Differential Calculus - Revision Notes
Differential calculus is one of the highest-yield topics in the MSTE (Mathematics, Surveying, and Transportation Engineering) portion of the PRC Civil Engineer Licensure Examination. Expect 8–12 items per board exam covering limits, differentiation rules, optimization (maxima-minima), and related rates. This chapter equips you to (1) evaluate limits including indeterminate forms, (2) apply all standard derivative rules fluently, (3) set up and solve optimization problems using a single-variable model, (4) differentiate implicit and composite functions, and (5) solve related-rates problems by differentiating geometric or physical relations with respect to time. Master the step-by-step strategies below and drill the worked examples until the procedures become automatic.
1. Limits and Continuity
Formulas
lim[x→a] f(x)/g(x) = lim[x→a] f'(x)/g'(x)
Example
lim[x→0] (sin 3x)/(2x): Apply L'Hôpital — (3 cos 3x)/2 → 3/2 as x→0. Answer: 3/2
Variables
f and g are differentiable; g'(a) ≠ 0 after applying the rule; form must be 0/0 or ∞/∞
Application
Evaluate indeterminate-form limits that cannot be resolved by direct substitution or algebraic simplification
lim[x→0] (sin x)/x = 1
Example
lim[x→0] (sin 5x)/(3x) = (5/3) · lim[x→0] (sin 5x)/(5x) = 5/3
Variables
x in radians
Application
Trigonometric limits; often appears disguised as lim (sin kx)/(kx) = 1
Exam tips
- Always attempt direct substitution first; only reach for L'Hôpital if an indeterminate form results.
- Factor and cancel common factors before differentiating — it is often faster than L'Hôpital.
- Memorize lim (sin x)/x = 1 and lim (1−cos x)/x = 0; they appear in disguised forms every exam.
- For limits at infinity of rational functions: divide every term by the highest power of x in the denominator.
Key points
- The limit lim[x→a] f(x) = L means f(x) approaches L as x approaches a, regardless of f(a) itself.
- For a limit to exist, the left-hand limit and right-hand limit must be equal: lim[x→a⁻] f(x) = lim[x→a⁺] f(x).
- f is continuous at x = a if: (1) f(a) exists, (2) lim[x→a] f(x) exists, and (3) lim[x→a] f(x) = f(a).
- Indeterminate forms requiring further work: 0/0, ∞/∞, 0·∞, ∞−∞, 0⁰, 1^∞, ∞⁰.
- L'Hôpital's Rule: If the limit gives 0/0 or ∞/∞, differentiate numerator and denominator separately, then re-evaluate.
- Squeeze Theorem: If g(x) ≤ f(x) ≤ h(x) near a and lim g = lim h = L, then lim f = L.
- Standard limits to memorize: lim[x→0] (sin x)/x = 1; lim[x→0] (1 − cos x)/x = 0; lim[x→∞] (1 + 1/n)^n = e.
Definitions
Limit
The value that f(x) approaches as x gets arbitrarily close to a, written lim[x→a] f(x) = L.
Importance
Foundation of all calculus; needed to define the derivative rigorously.
Continuity
f is continuous at a if it is defined at a, the limit exists at a, and both values agree.
Importance
Differentiability requires continuity; discontinuous points are endpoints of optimization domains.
Indeterminate Form
An expression whose limit cannot be determined by direct substitution because the result is 0/0, ∞/∞, etc.
Importance
Triggers the use of L'Hôpital's Rule or algebraic manipulation on board-exam problems.
Common mistakes
- Applying L'Hôpital's Rule when the form is NOT indeterminate — always check the form first.
- Differentiating the whole fraction as a quotient instead of differentiating numerator and denominator separately.
- Using degrees instead of radians when evaluating trigonometric limits.
- Confusing 'limit does not exist' with 'limit equals infinity' — infinity is not a real number.
