for the full chapter-by-chapter notes, see my TCX2101 notebook. for CT1 formulas (Chapters 1.1–3.4), see the CT1 cheatsheet.
3: Applications of Derivatives
3.5 Mean Value Theorem
Rolle’s Theorem
If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a) = f(b)$:
$$\exists, c \in (a,b) \text{ such that } f’(c) = 0$$
Intuition: curve starts and ends at same height → must have a flat spot somewhere.
Mean Value Theorem (MVT) ⭐️
If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$:
$$\exists, c \in (a,b) \text{ such that } f’(c) = \frac{f(b) - f(a)}{b - a}$$
Intuition: instantaneous speed must equal average speed at some point.
MVT Corollary
$$f’(x) = g’(x) ;\forall, x \in (a,b) \implies f(x) = g(x) + C$$
Same derivative = same function (up to a constant).
⚠️ Conditions are the trap. Must verify: (1) continuous on closed $[a,b]$, (2) differentiable on open $(a,b)$. Miss either → theorem doesn’t apply.
3.6 Derivative Tests
Monotonicity (First Derivative)
| $f’(x)$ on $(a,b)$ | $f$ on $[a,b]$ |
|---|---|
| $f’(x) > 0$ | Increasing |
| $f’(x) < 0$ | Decreasing |
First Derivative Test for Local Extrema
At critical point $c$ (where $f’(c) = 0$ or DNE), check sign change of $f’$:
| $f’$ changes | Result at $c$ |
|---|---|
| $- \to +$ | Local minimum |
| $+ \to -$ | Local maximum |
| No sign change | Not an extremum |
Concavity (Second Derivative)
| $f’’$ on $I$ | Graph shape | $f’$ is… |
|---|---|---|
| $f’’ > 0$ | Concave up (cup) | Increasing |
| $f’’ < 0$ | Concave down (cap) | Decreasing |
Inflection point: where concavity changes. At $(a, f(a))$: either $f’’(a) = 0$ or $f’’$ DNE.
Second Derivative Test for Local Extrema ⭐️
At critical point $c$ where $f’(c) = 0$:
| $f’’(c)$ | Result |
|---|---|
| $f’’(c) < 0$ | Local maximum (concave down = peak) |
| $f’’(c) > 0$ | Local minimum (concave up = valley) |
| $f’’(c) = 0$ | Test fails — use first derivative test |
⚠️ $f’’(c) = 0$ does NOT mean inflection. It means the test is inconclusive. Fall back to first derivative test.
3.7 L’Hôpital’s Rule
The Rule
If $\frac{f(x)}{g(x)} \to \frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ as $x \to a$:
$$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f’(x)}{g’(x)}$$
(assuming the right-hand limit exists). Works for $a = \pm\infty$ too.
When to Use
| Form | Indeterminate? | Use L’Hôpital? |
|---|---|---|
| $\frac{0}{0}$ | Yes | Yes |
| $\frac{\pm\infty}{\pm\infty}$ | Yes | Yes |
| $\frac{0}{\infty}$, $\frac{\infty}{0}$, $\frac{5}{0}$ | No | No — evaluate directly |
⚠️ Three traps:
- Must be indeterminate — check BEFORE differentiating
- Stop when the form is no longer indeterminate — don’t keep differentiating
- Differentiate top and bottom separately — this is NOT the quotient rule
3.8 Optimization
Algorithm
- Read — identify what to maximize/minimize
- Draw — label variables, note constraints
- Write — express objective as function of one variable (use constraint to eliminate)
- Differentiate — find critical points ($f’(x) = 0$)
- Test — check critical points AND endpoints (if closed interval)
- Answer — substitute back to get all requested values
Endpoint Check
| Domain | What to check |
|---|---|
| Closed $[a,b]$ | Critical points and $f(a)$, $f(b)$ |
| Open $(0, \infty)$ | Critical points + behavior as $x \to 0^+$, $x \to \infty$ |
| All $\mathbb{R}$ | Critical points + behavior as $x \to \pm\infty$ |
⚠️ Don’t forget endpoints. A critical point can be a local extremum but NOT the global one if an endpoint beats it.
⚠️ “Minimize cost” → objective is cost, constraint is volume (or similar). Read carefully which is which.
4: Integration
4.1 Definite Integrals & Riemann Sums
Integrability
$$\text{f integrable on } [a,b] \iff \lim_{n \to \infty} L_n = \lim_{n \to \infty} U_n$$
Upper and lower Riemann sums squeeze to the same value.
