Further Differentiation

— Pure Mathematics 1 · AS & A Level (9709) · Chapter 8 —

★ Exam Tip: Stationary points, max / min word problems and connected rates of change turn up in almost every Paper 1 (see the 9709 questions at the end of the chapter). Know the second-derivative test and the chain-rule set-up cold.

Learn how to: ① use dy/dx to find where a function is increasing / decreasing ② locate stationary points & decide their nature ③ solve practical max / min problems ④ solve rates-of-change problems.

Before you start (Prerequisites): solve quadratic inequalities (Ch 1) · find dy/dx and d²y/dx² of xⁿ (Ch 7) · differentiate composite functions with the chain rule (Ch 7).
e.g. x² − 2x − 3 > 0 ⇒ (x−3)(x+1) > 0 ⇒ x < −1 or x > 3 | d/dx (2x−1)⁵ = 10(2x−1)⁴.

Why study this? Cans: use the least metal for a given volume · doctors: when is a drug concentration rising? · economists: pricing / profit · scientists: how fast is an oil slick spreading?

1. Increasing & Decreasing Functions

Explore 8.1 – answers: increasing graph → as x ↑ the value of y ↑ and the gradient is always positive (+). Decreasing graph → as x ↑, y ↓ and the gradient is always negative (−).

Definition: Increasing function f(x): f(a) < f(b) whenever a < b.
Decreasing function f(x): f(a) > f(b) whenever a < b.

A function can also be increasing / decreasing at a point: gradient > 0 there ⇒ increasing; gradient < 0 there ⇒ decreasing.

Key idea (Checklist)dy/dx > 0 throughout an interval ⇒ increasing
dy/dx < 0 throughout an interval ⇒ decreasing

increasing

decreasing

h(x): ↓ for x < a, ↑ for x > a

Graph of y = h(x): h(x) decreasing for x < a (dy/dx < 0) · increasing for x > a (dy/dx > 0).

Method: ① find dy/dx ② solve dy/dx > 0 (increasing) or dy/dx < 0 (decreasing) ③ if it is a quadratic inequality → find the critical values (roots) and sketch the graph of dy/dx to read off the answer.

WE 8.1decreasing · linear gradient

Find the set of values of x for which y = 8 − 3x − x² is decreasing.

y = 8 − 3x − x²dy/dx = −3 − 2xdecreasing ⇒ dy/dx < 0−3 − 2x < 0−3 < 2xx > −3/2

WE 8.2cubic · quadratic inequality

f(x) = 4x³ − 15x² − 72x − 8. (a) find f′(x) (b) f increasing for? (c) f decreasing for?

f′(x) = 12x² − 30x − 72increasing ⇒ f′(x) > 012x² − 30x − 72 > 02x² − 5x − 12 > 0 (÷6)(2x + 3)(x − 4) > 0critical values: x = −3/2 and x = 4b) x < −3/2 and x > 4c) f′(x) < 0 ⇒ −3/2 < x < 4

sketch f′(x): + outside the roots, − between ⇒ f ↑ ↑, ↓ in the middle

Common mistake: forgetting to flip the inequality sign when you divide by a negative number. Also, for a quadratic ≥ / ≤ question don’t just write “x = −3/2, 4” — give the ranges using the sketch.

WE 8.3reciprocal · chain rule

f(x) = 5/(2x − 3) for x > 3/2. Find f′(x); is f increasing, decreasing or neither?

f(x) = 5(2x − 3)⁻¹ (write ready to differentiate)f′(x) = −5(2x − 3)⁻²(2) (chain rule)f′(x) = −10/(2x − 3)²(2x − 3)² > 0 for all x ⇒ f′(x) < 0

∴ f is a decreasing function (for all x in its domain)

Trick: a squared bracket is never negative, so −(…)² or −k/(…)² is always negative ⇒ always decreasing. Quote this reason in answers!

✓ Quick Recap — Inc / Dec

dy/dx > 0 ⇒ increasing, dy/dx < 0 ⇒ decreasing
quadratic inequality: critical values + sketch
“is f increasing/decreasing/neither?” → find f′ and give a reason for its sign

2. Stationary Points

Definition: a point where the gradient is zero is a stationary point (turning point): dy/dx = 0.

red = gradient negative (f decreasing), blue = gradient positive (f increasing). Gradient is 0 at P, Q, R

Three kinds of stationary point:

Maximum (Q): y is greater than at nearby points.
Minimum (P, R): y is smaller than at nearby points.
Point of inflexion: gradient 0 but the curve keeps going the same way.