2. Differentiation Rules and Techniques
Formulas
d/dx [xⁿ] = nxⁿ⁻¹
Example
y = 4x³ − 2x⁻² + 7 → y' = 12x² + 4x⁻³ = 12x² + 4/x³
Variables
n = any real number
Application
Polynomial, radical, and negative-power functions
(uv)' = u'v + uv'
Example
y = x² sin x → y' = 2x sin x + x² cos x
Variables
u = u(x), v = v(x), both differentiable
Application
Product of two separate differentiable functions
(u/v)' = (u'v − uv') / v²
Example
y = (x²+1)/(2x−3) → y' = [(2x)(2x−3) − (x²+1)(2)] / (2x−3)² = (2x²−6x−2)/(2x−3)²
Variables
v ≠ 0
Application
Rational and trigonometric quotient functions
d/dx [f(g(x))] = f'(g(x)) · g'(x)
Example
y = (3x²+1)⁵ → y' = 5(3x²+1)⁴ · 6x = 30x(3x²+1)⁴
Variables
f = outer function, g = inner function
Application
Composite functions: powers of polynomials, trig of trig, exponentials with variable exponents
d/dx [sin u] = cos u · u'; d/dx [cos u] = −sin u · u'
Example
y = sin(x³) → y' = cos(x³) · 3x²
Variables
u = u(x)
Application
All trigonometric differentiation; chain rule automatically included
d/dx [eᵘ] = eᵘ · u'; d/dx [ln u] = u'/u
Example
y = e^(2x) → y' = 2e^(2x); y = ln(x²+1) → y' = 2x/(x²+1)
Variables
u = u(x), u > 0 for ln
Application
Exponential growth/decay models, logarithmic differentiation
d/dx [tan u] = sec²u · u'; d/dx [sec u] = sec u tan u · u'
Example
y = tan(3x) → y' = 3sec²(3x)
Variables
u = u(x)
Application
Tangent and secant differentiation in trig optimization problems
Exam tips
- Board-exam composite functions like y = (2x+1)^10 are solved in seconds with the chain rule — never expand.
- Logarithmic differentiation simplifies products and quotients raised to powers: take ln of both sides first.
- For d/dx [aˣ] = aˣ ln a — do NOT write aˣ alone; the ln a factor is always present.
- When differentiating implicitly, group all dy/dx terms on the left, then factor and divide.
Key points
- The derivative f'(x) gives the instantaneous rate of change and the slope of the tangent line at any point.
- Power Rule: d/dx [xⁿ] = nxⁿ⁻¹ — the single most frequently used rule on board exams.
- Product Rule: (uv)' = u'v + uv' — use whenever two functions are multiplied.
- Quotient Rule: (u/v)' = (u'v − uv')/v² — use whenever a function is divided by another.
- Chain Rule: d/dx [f(g(x))] = f'(g(x)) · g'(x) — the most commonly missed rule, especially in composite trig/exponential functions.
- Implicit differentiation: differentiate both sides w.r.t. x, treating y as a function of x; collect dy/dx terms.
- Higher-order derivatives: y'', y''' are successive derivatives; used in classifying critical points and curvature.
- Partial derivatives: ∂f/∂x treats all other variables as constants; ∂f/∂y treats x as constant.
Definitions
Derivative
The derivative f'(x) = lim[h→0] [f(x+h) − f(x)]/h; it measures the instantaneous rate of change of f at x.
Importance
Core concept underpinning all differentiation; understanding this definition helps in deriving rules.
Chain Rule
The rule for differentiating a function of a function: d/dx [f(g(x))] = f'(g(x)) · g'(x).
Importance
Most frequently omitted step on board exams — always identify the inner function and multiply by its derivative.
Implicit Differentiation
Technique for finding dy/dx when y is not isolated; differentiate both sides w.r.t. x, applying the chain rule to y-terms.
Importance
Used in related-rates problems, tangent lines to curves like circles and ellipses.
Partial Derivative
∂f/∂x: derivative of f(x,y) with respect to x, treating y as a constant.
Importance
Required for multivariable optimization and MSTE problems involving surfaces.
Common mistakes
- Forgetting to apply the chain rule on composite functions, e.g., writing d/dx [sin(x²)] = cos(x²) instead of 2x cos(x²).
- Sign errors in the quotient rule — memorize: 'lo d-hi minus hi d-lo, over lo-squared'.
- Differentiating constants as zero but failing to recognize that e³ is a constant (not e^(3x)).
- In implicit differentiation, forgetting to multiply by dy/dx when differentiating y-terms.
- Treating y' as a separate variable rather than dy/dx when collecting terms.