Upper & Lower Sums
| $f$ on $[a,b]$ | Upper sum $U_n$ | Lower sum $L_n$ |
|---|---|---|
| $f$ increasing | Right endpoints | Left endpoints |
| $f$ decreasing | Left endpoints | Right endpoints |
$$\Delta x = \frac{b-a}{n}$$
Properties of Definite Integrals
| Rule | Formula | Intuition |
|---|---|---|
| Max-min inequality | $m(b-a) \leq \int_a^b f \leq M(b-a)$ | Rectangle bounds $\times$ interval width |
| Additive | $\int_a^c = \int_a^b + \int_b^c$ | Split interval = split integral |
| Reverse limits | $\int_a^b = -\int_b^a$ | Backwards = negate ($\Delta x$ flips) |
| Zero-width | $\int_a^a = 0$ | No width = no area |
| Comparison | $f \leq g \Rightarrow \int f \leq \int g$ | Smaller function = smaller area |
| Non-negative | $f \geq 0 \Rightarrow \int f \geq 0$ | Positive function = positive area |
⚠️ Max-min inequality: Don’t forget to multiply by $(b-a)$! Bounds alone are not enough.
Integrability & Discontinuities
| Situation | Integrable? |
|---|---|
| Continuous on $[a,b]$ | Yes |
| Finitely many discontinuities | Yes |
| Dirichlet function (everywhere discontinuous) | No ($L_n = 0$, $U_n = 1$, never equal) |
⚠️ Finite bad points = still integrable. Only “everywhere discontinuous” breaks it.
Reversed Limits + Inequality (Double Trap)
$$f(x) < 0 \text{ on } [b,a] \text{ with } a > b$$
- Normal direction: $\int_b^a f < 0$ (negative function → negative integral)
- Reversed: $\int_a^b f = -\int_b^a f > 0$ (negate → positive!)
⚠️ Two negatives: $f < 0$ gives $\int_b^a < 0$, reversed $\int_a^b = -(\text{negative}) > 0$
4.2 Fundamental Theorem of Calculus
FTC I — The 4 Cases ⭐️
Step 1: Look at the limits
| Lower | Upper | Case | |
|---|---|---|---|
| Case 1 | constant | $x$ | Basic |
| Case 2 | constant | $g(x)$ | Chain rule |
| Case 3 | $x$ | constant | Minus sign |
| Case 4 | $a(x)$ | $b(x)$ | Both move |
Step 2: Apply formula
| Case | Formula | Mnemonic |
|---|---|---|
| 1 | $f(x)$ | Just substitute $t \to x$ |
| 2 | $f(g(x)) \cdot g’(x)$ | Substitute, multiply by derivative |
| 3 | $-f(x)$ | Add minus sign |
| 4 | $f(b(x)) \cdot b’(x) - f(a(x)) \cdot a’(x)$ | Upper minus lower, each $\times$ derivative |
Examples:
| Integral | Case | $F’(x)$ |
|---|---|---|
| $\int_0^x e^t,dt$ | 1 | $e^x$ |
| $\int_0^{x^2} e^t,dt$ | 2 | $e^{x^2} \cdot 2x$ |
| $\int_x^5 e^t,dt$ | 3 | $-e^x$ |
| $\int_x^{3x} \ln t,dt$ | 4 | $\ln(3x) \cdot 3 - \ln(x) \cdot 1$ |
⚠️ Chain rule is the #1 mistake. See $x^2$, $\sqrt{x}$, $x^3$ as upper limit? MUST multiply by its derivative.
FTC I — Fine Print
| Point | Detail |
|---|---|
| Domain | $F’(x) = f(x)$ on open $(a,b)$, NOT closed $[a,b]$ |
| Continuity | $F’(c) = f(c)$ requires $f$ continuous at $c$ |
| Discontinuity | If $f$ has removable discontinuity at $c$: $F’(c)$ may exist but $\neq f(c)$ |
| Not integrable | Dirichlet $\to$ $F(x)$ undefined $\to$ can’t discuss $F’(x)$ |
⚠️ See $[a,b]$ square brackets in “FTC I gives $F’(x) = f(x)$ for $x \in [a,b]$”? That’s WRONG. Must be $(a,b)$ round brackets.
FTC II (Evaluation Theorem)
$$\int_a^b f(x),dx = F(b) - F(a)$$
where $F$ is any antiderivative of $f$.
⚠️ Don’t forget $F(a)$! Result is $F(b) - F(a)$, not just $F(b)$.
Antiderivatives
Two antiderivatives of the same function differ by a constant:
$$F’(x) = G’(x) = f(x) \implies F(x) - G(x) = C$$
Therefore: $F(b) - F(a) = G(b) - G(a)$ (constant cancels).
