Difficulty: Intermediate | Prerequisites: Limits, continuity and the IVT (see Part 1 notes).
The derivative measures the instantaneous rate of change of a function, which geometrically is the slope of the tangent line. Differentiation rules (power, product, quotient, chain) let you compute derivatives efficiently, and the Mean Value Theorem connects a function's average and instantaneous rates of change. Applications range from finding when a particle speeds up or slows down, to locating absolute extrema on closed intervals, to optimising real-world quantities like area or cost.
Derivative
The limit f'(x) = lim(h→0) [f(x+h) – f(x)]/h. It gives the instantaneous rate of change of f at x. Think of it as: the speedometer reading at one exact moment, rather than your average speed over a trip.
Instantaneous velocity
The derivative of the position function: v(t) = s'(t). Positive means moving forward; negative means moving backward. In simple terms, this means: how fast and in which direction you are going right now.
Tangent line
The line that just touches the curve at a point and has slope equal to the derivative there. Its equation is y = f(a) + f'(a)(x – a). Think of it as: the best straight-line approximation to the curve near that point.
Chain rule
If h(x) = f(g(x)), then h'(x) = f'(g(x)) · g'(x). Differentiate the outer function, evaluate it at the inner function, then multiply by the derivative of the inner function. In simple terms, this means: peel off one layer at a time, from outside in, and multiply the results together.
Mean Value Theorem (MVT)
If f is continuous on [a, b] and differentiable on (a, b), there exists some c in (a, b) where f'(c) = [f(b) – f(a)]/(b – a). Think of it as: at some moment during the journey, your instantaneous speed exactly equals your average speed.
Critical point
A point where f'(c) = 0 or f'(c) does not exist. These are the only candidates for local maxima and minima. In simple terms, this means: the places where the curve flattens out or has a sharp corner.
Absolute maximum / minimum on a closed interval
The largest or smallest value f takes on [a, b]. Found by evaluating f at every critical point in (a, b) and at both endpoints, then comparing. Think of it as: checking every hilltop, valley and fence post, then picking the highest and lowest.
Related rates
Problems where two or more quantities change with time and are linked by an equation. You differentiate both sides with respect to t using the chain rule. In simple terms, this means: if two measurements are connected by a formula and one is changing, you can work out how fast the other changes.
For a position function like h(t) = 10t – 2t², the instantaneous velocity is h'(t) = 10 – 4t.
At t = 1: h'(1) = 10 – 4 = 6 metres per second.
Average velocity from t = 1 to t = 2 uses the secant formula: [h(2) – h(1)]/(2 – 1) = [12 – 8]/1 = 4 metres per second. (h(1) = 8, h(2) = 12.)
The tangent line at x = a is y = f(a) + f'(a)(x – a).
For h(t) = 10t – 2t² at t = 1: h(1) = 8, h'(1) = 6, so the tangent line is y = 8 + 6(t – 1).
Approximation at t = 1.1: y = 8 + 6(0.1) = 8.6. This is the tangent-line approximation.
For f(x) = x² – 4x, f'(x) = 2x – 4. At x = 4, the slope is f'(4) = 4.
Power rule: d/dx [xⁿ] = nxⁿ⁻¹.
Chain rule: If h(x) = sin(x² + x), then h'(x) = cos(x² + x) · (2x + 1). The outer function is sin, whose derivative is cos, evaluated at the inner function x² + x. Then multiply by the derivative of the inner function, which is 2x + 1.
At x = 0: h'(0) = cos(0) · 1 = 1.
The derivatives cycle with period 4: sin x → cos x → –sin x → –cos x → sin x → ...
To find the 83rd derivative of sin x: 83 = 4 × 20 + 3, so the 83rd derivative is the same as the 3rd, which is –cos x.
At x = 0: –cos(0) = –1.
Hypotheses: f continuous on [a, b], differentiable on (a, b).
Conclusion: there exists c in (a, b) with f'(c) = [f(b) – f(a)]/(b – a).
For f(x) = x² – 3x + 5 on [0, 3]: f(0) = 5, f(3) = 5, so the right side is 0/3 = 0. Set f'(x) = 2x – 3 = 0, giving c = 3/2.
The MVT can also prove uniqueness: if f'(x) > 0 everywhere, f is strictly increasing and can cross a horizontal line at most once.
Given s(t) = t³ – 3t² + 5t, velocity is v(t) = 3t² – 6t + 5 and acceleration is a(t) = 6t – 6.
At t = 2: v(2) = 12 – 12 + 5 = 5 (positive), a(2) = 12 – 6 = 6 (positive). Same sign means the particle is speeding up.
Rule: speeding up when v and a share the same sign; slowing down when they have opposite signs.
For f(x) = x³ – 12x + 1 on [0, 4]: f'(x) = 3x² – 12 = 0 gives x = 2 (only critical point in [0, 4]).
Evaluate: f(0) = 1, f(2) = 8 – 24 + 1 = –15, f(4) = 64 – 48 + 1 = 17.
Absolute max = 17 at x = 4. Absolute min = –15 at x = 2.
A Norman window: rectangle (width 2r, height h) topped by a semicircle of radius r. Perimeter = 2h + 2r + πr = 46.
Express h in terms of r: h = (46 – 2r – πr)/2.
Area = 2rh + (πr²)/2. Substitute h, differentiate, set A'(r) = 0, and solve for r = 46/(4 + π).
Justify it is a maximum: A''(r) < 0 at that point (or note that A is a downward-opening quadratic in r).
Example: a sphere's volume increases at dV/dt = 10 in³/s. Find dS/dt when r = 2.