3. Maxima, Minima, and Optimization
Formulas
f'(x₀) = 0 and f''(x₀) < 0 ⟹ local maximum at x₀
Example
f(x) = −x² + 4x; f'(x) = −2x+4 = 0 → x=2; f''(2) = −2 < 0 → maximum. f(2) = 4
Variables
x₀ = critical point; f' and f'' must exist
Application
Classify critical points without sign charts — faster for board exams
A_rect = xy, constraint x+y = P/2 = k → maximize A = x(k−x) → x = k/2
Example
Perimeter = 40 m → k=20; A = x(20−x); A' = 20−2x = 0 → x=10; A_max = 100 m²
Variables
x, y = dimensions; P = perimeter; k = semi-perimeter
Application
Maximum-area rectangle for a given perimeter; result: square is optimal
V_box = x(a−2x)(b−2x) where x = corner cut
Example
30×30 sheet: V = x(30−2x)²; V' = (30−2x)² + x·2(30−2x)(−2) = 0 → 30−2x−4x = 0 → x=5 cm; V_max = 5(20)² = 2000 cm³
Variables
a, b = sheet dimensions; x = height of box
Application
Open box optimization from rectangular sheet with square corners removed
Exam tips
- Board-exam optimization problems always have a constraint equation — identify it first and use it to eliminate one variable.
- For symmetric shapes (square, cube, equilateral), the optimal solution is almost always the symmetric case — use this as a quick check.
- In fencing/perimeter problems along a river (one side free), the optimal rectangle has width = half the length.
- After solving x from f'(x) = 0, substitute back to find the optimized quantity — partial credit is lost if you stop at x.
- If the constraint gives two solutions, check both — one may be outside the physical domain.
Key points
- A critical point occurs where f'(x) = 0 or f'(x) is undefined.
- First Derivative Test: f' changes + to − at x₀ → local maximum; f' changes − to + → local minimum.
- Second Derivative Test: f''(x₀) < 0 → local max; f''(x₀) > 0 → local min; f''(x₀) = 0 → inconclusive.
- Absolute (global) extrema on a closed interval [a, b]: evaluate f at all critical points AND endpoints; the largest/smallest value is the answer.
- Inflection point: f'' changes sign; curvature changes from concave up to concave down (or vice versa).
- Optimization strategy: identify the quantity to optimize, write it as a function of ONE variable using the constraint, differentiate, set equal to zero, solve, then verify.
- Common optimization shapes on board exams: rectangle of maximum area for fixed perimeter (square), cylinder of maximum volume, open box from a flat sheet.
Definitions
Critical Point
A value x₀ in the domain of f where f'(x₀) = 0 or f'(x₀) is undefined.
Importance
All maxima and minima must occur at critical points or endpoints — the search begins here.
Inflection Point
A point where the concavity of f changes; f''(x) = 0 is necessary but not sufficient.
Importance
Sometimes confused with extrema on board exams; inflection points are not necessarily extrema.
Absolute vs. Local Extremum
A local extremum is the largest/smallest in a neighborhood; an absolute extremum is the largest/smallest on the entire domain.
Importance
On closed-interval problems, endpoints must be checked — a critical-point maximum may not be the global maximum.
Common mistakes
- Forgetting to check endpoints when the problem specifies a closed interval or physical constraint (e.g., x > 0).
- Setting up the area/volume formula incorrectly before differentiating — always sketch the problem.
- Not verifying whether a critical point is a max or min — use either the first or second derivative test.
- Stopping after finding x without computing the maximum value of the quantity (the question often asks for the value, not x).
- Using the second-derivative test when f'' = 0 — switch to the first-derivative test in that case.
4. Related Rates
Formulas
dV/dt = 4πr² · dr/dt (sphere)
Example
dV/dt = 10 m³/s, r = 2 m: dr/dt = 10/(4π·4) = 10/50.27 ≈ 0.199 m/s
Variables
V = volume of sphere; r = radius; t = time
Application
Inflating balloons, expanding bubbles — find dr/dt given dV/dt or vice versa
dA/dt = 2πr · dr/dt (circle)
Example
r = 5 m, dr/dt = 0.3 m/s: dA/dt = 2π(5)(0.3) = 3π ≈ 9.42 m²/s
Variables
A = area of circle; r = radius
Application
Ripple problems, circular pool expanding
2x·(dx/dt) + 2y·(dy/dt) = 0 (ladder/Pythagorean)
Example
L=5 m, x=3 m → y=4 m; dx/dt=0.5 m/s: dy/dt = −x(dx/dt)/y = −3(0.5)/4 = −0.375 m/s (top slides down)
Variables
x = horizontal distance; y = height on wall; x² + y² = L² (L = fixed ladder length)
Application
Sliding ladder, receding boat, kite problems
dV/dt = (1/3)π[r²(dh/dt) + h·2r(dr/dt)] (cone: general); dV/dt = (πr²/3)·dh/dt (similar-cone: r/h = constant)
Example
Conical tank, r/h = 1/2, so r = h/2. V = (π/12)h³; dV/dt = (π/4)h²·(dh/dt). At h=6m, dV/dt=−2m³/min: dh/dt = −2·4/(π·36) = −8/(36π) ≈ −0.0707 m/min
Variables
r = base radius; h = height of cone
Application
Draining conical tanks — most frequent board-exam related-rates problem type
Exam tips
- Always draw a diagram and assign variable names before writing any equation.