Indefinite Integral
$$\int f(x),dx = F(x) + C \qquad \text{where } F’(x) = f(x)$$
4.3 Area Under and Between Graphs
Area Under a Curve
$$\text{Area} = \int_a^b |f(x)|,dx$$
Algorithm:
- Find zeros of $f$ in $[a,b]$
- Determine sign of $f$ on each sub-interval
- Where $f < 0$, negate: $\int(-f),dx$
- Sum all pieces
⚠️ $\int_a^b f(x),dx$ is NOT always area. If $f < 0$ on part of $[a,b]$, the integral gives signed area (negative parts cancel). Use $|f(x)|$ for actual area.
Area Between Two Curves
$$\text{Area} = \int_a^b |f(x) - g(x)|,dx$$
Algorithm:
- Find intersections: solve $f(x) = g(x)$ in $[a,b]$
- On each sub-interval, determine which function is on top
- Integrate (top $-$ bottom) on each piece
- Sum all pieces
⚠️ Always subtract in the right order. If you get a negative area for a piece, you subtracted the wrong way — flip it.
4.4 Integration by Substitution
$u$-Substitution (Indefinite)
If you spot $f(u(x)) \cdot u’(x)$ in the integrand:
$$\int f(u(x)) \cdot u’(x),dx = F(u(x)) + C$$
Steps: Let $u = $ (inside function), $du = u’(x),dx$, rewrite, integrate, substitute back.
$u$-Substitution (Definite) ⭐️
$$\int_a^b f(u(x)) \cdot u’(x),dx = \int_{u(a)}^{u(b)} f(u),du$$
⚠️ Change the limits! $x = a \to u = u(a)$, $x = b \to u = u(b)$. If you forget, you’ll integrate with wrong bounds.
Trigonometric Substitution
| Expression | Substitution | Identity | Simplifies to |
|---|---|---|---|
| $\sqrt{a^2 + x^2}$ | $x = a\tan\theta$ | $1 + \tan^2 = \sec^2$ | $a\sec\theta$ |
| $\sqrt{a^2 - x^2}$ | $x = a\sin\theta$ | $1 - \sin^2 = \cos^2$ | $a\cos\theta$ |
| $\sqrt{x^2 - a^2}$ | $x = a\sec\theta$ | $\sec^2 - 1 = \tan^2$ | $a\tan\theta$ |
Memory trick — right triangle:
- $a^2 + x^2$: $a$, $x$ are legs → hypotenuse = $\sqrt{a^2+x^2}$ → $\tan\theta$
- $a^2 - x^2$: $a$ is hypotenuse → $\sin\theta$
- $x^2 - a^2$: $x$ is hypotenuse → $\sec\theta$
⚠️ After trig sub, convert back. Draw the right triangle to express $\theta$ in terms of $x$.
Ch4 Mistakes ⚠️
| Mistake | Times | Fix |
|---|---|---|
| Chain rule — forgot $g’(x)$ | 3$\times$ | Upper limit $\neq x$? MUST multiply by derivative |
| Open interval $()$ vs closed $[]$ | 4$\times$ | FTC I output = (a,b) always. See $[$ → wrong |
| Continuity required for $F’(c) = f(c)$ | 2$\times$ | “Bounded” or “integrable” not enough |
| Forgot minus sign (lower limit) | 2$\times$ | $x$ in lower = flip = negative |
| Basic FTC I overcomplicated | 1$\times$ | Upper limit = $x$? Just substitute. Done. |
| Variable inside integral | 2$\times$ | $1/h$ is variable, can’t move into $\int$ |
Quick Reference
FTC I Decision Tree
1Given: d/dx ∫_?^? f(t) dt
2
3Step 1: What are the limits?
4 │
5 ┌────┴────┬──────────┬──────────┐
6 │ │ │ │
7 ∫_a^x ∫_a^g(x) ∫_x^b ∫_a(x)^b(x)
8 │ │ │ │
9 f(x) f(g(x))·g' -f(x) f(b)·b' - f(a)·a'
10 ↑ ↑
11 DON'T FORGET UPPER MINUS LOWER
Derivative Test Decision Tree
1Found critical point c (f'(c) = 0)?
2 │
3 Can you compute f''(c)?
4 ┌────┴────┐
5 Yes No → use First Derivative Test
6 │
7 f''(c) = ?
8 ┌───┼───┐
9 <0 =0 >0
10 │ │ │
11 MAX FAIL MIN
12 │
13 Fall back to
14 First Deriv Test
Trig Sub Cheat
1See √(___) ?
2 │
3 ┌────┴────┬──────────┐
4 │ │ │
5 a² + x² a² - x² x² - a²
6 │ │ │
7 x=a·tanθ x=a·sinθ x=a·secθ
8 │ │ │
9 a·secθ a·cosθ a·tanθ
- Coverage: TCX2101 Chapters 3.5–4.11 (CT2 scope)
- Purpose: Class Test 2 reference (16 Mar)
- Format: Closed book — memorise this!
- Source: NUS TCX2101 Calculus and Linear Algebra, CT2 prep
- GitHub: Repository