Calculus I › Differentiation rules › full formula sheet
The definition of the derivative
The one limit every other differentiation rule is built from — what it measures, how to compute with it, and why h can never be zero.
Notation: f′(x) means df/dx. h is a nonzero step that we send toward 0; limh→0 means “approach 0, never arrive.”
Before this lesson: Limit laws · Squeeze theorem
Where it comes from
The problem: a car's speedometer shows your speed right now, but distance divided by time only gives an average over the trip. A curve has the same issue: between two points you can compute rise over run, but that is the slope of the secant line — the average rate — not the slope at a single point.
Watch what happens for f(x) = x² near x = 2. The average rate over the step from 2 to 2+h is:
Before reading on: the average rate from 2 to 2+h is 4+h. What does that average approach as h shrinks — and why can’t you just set h = 0 to find out?
At h = 0.1 the average is 4.1; at h = 0.01 it is 4.01. Every finite step overshoots the true instantaneous rate 4 a little. The naive fix — just plug in h = 0 — gives 0/0, which is undefined. You cannot arrive; you can only approach. That is exactly what the limit does:
Every rule in this chapter — power, product, quotient, chain — is just this limit with the algebra done once and for all. Learn the definition and you own the source code of calculus.
Derivation
We compute the definition directly for f(x) = x². Four lines, no tricks — this is the template every “from first principles” problem follows.
Before reading on: after expanding (x+h)² and canceling, every term still carries an h except one. Which term is it — and what does that predict for f′(x)?
The same four steps work for f(x) = x³: ((x+h)³ − x³)/h = (3x²h + 3xh² + h³)/h = 3x² + 3xh + h² → 3x². Notice the pattern: after dividing by h, every surviving term except the first still contains an h — and every one of those dies in the limit. The power rule is this observation, generalized.
How to use it
The procedure, every time — “first principles” means these five steps, in this order:
- Write f(x+h). Substitute x+h for every x. Expand powers; get common denominators for fractions.
- Subtract f(x). The leading terms should cancel — if nothing cancels, recheck step 1.
- Divide by h. You now hold the difference quotient (the secant slope).
- Simplify until h cancels. Factor h out of the numerator. If h will not cancel, you have an algebra error upstream — go back, don’t force it.
- Take the limit as h → 0. Only now may h vanish. What remains is f′(x).
When to use it — and when not to
Use the definition when the problem says “from first principles,” “using the definition,” or asks you to prove a rule. Use the differentiation rules for everything routine: computing d/dx [xⁿ] from the definition on an exam is correct but costs five minutes the power rule does in five seconds. Judgment call: the definition is a foundation tool, not a daily driver.
Worked examples
Four problems, easiest first. Read every step — the why of each move is the lesson.
Example 1 — from the definition: f′(2) for f(x) = x²
- Write the quotient at a = 2. [f(2+h) − f(2)]/h = [(2+h)² − 4]/h.
- Expand. = [4 + 4h + h² − 4]/h = (4h + h²)/h.
- Cancel h (h ≠ 0): = 4 + h.
- Limit: h → 0 gives 4.
- Check with the power rule: f′(x) = 2x, so f′(2) = 4. Matches ✓
Your turn: Find f′(3) for f(x) = x², from the definition.
Answer: 6
[(3+h)² − 9]/h = (6h + h²)/h = 6 + h; letting h → 0 gives 6. Same shape as the example — only the point moved.
Example 2 — f(x) = x³
- Quotient: [(x+h)³ − x³]/h.
- Expand (x+h)³ = x³ + 3x²h + 3xh² + h³; subtract x³: = (3x²h + 3xh² + h³)/h.
- Cancel h: = 3x² + 3xh + h².
- Limit h → 0: the last two terms die, leaving 3x².
Your turn: Find f′(x) for f(x) = x⁴, from the definition.
Answer: 4x³
(x+h)⁴ = x⁴ + 4x³h + 6x²h² + 4xh³ + h⁴. Subtract x⁴, divide by h: 4x³ + 6x²h + 4xh² + h³ → 4x³ as h → 0.