- In cone tank problems, write r in terms of h using r/h = R/H before differentiating — eliminates one variable.
- Label all given rates with their signs: filling = positive dV/dt; draining = negative dV/dt.
- Finish by checking units — if dV/dt is in m³/s and r is in m, then dr/dt should be in m/s.
- Ladder problems: once x and dx/dt are given, find y from the Pythagorean theorem before substituting.
Key points
- Related rates problems involve two or more quantities that both change with time; you differentiate their relationship with respect to t.
- Step 1: Draw and label a diagram identifying all time-varying quantities.
- Step 2: Write an equation relating the variables (geometric formula, Pythagorean theorem, similar triangles, etc.).
- Step 3: Differentiate both sides with respect to t using the chain rule.
- Step 4: Substitute the given numerical values (including the given rates) AFTER differentiating — NEVER before.
- Step 5: Solve for the unknown rate.
- Common geometric relations: sphere V = (4/3)πr³; circle A = πr²; cone V = (1/3)πr²h; right triangle: x² + y² = L².
- Rates are derivatives: dV/dt = rate of volume change; dr/dt = rate of radius change; dx/dt = speed of a moving point.
Definitions
Rate of Change
The derivative of a quantity with respect to time, e.g., dV/dt = rate of volume change in m³/s.
Importance
Related rates problems are entirely about connecting rates through a differentiating equation.
Similar Triangle Proportion
In cone problems, the ratio r/h = R/H (constant) allows expressing r in terms of h alone, reducing to one variable.
Importance
Without this substitution, cone problems require implicit differentiation of two unknowns — use the proportion.
Common mistakes
- Substituting numerical values BEFORE differentiating — the most common and costly error in related-rates problems.
- Getting the sign of rates wrong — decreasing quantities have negative rates; always assign signs consistently.
- Not using the similar-triangle ratio in cone problems, leading to two unknowns after differentiation.
- Confusing 'the shadow lengthening' rate with 'the tip of the shadow moving' rate in lamppost problems — these are different quantities.
- Forgetting the chain rule factor (dr/dt or dy/dt) when differentiating geometric formulas with respect to t.
5. Tangent Lines, Curvature, and Other Applications
Formulas
y − y₀ = f'(x₀)(x − x₀)
Example
y = x³ at x₀=1: y₀=1, y'=3x²→3; tangent: y−1=3(x−1) → y = 3x−2
Variables
x₀ = point of tangency; y₀ = f(x₀); f'(x₀) = slope
Application
Finding tangent line equation to any differentiable curve
R = [1 + (y')²]^(3/2) / |y''|
Example
y = x² at origin: y'(0)=0, y''(0)=2; R = [1+0]^(3/2)/2 = 1/2
Variables
y' = first derivative; y'' = second derivative; R = radius of curvature
Application
Road/rail alignment design; finding the tightest bend of a curve
f'(c) = [f(b) − f(a)] / (b − a)
Example
f(x)=x² on [1,3]: MVT gives 2c = (9−1)/(3−1) = 4 → c = 2 ∈ (1,3) ✓
Variables
c ∈ (a, b); f continuous on [a,b] and differentiable on (a,b)
Application
Mean Value Theorem; proving existence of specific derivative values
Exam tips
- For radius of curvature at the origin or vertex of a parabola, y'=0, so R simplifies to 1/|y''| — very fast to compute.
- Normal line slope is −1/f'(x₀); if f'(x₀)=0 (horizontal tangent), the normal is vertical (undefined slope — write x = x₀).
- Curve analysis: always find critical points AND inflection points before sketching — board multiple-choice options differ by concavity.
- MVT problems: set f'(x) = average rate, solve for x, and verify x is inside the open interval.
Key points
- Tangent line at (x₀, y₀): slope m = f'(x₀); equation: y − y₀ = m(x − x₀).
