Study · Mathematics
Calculus
Calculus
Limits, derivative rules, and basic applications: tangents, monotonicity, extrema and simple optimization.
3
lessons in this chapter
0
previous exam questions
6 hr
suggested study time
Hard
chapter difficulty
Last reviewed Sep 2026 · GetCSCA editors
Study order
Work the lessons top to bottom — each ends with practice, a timed set and a 5-stage drill. Then prove it in a full 48Q mock.
Limits
Lesson 1 · target ~75s per question
Intuitive limits, 0/0 forms by factoring and rationalizing, and the two classic special limits.
Why CSCA tests this
Key points — memorize
- Try direct substitution first; only the indeterminate forms (0/0, ∞/∞) need algebra, while the rest evaluate immediately.
- For 0/0 with polynomials, factor and cancel the common zero factor; with roots, rationalize by the conjugate, then substitute.
- Scale to the memorized specials: sin u/u → 1 and (1 − cos u)/u → 0 require the inside argument u → 0.
- A two-sided limit exists only if both one-sided limits exist and agree — piecewise joints and absolute values need both sides checked.
- Continuity at c means three things coincide: f(c) defined, the limit exists, and the limit equals f(c).
Formula sheet
- lim(x→0) sin x / x = 1
- lim(n→∞)(1 + 1/n)ⁿ = e
- lim(x→c)(f ± g) = lim f ± lim g
- lim(x→c)(f · g) = lim f · lim g
- 0/0: factor-cancel or rationalize first
- lim(x→0)(1 − cos x)/x = 0; lim(x→0)(eˣ − 1)/x = 1
- Two-sided limit exists ⇔ left and right limits agree
- Continuous at c ⇔ lim(x→c) f(x) = f(c)
Classic traps
- Cancelling (x − 2) while forgetting x ≠ 2 matters only for the function value, not the limit — then refusing to substitute.
- Applying lim sin x/x = 1 when the argument does not tend to 0, e.g. sin 3x/x without rescaling to 3·sin(3x)/(3x).
- Checking only one side at a piecewise joint or cusp and declaring the limit exists.
- Concluding a limit equals f(c) for a discontinuous function without verifying continuity.
Exam tactic
Worked examples
Example 1: Compute lim(x→2) (x² − 4)/(x − 2).
- Direct substitution gives 0/0 — factor: (x − 2)(x + 2)/(x − 2).
- Cancel (x ≠ 2 near the limit): limit = 2 + 2.
Answer: 4
Example 2: Compute lim(x→0) sin 3x / x.
- Rewrite: 3 · (sin 3x)/(3x).
- As x → 0, sin 3x/(3x) → 1, so the limit is 3.
Answer: 3
Derivatives
Lesson 2 · target ~70s per question
Derivative meaning (slope + rate of change), power/product/quotient/chain rules, and derivatives of key functions.
Why CSCA tests this
Key points — memorize
- Classify the structure before differentiating: single power → power rule, product → product rule, quotient → quotient rule, nested → chain rule.
- Chain rule multiplies rates layer by layer: derivative of the outside at the inside, times the derivative of the inside — repeat for triple nesting.
- Memorize the transcendental five: eˣ reproduces, ln x gives 1/x, sin ↔ cos with the minus on (cos)′, plus aˣ ln a and tan → sec².
- Simplify first (expand products, split quotients, convert roots to fractional powers) so the power rule handles more terms directly.
- Read f′(a) geometrically as the tangent slope at a and physically as the instantaneous rate of change of f at a.
Formula sheet
- (xⁿ)′ = n·xⁿ⁻¹
- (uv)′ = u′v + uv′
- (u/v)′ = (u′v − uv′)/v²
- [f(g(x))]′ = f′(g(x)) · g′(x)
- (eˣ)′ = eˣ; (ln x)′ = 1/x; (sin x)′ = cos x; (cos x)′ = −sin x
- (c)′ = 0; (cu)′ = cu′; (u ± v)′ = u′ ± v′
- (tan x)′ = sec²x; (aˣ)′ = aˣ ln a; (log_a x)′ = 1/(x ln a)
- f′(a) = lim(h→0)(f(a+h) − f(a))/h — slope of tangent at a
Classic traps
- Dropping the inner derivative in the chain rule, e.g. writing (sin(x²))′ = cos(x²).
