(2x)ex + C(10,2)(2)ex. At x = 0 that's 0 + 0 + 90 = 90. Doing this by ten successive product rules is an afternoon.
Unit 2 — Definition and Fundamental Properties
Definition
- f′(a) = limh→0 [f(a+h) − f(a)]/h = limx→a [f(x) − f(a)]/(x−a)
- Equivalently f(a+h) = f(a) + f′(a)h + error, error/h → 0 (best linear approximation)
- Differentiable ⇒ continuous, not conversely. Fails at corners, cusps, vertical tangents, discontinuities.
Rules
- (c)′ = 0 · (cf)′ = cf′ · (f ± g)′ = f′ ± g′
- (xn)′ = n xn−1 — all real n; rewrite roots and reciprocals as exponents first
- (fg)′ = f′g + fg′ — the rectangle picture; the corner term vanishes
- (f/g)′ = (f′g − fg′)/g² — reconstruct from the product rule if the order deserts you
Assembly order
- Name the outermost structure first; nest the other rules inside it.
- Rewrite roots and reciprocals as exponents before differentiating.
- Don’t simplify unless asked.
Run the quotient rule on something you could have simplified instead. For x²/x: (2x·x − x²·1)/x² = x²/x² = 1. Simplifying first, x²/x = x, whose derivative is 1 ✓.
The two must agree. When they do not, it is almost always the order of the numerator — the minus sign belongs with the second term, not the first.
Practise Unit 2 → Fresh problems from this unit, with worked solutions.