Type	dy/dx	gradient left → right
Maximum	= 0	+ → 0 → −
Minimum	= 0	− → 0 → +
Inflexion	= 0	+ → 0 → + or − → 0 → −

maximum

minimum

inflexion (rising)

inflexion (falling)

Remember: a max looks like a hill ⌒ (+ 0 −), a min like a valley ⌣ (− 0 +). Inflexion: the sign on both sides is the same.

First derivative test

Check the sign of dy/dx just either side of the stationary point (use x-values very close to it) — see the table in WE 8.4.

WE 8.4cubic · first derivative test + sketch

Find the stationary points of y = x³ − 12x + 5, their nature, and sketch the graph.

dy/dx = 3x² − 12dy/dx = 0 ⇒ 3x² − 12 = 0x² − 4 = 0(x + 2)(x − 2) = 0 ⇒ x = −2 or x = 2x = −2: y = (−2)³ − 12(−2) + 5 = 21x = 2: y = (2)³ − 12(2) + 5 = −11

Stationary points: (−2, 21) and (2, −11)

x	−2.1	−2	−1.9
dy/dx = 3x²−12	3(−2.1)²−12 = +	0	3(−1.9)²−12 = −
tangent	╱ rising	— flat	╲ falling
shape	⌒ (+ 0 −) ⇒ max

x	1.9	2	2.1
dy/dx = 3x²−12	3(1.9)²−12 = −	0	3(2.1)²−12 = +
tangent	╲ falling	— flat	╱ rising
shape	⌣ (− 0 +) ⇒ min

y = x³ − 12x + 5 : max at (−2, 21), min at (2, −11), y-intercept 5

Second derivative test

Moving left → right through a max: gradient dy/dx goes + → 0 → −, so it is decreasing ⇒ the rate of change of dy/dx is negative.
Through a min: gradient goes − → 0 → +, increasing ⇒ rate of change of dy/dx is positive.

Second derivatived²ydx² = ddx(dydx) = rate of change of the gradient

Key Point 8.1if dy/dx = 0 and d²ydx² < 0 ⇒ maximum point
Key Point 8.2if dy/dx = 0 and d²ydx² > 0 ⇒ minimum point

Careful: if dy/dx = 0 and d²y/dx² = 0 the second derivative test tells you nothing (could be max, min or inflexion) ⇒ go back to the first derivative test (signs either side).

Memory: d²y/dx² negative → sad face ⌒ → max. d²y/dx² positive → smile ⌣ → min.

WE 8.5second derivative test

Find the stationary points of y = (x² + 9)/x and use d²y/dx² to find their nature.

y = x + 9x⁻¹dy/dx = 1 − 9x⁻² = 1 − 9/x²dy/dx = 0 ⇒ 1 − 9/x² = 0x² − 9 = 0 ⇒ (x+3)(x−3) = 0x = −3 : y = ((−3)²+9)/(−3) = −6x = 3 : y = (3²+9)/3 = 6d²y/dx² = 18x⁻³ = 18/x³x = −3 : 18/(−3)³ < 0 ⇒ maxx = 3 : 18/3³ > 0 ⇒ min

(−3, −6) is a maximum · (3, 6) is a minimum

Note: a “maximum” is only a local maximum — the y-value at (−3, −6) is lower than at (3, 6)! It is higher than the points near it only.

Explore 8.2reading f′ and f″ graphs

y = f(x) passes through (1, −35) and (6, 90). The graph of f′(x) crosses the x-axis at x = 1 and x = 6 (a ∩ shape); f″(x) is a straight line with negative gradient, crossing the axis at x = 3.5.

f′(x) = 0 at x = 1 and x = 6 ⇒ stationary points theref″(1) > 0 (line above axis) ⇒ min at (1, −35)f″(6) < 0 (line below axis) ⇒ max at (6, 90)f′ < 0 for x < 1 and x > 6, f′ > 0 for 1 < x < 6 ⇒ ↓ ↑ ↓

sketch of f: decreasing → increasing → decreasing

3. Practical Maximum & Minimum Problems

Use: least metal for a can, biggest area for a fence, cheapest cost, etc. Always the same plan:

method for every optimisation question

Exam Tip: Many are “Show that A = …” — the constraint gives you one variable in terms of the other; substitute it into the formula. Then state max/min with the d²y/dx² reason. Finally answer the actual question (value of x? the max value? the dimensions?).

WE 8.6cuboid x × x × h · max volume

Solid cuboid with square base x cm × x cm, height h cm. Surface area = 100 cm², volume V cm³. (a) h in terms of x (b) show V = 25x − ½x³ (c) stationary value of V, max or min?