- Normal line: perpendicular to tangent; slope = −1/m.
- Radius of curvature R = [1 + (y')²]^(3/2) / |y''|; large R means nearly flat, small R means sharply curved.
- Center of curvature (evolute): the center of the osculating circle at a point.
- Curve sketching sequence: domain → intercepts → symmetry → asymptotes → critical points → inflection points → sketch.
- Rolle's Theorem: If f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then ∃ c ∈ (a,b) where f'(c) = 0.
- Mean Value Theorem (MVT): ∃ c ∈ (a,b) such that f'(c) = [f(b)−f(a)]/(b−a) — the derivative equals the average rate of change.
Definitions
Radius of Curvature
R = [1+(y')²]^(3/2)/|y''|; the radius of the circle that best fits the curve at a given point.
Importance
Appears in engineering design (road curves, beam deflection) and board-exam problems directly.
Inflection Point
A point where y'' = 0 AND y'' changes sign — the curve changes concavity.
Importance
Distinguish from critical points: inflection points are not extrema unless f' = 0 also holds.
Mean Value Theorem
Guarantees a point c where the instantaneous slope equals the average slope over an interval.
Importance
Foundation for many calculus proofs; occasionally tested directly on the licensure exam.
Common mistakes
- Confusing tangent slope with normal slope — they are negative reciprocals, not equal.
- Applying R = [1+(y')²]^(3/2)/|y''| with y' and y'' evaluated at different x values.
- Declaring an inflection point at x where y''=0 without verifying a sign change in y''.
- Mixing up Rolle's Theorem (requires f(a)=f(b)) with the general MVT.
Connections
- Differential Calculus → Integral Calculus: The Fundamental Theorem of Calculus links differentiation and integration — derivatives of integrals and antiderivatives of functions are board-exam bridge topics.
- Differential Calculus → Structural Analysis: Slope and deflection of beams involve EI·y'' = M(x); the bending moment diagram is derived by integrating the load function and the slope by differentiating the deflection.
- Differential Calculus → Fluid Mechanics: Rate-of-flow problems (dV/dt, dQ/dt) use the same related-rates framework applied to pipe flow and storage tanks.
- Differential Calculus → Physics / Engineering Mechanics: velocity = ds/dt (first derivative of displacement), acceleration = d²s/dt² (second derivative) — directly used in kinematics problems on the board exam.
- Differential Calculus → Optimization in Engineering Design: Minimum cost/maximum capacity problems in surveying (route alignment), geotechnical (earth volume), and structural engineering (minimum steel area) all reduce to f'(x)=0.
- Maxima-Minima → NSCP 2015 Load Combinations: While not calculus per se, optimization of load effects and structural efficiency uses the concept of extreme values.
- Differential Calculus → Numerical Methods: Newton-Raphson method for root-finding uses f'(x) iteratively — a direct application of differentiation reviewed in the Engineering Sciences portion.
Quick review questions
Evaluate: lim[x→0] (tan 3x) / (sin 5x)
Answer
3/5
Both numerator and denominator → 0 as x→0 (form 0/0). Use small-angle approximations: tan 3x ≈ 3x and sin 5x ≈ 5x near zero, so the limit = 3x/5x = 3/5. Alternatively, L'Hôpital: (3sec²3x)/(5cos5x) → 3(1)/5(1) = 3/5.
Find y' for y = (x² + 3)⁴ · e^(2x).
Answer
y' = e^(2x) · (x²+3)³ · [8x + 2(x²+3)] = 2e^(2x)(x²+3)³(4x + x²+3)
Product rule: u=(x²+3)⁴, v=e^(2x). u'=4(x²+3)³·2x=8x(x²+3)³; v'=2e^(2x). y' = 8x(x²+3)³·e^(2x) + (x²+3)⁴·2e^(2x) = 2e^(2x)(x²+3)³[4x + (x²+3)].
A farmer has 120 m of fencing to enclose a rectangular area with one side against a barn (no fencing needed on that side). Find the dimensions for maximum area.
Answer
Width = 30 m, Length = 60 m, Maximum Area = 1800 m²
Let width = x (two sides), length = y (one side free). Constraint: 2x + y = 120 → y = 120−2x. A = xy = x(120−2x). dA/dx = 120−4x = 0 → x=30 m; y=60 m. A_max = 30×60 = 1800 m². Check: d²A/dx²=−4<0 ✓ (maximum).