V = (4/3)πr³, so dV/dt = 4πr² · dr/dt. At r = 2: 10 = 16π · dr/dt, so dr/dt = 10/(16π) = 5/(8π).
S = 4πr², so dS/dt = 8πr · dr/dt = 8π(2) · 5/(8π) = 10 in²/s.
Shortcut: since dS/dt = (2/r) · dV/dt, you get (2/2)(10) = 10 directly.
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\frac{d}{dx}[x^n] = n x^{n-1} \quad \text{(power rule)}\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \quad \text{(chain rule)}\text{Tangent line: } y = f(a) + f'(a)(x - a)\text{MVT: } f'(c) = \frac{f(b) - f(a)}{b - a} \quad \text{for some } c \in (a, b)\text{Derivatives of } \sin x \text{ cycle: } \sin x,\; \cos x,\; -\sin x,\; -\cos x,\; \sin x, \ldotsStudents often forget the chain rule when differentiating compositions. If you see sin(x² + x), you must multiply by (2x + 1). Writing just cos(x² + x) is incomplete.
Students confuse "velocity" with "speed." Velocity is signed (positive or negative); speed is its absolute value. A particle with negative velocity and negative acceleration is speeding up, not slowing down.
When finding absolute extrema, students sometimes check only the critical points and forget to evaluate f at the endpoints. On a closed interval, the endpoints can hold the absolute max or min.
Students sometimes apply the MVT without confirming the hypotheses. The function must be continuous on the closed interval and differentiable on the open interval.
⚠️ The chain rule appears in multiple questions. Practise it with compositions like sin(x² + x) and be ready to evaluate at a specific point.
⚠️ The 83rd derivative of sin x question tests whether you know the four-step cycle. Divide the derivative order by 4 and use the remainder.
⚠️ Speed questions ("is the particle speeding up or slowing down?") require you to check the signs of both velocity and acceleration. Same sign = speeding up, opposite signs = slowing down.
⚠️ MVT questions come in two forms: finding the value of c that satisfies the conclusion, and using the theorem to prove uniqueness of solutions (long answer). Know both.
⚠️ The Norman window optimisation is a classic long-answer problem. The perimeter constraint, area formula, and justification that the critical point is a maximum are all required for full marks.
⚠️ Related rates: set up the equation linking the quantities before differentiating. A common error is substituting numerical values too early, before differentiating.
True or false: If velocity is negative and acceleration is negative, the particle is slowing down. (False. Same sign means speeding up.)
Fill in the blank: The derivative of sin(x² + x) is cos(x² + x) · ___. (2x + 1)
True or false: The Mean Value Theorem requires f to be differentiable on the closed interval [a, b]. (False. Differentiable on the open interval (a, b) and continuous on [a, b].)
Fill in the blank: The 83rd derivative of sin x equals the same function as the ___rd derivative. (3rd, which is –cos x.)
True or false: When finding absolute extrema on [a, b], you only need to check critical points. (False. You must also check the endpoints.)
Q: For h(t) = 10t – 2t², what is the instantaneous velocity at t = 1?
A: h'(t) = 10 – 4t. At t = 1, h'(1) = 6 metres per second.
Q: For f(x) = x² – 3x + 5 on [0, 3], find the value of c satisfying the Mean Value Theorem.
A: f(3) = 5, f(0) = 5, so [f(3) – f(0)]/3 = 0. Set f'(x) = 2x – 3 = 0, giving c = 3/2.
Q: For h(x) = sin(x² + x), calculate h'(0).
A: h'(x) = (2x + 1)cos(x² + x). At x = 0: (0 + 1)cos(0) = 1.
Q: What is the value at x = 0 of the 83rd derivative of sin x?
A: 83 mod 4 = 3, so the 83rd derivative is –cos x. At x = 0: –cos(0) = –1.
Q: A particle has position s(t) = t³ – 3t² + 5t. At t = 2, is it speeding up or slowing down?
A: v(2) = 12 – 12 + 5 = 5, a(2) = 12 – 6 = 6. Both positive, so it is speeding up.
Q: Find the absolute maximum and minimum of f(x) = x³ – 12x + 1 on [0, 4].
A: f'(x) = 3x² – 12 = 0 gives x = 2. f(0) = 1, f(2) = –15, f(4) = 17. Absolute max = 17 at x = 4, absolute min = –15 at x = 2.
Q: Use the MVT to show that 3x + 2cos x – 5 = 0 has at most one real solution.
A: Let f(x) = 3x + 2cos x – 5. Then f'(x) = 3 – 2sin x. Since –1 ≤ sin x ≤ 1, we have f'(x) ≥ 3 – 2 = 1 > 0 for all x. A function with a strictly positive derivative is strictly increasing, so it can cross the x-axis at most once.
Q: A sphere’s volume increases at 10 in³/s. At r = 2 inches, find the rate of change of surface area.
A: dS/dt = (2/r) · dV/dt = (2/2)(10) = 10 in²/s.
Derivatives connect back to limits (the derivative is defined as a limit) and forward to integration (the Fundamental Theorem of Calculus links the two). The MVT underpins several results in integration, including the proof that two antiderivatives of the same function differ by a constant. Optimisation problems combine derivative techniques with real-world modelling, and related rates extend the chain rule to time-dependent scenarios.
derivative, differentiation, instantaneous rate of change, tangent line, linear approximation, chain rule, composition of functions, power rule, product rule, quotient rule, Mean Value Theorem, MVT, Rolle's theorem, critical point, absolute maximum, absolute minimum, closed interval method, optimisation, Norman window, related rates, velocity, acceleration, speed, speeding up, slowing down, higher-order derivatives, successive derivatives, sin x derivative cycle, MTH 132, Calculus I, Michigan State University