- Swapping quotient-rule order to (uv′ − u′v)/v² or forgetting to square the denominator.
- Writing (ln x)′ = 1/x without restricting x > 0, or (xⁿ)′ = nxⁿ without lowering the exponent.
- Applying product/quotient rules to compositions (or vice versa) instead of naming the outer structure first.
Exam tactic
Worked examples
Example 1: Differentiate f(x) = x³·eˣ.
- Product rule: f′ = 3x²·eˣ + x³·eˣ.
- Factor: x²eˣ(x + 3).
Answer: f′(x) = x²eˣ(x + 3)
Example 2: Differentiate f(x) = sin(x²).
- Chain rule: outside sin → cos(x²).
- Multiply by inside derivative 2x.
Answer: f′(x) = 2x·cos(x²)
Basic Applications
Lesson 3 · target ~75s per question
Tangent lines, increasing/decreasing intervals, local extrema, and one-variable optimization word problems.
Why CSCA tests this
Key points — memorize
- Tangent lines need exactly two numbers: the point value f(x₀) and the slope f′(x₀), assembled as y = f(x₀) + f′(x₀)(x − x₀).
- Collect all critical points (f′ = 0 or undefined but f defined), then run a sign chart: − → + is a min, + → − is a max, no change is neither.
- On a closed interval, the global max/min is the largest/smallest among f at the endpoints and at interior critical points — never skip endpoints.
- Optimization word problems reduce to one variable via the constraint (e.g. perimeter), then differentiate, solve f′ = 0, and verify with sign or f′′.
- Respect the word-problem domain (lengths, counts positive): a critical point outside the feasible interval is discarded.
Formula sheet
- Tangent: y = f(x₀) + f′(x₀)(x − x₀)
- Critical points: f′(x) = 0 or undefined
- Max on [a, b]: max{f(a), f(b), f(critical)}
- f′: − → + ⇒ local min; + → − ⇒ local max
- Second-derivative check: f′′(x₀) > 0 ⇒ local min
- f′ > 0 increasing; f′ < 0 decreasing — sign chart decides
- Optimization: single-variable f → f′ = 0 → verify → check endpoints/domain
- Second-derivative check: f′′(x₀) < 0 ⇒ local max; = 0 inconclusive
Classic traps
- Reporting a local extremum as the global one without evaluating endpoints on [a, b].
- Treating f′(x₀) = 0 as sufficient (x³ at 0) without a sign or second-derivative check.
- Optimizing a two-variable expression without substituting the constraint first.
- Giving the extremal x-value when the question asks for the maximum value f(x) (or vice versa).
Exam tactic
Worked examples
Example 1: Find the tangent to f(x) = x² at x = 3.
- f(3) = 9, f′(x) = 2x so f′(3) = 6.
- y = 9 + 6(x − 3) = 6x − 9.
Answer: y = 6x − 9
Example 2: Find the local extrema of f(x) = x³ − 3x.
- f′(x) = 3x² − 3 = 0 ⇒ x = ±1.
- f′ changes + → − at x = −1 (max, f = 2) and − → + at x = 1 (min, f = −2).
Answer: Local max 2 at x = −1; local min −2 at x = 1
Previous exam questions
Real CSCA-style questions tagged to Calculus — answer right here.
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Related guides
- 30-day study plan → fit this chapter into a week-by-week system
- 75-second pacing rule → hold 75s/question once the content clicks
- Top 10 mistakes → the error patterns that cost the most points
3 lessons · one chapter
Finished Calculus? Prove it.
Run a timed set, clear every mistake, then take a full 48Q mock.