The radius of a circle is increasing at 2 cm/s. How fast is the area increasing when r = 5 cm?
Answer
dA/dt = 20π ≈ 62.83 cm²/s
A = πr². Differentiate w.r.t. t: dA/dt = 2πr·(dr/dt). Substitute r=5, dr/dt=2: dA/dt = 2π(5)(2) = 20π cm²/s.
For y = 3x³ − 5x² + 2x, find y'(2).
Answer
y'(2) = 18
y' = 9x² − 10x + 2. At x=2: y'(2) = 9(4) − 10(2) + 2 = 36 − 20 + 2 = 18.
Find the radius of curvature of y = x² at the origin.
Answer
R = 0.5
y' = 2x → y'(0) = 0. y'' = 2 → y''(0) = 2. R = [1+(y')²]^(3/2)/|y''| = [1+0]^(3/2)/2 = 1/2 = 0.5.
A 5-m ladder leans against a wall. The base slides away at 0.5 m/s. How fast is the top sliding down when the base is 3 m from the wall?
Answer
dy/dt = −0.375 m/s (the top slides down at 0.375 m/s)
x²+y²=25. At x=3: y=√(25−9)=4 m. Differentiate: 2x(dx/dt)+2y(dy/dt)=0. Substitute x=3, y=4, dx/dt=0.5: 2(3)(0.5)+2(4)(dy/dt)=0 → 3+8(dy/dt)=0 → dy/dt=−3/8=−0.375 m/s.
Find all critical points of f(x) = x³ − 3x² − 9x + 5 and classify them.
Answer
Local maximum at x = −1 [f(−1) = 10]; Local minimum at x = 3 [f(3) = −22]
f'(x) = 3x²−6x−9 = 3(x²−2x−3) = 3(x−3)(x+1) = 0 → x=3 or x=−1. f''(x)=6x−6. f''(−1)=−12<0 → local max; f''(3)=12>0 → local min. f(−1)=(−1)³−3(1)−9(−1)+5=−1−3+9+5=10; f(3)=27−27−27+5=−22.
Diagrams
Differential Calculus — Complete Topic Overview
A hierarchical mind map showing all major topic branches of Differential Calculus: Limits, Derivatives (with all rules), Extrema/Optimization, Related Rates, and other curve analysis applications. Use this as a navigation guide before diving into each section.
Optimization Problem-Solving Flowchart
A step-by-step decision flowchart for solving any optimization problem on the board exam. Follow from 'Start' through diagram drawing, constraint identification, single-variable reduction, differentiation, critical point classification, and endpoint checking to arrive at the final answer.
Related Rates — 5-Step Solution Procedure
A five-step flowchart for related-rates problems emphasizing the critical rule: substitute numerical values ONLY after differentiation (Step 4, highlighted in red). The procedure works for all related-rates problem types including spheres, circles, ladders, and cones.
Choosing the Correct Differentiation Rule
A decision-tree flowchart to help students quickly identify which differentiation rule to apply. Start at the top and follow the branches based on the form of the function. This eliminates the common mistake of applying the wrong rule under exam time pressure.
Approximate Distribution of Differential Calculus Topics in PRC MSTE Board Exams
Pie chart showing the estimated percentage distribution of differential calculus items in recent PRC Civil Engineer board exams. Optimization (30%) and differentiation rules (25%) account for more than half of all items — prioritize these topics in your review.
Exam strategy
For the MSTE board-exam differential calculus section, allocate time as follows: (1) Limits — 1 min each; direct substitution first, then L'Hôpital if needed. (2) Differentiation — 1.5 min each; identify the rule type (power, product, quotient, chain) before writing anything. (3) Maxima-minima — 3–4 min each; spend the first 30 seconds drawing the diagram and labeling variables, then set up the constraint and objective function before differentiating. (4) Related rates — 3–4 min each; write the geometric equation, differentiate w.r.t. t, THEN substitute numbers. High-yield topics based on past board exams: (a) Optimization of rectangles, boxes, and cylinders — appears almost every exam; (b) Sliding-ladder and conical-tank related rates — classic recurring items; (c) Chain-rule composite functions — 2–3 items every exam; (d) Radius of curvature — 1 item, very formula-direct. Use the process of elimination on multiple-choice: if the problem has a symmetric constraint (fixed perimeter, fixed surface area), the answer is almost always the most symmetric shape (square, cube, hemisphere). Always verify your critical point using the second-derivative test before selecting the final answer.