S.A. = 2x² + 4xh (2 square ends + 4 sides)2x² + 4xh = 100h = (100 − 2x²)/4x = 25/x − ½x (a)V = x²h (substitute for h)V = x²(25/x − ½x) = 25x − ½x³ (b) shown

cuboid: base x by x, height h

dV/dx = 25 − (3/2)x²dV/dx = 0 ⇒ 25 − (3/2)x² = 0x² = 50/3 ⇒ x = 5√6/3 (≈ 4.08)V = 25(5√6/3) − ½(5√6/3)³ = 68.04 (2 d.p.)d²V/dx² = −3xat x = 5√6/3 : d²V/dx² = −5√6 < 0

Stationary value of V is 68.04 and it is a maximum.

WE 8.7cylinder inside sphere (radius 5 cm) · Pythagoras

Solid cylinder of radius r cm and height 2h cm cut from a sphere of radius 5 cm. Volume V cm³. (a) r in terms of h (b) show V = 50πh − 2πh³ (c) h for a stationary V (d) nature.

r² + h² = 5² (Pythagoras: half-height h, radius r, hypotenuse 5)r = √(25 − h²) (a)V = πr²(2h) (substitute for r)V = π(25 − h²)(2h) = 50πh − 2πh³ (b)

right-angled triangle: r, h and 5 ⇒ r² = 25 − h²

dV/dh = 50π − 6πh²50π − 6πh² = 0h² = 50π/6π = 25/3h = 5√3/3 (c) (≈ 2.89)d²V/dh² = −12πhat h = 5√3/3 : d²V/dh² < 0 ⇒ maximum (d)

Note: the printed worked example writes “d²V/dx²” in the last lines of 8.7 and 8.8 — it should be d²V/dh² and d²V/dr² (the variable you differentiated with respect to).

WE 8.8cylinder inside cone · similar triangles

Hollow cone: base radius 12 cm, height 24 cm. A solid cylinder (radius r cm, height h cm, volume V cm³) stands on the base, upper edge touching the cone. (a) h in terms of r (b) show V = 24πr² − 2πr³ (c) volume of the largest cylinder.

r/12 = (24 − h)/24 (similar triangles)2r = 24 − hh = 24 − 2r (a)V = πr²h = πr²(24 − 2r) = 24πr² − 2πr³ (b)

small triangle on top: radius r out of 12, height 24 − h out of 24

dV/dr = 48πr − 6πr²48πr − 6πr² = 06πr(8 − r) = 0 ⇒ r = 8 (r ≠ 0)V = 24π(8)² − 2π(8)³ = 512πd²V/dr² = 48π − 12πrat r = 8 : 48π − 96π < 0 ⇒ max

Largest cylinder has volume 512π cm³ (r = 8, h = 8).

Did you know? Differentiation is used in business to maximise profit and minimise production costs.

✓ Quick Recap — Optimisation

get ONE variable (use the constraint) → differentiate → set = 0
d²y/dx² < 0 max, > 0 min — say it
similar triangles, Pythagoras, S.A. / volume formulas are the usual helpers

4. Rates of Change

Idea: rate of change = gradient = derivative. For h against t: dh/dt = rate of change of height with time.

Explore 8.3: pour water at a constant 10 cm³ s⁻¹ into 3 containers: A straight-sided (cylinder), B narrowing upward (cone-like), C widening upward (bowl).

A: constant width ⇒ h rises at a constant rate (gradient constant)

B: gets narrower ⇒ h rises faster and faster

C: gets wider ⇒ h rises slower and slower

A: height increases at a constant rate (gradient of the straight line). B and C: not constant — rate at time t = gradient of the tangent (estimate by drawing it) or, more accurately, differentiate if you know the equation.

WE 8.9rate of change of h

h = t²/5. Find the rate of change of h with respect to t when t = 2.

h = (1/5)t²dh/dt = (2/5)tt = 2 : dh/dt = (2/5)(2) = 4/5

WE 8.10rate of change of V

V = 2t² − 3t + 8. Find the rate of change of V with respect to t when t = 4.

dV/dt = 4t − 3t = 4 : dV/dt = 4(4) − 3 = 13

Connected rates of change

When x and y both change with a third variable t, link them with the chain rule:

Key Point 8.3 (chain rule)dydt = dydx × dxdt

Key Point 8.4dxdy = 1dydx (flip the fraction)

Why? Put t = y in the chain rule: dy/dy = (dy/dx)(dx/dy) and dy/dy = 1 ⇒ dx/dy = 1 ÷ (dy/dx).

Trick: chain rule = “fractions cancel”: dy/dt = dy/dx × dx/dt. Write what you want on the left, then fill the gaps.