Example 3 — f(x) = 1/x (fractions need a common denominator)
- Quotient: [1/(x+h) − 1/x]/h.
- Combine the top: = [(x − (x+h)) / (x(x+h))]/h = [−h / (x(x+h))]/h.
- Cancel h: = −1 / [x(x+h)]. (Why legal? h ≠ 0, and we assume x ≠ 0, x+h ≠ 0.)
- Limit: x+h → x, so the result is −1/x², for x ≠ 0.
Your turn: Find f′(x) for f(x) = 1/x², from the definition.
Answer: −2/x³, for x ≠ 0
[1/(x+h)² − 1/x²]/h = [(x² − (x+h)²) / (x²(x+h)²)]/h = (−2x − h) / [x²(x+h)²] → −2x/x⁴ = −2/x³.
Example 4 — f(x) = √x (the conjugate trick)
- Quotient: [√(x+h) − √x]/h. Direct expansion stalls — multiply top and bottom by the conjugate √(x+h) + √x.
- Multiply out: = [(x+h) − x] / [h(√(x+h) + √x)] = h / [h(√(x+h) + √x)]. (Why? (a−b)(a+b) = a²−b² kills the roots.)
- Cancel h: = 1 / [√(x+h) + √x].
- Limit: → 1/(2√x) = 1/(2√x), for x > 0.
Your turn: Find f′(x) for f(x) = √(2x), from the definition (conjugate again).
Answer: 1/√(2x), for x > 0
[√(2x+2h) − √(2x)]/h · conj/conj = 2h / [h(√(2x+2h)+√(2x))] = 2 / [√(2x+2h)+√(2x)] → 2/(2√(2x)) = 1/√(2x).
Memorization tips
- Say it as a sentence: “the derivative is the limit of the difference quotient as h goes to zero.” If you can say it, you can write it.
- Secant → tangent: before the limit you hold a secant slope (average); after the limit you hold a tangent slope (instantaneous). The limit is the bridge.
- h is never zero: you cancel h because h ≠ 0, then let h approach 0. “Cancel, then collapse” — in that order, always.
- The x² prototype: memorize the four-line x² computation cold. Every first-principles problem is the same four moves with different algebra.
- Watch for 0/0: if plugging h = 0 gives 0/0, that’s not a dead end — it’s the signal to simplify first. The definition only fails if simplification is impossible.
- Definition vs. rules: use the definition to prove, the rules to compute. On exams, “from first principles” is the keyword that demands the limit.
Final challenge
Five mixed questions — computations, concepts, and the non-differentiable trap. Score 5/5 and the definition is yours.
← Back to the Calculus I formula sheet
How to learn a formula here
- Read each section in order — every section ends with a short quiz. Take it before moving on; the questions test exactly what you just read.
- Work the examples with the answers covered, then uncover one step at a time and compare.
- Finish with the final challenge — five mixed questions including the classic traps.
- Retake what you miss — every quiz reshuffles each attempt, and every answer explains itself.
Frequently asked questions
What is the definition of the derivative?
The derivative of f at x is f′(x) = lim(h→0) [f(x+h) − f(x)] / h — the limit of the secant slopes (difference quotients) as the step h collapses to zero. It gives the slope of the tangent line, i.e. the instantaneous rate of change.
Why can’t we just set h = 0?
Plugging in h = 0 gives [f(x) − f(x)] / 0 = 0/0, which is undefined — you’d be dividing by zero. The limit asks what the quotient approaches as h gets arbitrarily small without ever being zero, which is a well-defined number.
What’s the difference between the difference quotient and the derivative?
The difference quotient [f(x+h)−f(x)]/h is the slope of a secant line — an average rate over a step. The derivative is the limit of those quotients as h→0 — the tangent slope, an instantaneous rate.
Does the derivative always exist?
No. The limit must exist (from both sides). For f(x) = |x| at x = 0, the left-hand quotient tends to −1 and the right-hand to +1, so the two-sided limit fails — the graph has a corner and no tangent there.
Why learn the definition if the rules are faster?
Every rule — power, product, quotient, chain — is proved from this limit. Exams also ask you to differentiate “from first principles,” which means this definition and nothing else.
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