WE 8.11point moving on a curve

Point (x, y) moves on y = x + √(2x + 3) so that x increases at a constant 0.06 units/s. Find the rate of increase of y when x = 3; is y increasing or decreasing?

given: dx/dt = 0.06y = x + (2x + 3)^½dy/dx = 1 + ½(2x + 3)^(−½)(2) = 1 + 1/√(2x + 3)x = 3 : dy/dx = 1 + 1/√9 = 4/3dy/dt = dy/dx × dx/dtdy/dt = (4/3) × 0.06 = 0.08

y increases at 0.08 units/s (dy/dt is positive ⇒ increasing).

Common mistake: forgetting the sign: dy/dt > 0 ⇒ increasing, dy/dt < 0 ⇒ decreasing — say which when asked. “Decreasing at 3 units/s” means dy/dt = −3.

5. Practical Applications of Connected Rates

Steps: ① write the rate given and the rate wanted as derivatives (with units) ② find a formula linking the quantities (A = πr², V = ⁴⁄₃πr³ …) ③ differentiate it w.r.t. its own variable ④ chain rule ⑤ substitute the value ⑥ answer with units & “increasing”.

oil patch: radius grows, area grows even faster

WE 8.12oil patch · area

A circular oil patch: radius increases at 2 m per hour. Find the rate the area is increasing when r = 25 m.

find dA/dt when r = 25 ; dr/dt = 2A = πr²dA/dr = 2πrr = 25 : dA/dr = 50πdA/dt = dA/dr × dr/dtdA/dt = 50π × 2 = 100π

Area increases at 100π m² per hour.

WE 8.13sphere · surface area & volume

Solid sphere radius r cm, surface area A cm², volume V cm³. Radius increases at 1/(5π) cm s⁻¹. (a) rate of increase of surface area when r = 3 (b) rate of increase of volume when r = 5.

(a) dr/dt = 1/5πA = 4πr²dA/dr = 8πr = 24π (r = 3)dA/dt = dA/dr × dr/dt = 24π × 1/5π = 4.8 cm² s⁻¹

(b) V = ⁴⁄₃πr³dV/dr = 4πr² = 100π (r = 5)dV/dt = 100π × 1/5π = 20 cm³ s⁻¹

water in a conical container (vertex down)

WE 8.14filling a cone · dh/dt

Water poured into a cone at 2π cm³ s⁻¹; volume of water V = (1/12)πh³, h = height of water. (a) rate of change of h when h = 5 (b) cone has radius 10, height 20: rate of change of h when it is half full (3 s.f.).

(a) dV/dt = 2π ; want dh/dtV = (1/12)πh³dV/dh = ¼πh² = 25π/4 (h = 5)dh/dt = dh/dV × dV/dtdh/dt = 4/25π × 2π = 0.32 cm s⁻¹

WE 8.14 (b)half full

full volume = (1/12)π(20)³ = 2000π/3half full: V = 1000π/3(1/12)πh³ = 1000π/3h³ = 4000 ⇒ h = 15.874…dV/dh = ¼π(15.874)² = 197.9dh/dt = 1/197.9 × 2π = 0.0317 cm s⁻¹ (3 s.f.)

Trick: when the question gives dV/dt and wants dh/dt use dh/dt = (dh/dV) × (dV/dt) — and dh/dV = 1 ÷ (dV/dh) (Key Point 8.4). Don’t forget a “half-full” means HALF THE VOLUME, not half the height.

6. Full Chapter Cheat-Sheet

✓ Increasing / decreasing

dy/dx > 0 throughout an interval ⇒ increasing; dy/dx < 0 ⇒ decreasing
f(a) < f(b) when a < b (increasing); f(a) > f(b) when a < b (decreasing)
quadratic f′: critical values + sketch → ranges

✓ Stationary points

stationary (turning) point ⇔ dy/dx = 0 → solve for x, then find y
First derivative test: max (+ 0 −) · min (− 0 +) · inflexion (same sign both sides)
Second derivative test: d²y/dx² < 0 max · > 0 min · = 0 ⇒ use first derivative test

✓ Practical max / min

diagram → formula → constraint → one variable → dV/dx = 0 → test → answer the question asked
sphere / cone / cylinder: Pythagoras, similar triangles

✓ Connected rates

dy/dt = dy/dx × dx/dt | dx/dy = 1 ÷ (dy/dx)
A = πr², dA/dr = 2πr · A = 4πr², dA/dr = 8πr · V = ⁴⁄₃πr³, dV/dr = 4πr²
sign of dy/dt tells increasing or decreasing; keep units (cm s⁻¹, cm² s⁻¹, cm³ s⁻¹)