📐 Differential Calculus
How fast is something changing right now? Differentiation answers exactly that — the gradient of a curve at a single instant — and underlies velocity, stress, optimisation and countless other engineering calculations.
2Topics
8Worked Examples
4Eng. Cases
12Practice Qs
📏Differentiation RulesPower, product, chain rules
📈Turning PointsMax/min, rates of change
⚙️ApplicationsReal engineering
🎮Quick-Fire GameTest your speed
📝PracticeGraded questions
🎯What you will learn
Apply the power, product, quotient and chain rules to differentiate polynomial, trigonometric and exponential functions; find and classify turning points using the second derivative; and interpret derivatives as rates of change in real engineering contexts.
💡 dy/dx is not a fraction — it's the limit of a ratio as the change in x shrinks to zero. But for calculation purposes, treating it like a fraction (as the chain rule does) works perfectly.
🧭How to use this resource
1. Read the rules banner at the top of each tab
2. Experiment with the interactive tool
3. Check the key vocabulary panel
4. Study the visual diagram, then the worked examples
📏
Differentiation Rules
A small toolkit of rules handles almost every function an engineer will meet.
🔧Power Rule DifferentiatorEnter a coefficient and power for axⁿ
Enter a and n for the term axⁿ, then press Differentiate to see the power rule applied step by step…
📖 Key Vocabulary
Derivative (dy/dx) — the instantaneous rate of change of y with respect to x; the gradient of the curve at a point.
Power rule — the fundamental rule for differentiating xⁿ, reducing the power by one and multiplying by the original power.
Chain rule — differentiates a "function of a function" by multiplying the outer and inner derivatives together.
Product rule — d/dx(uv) = u(dv/dx) + v(du/dx), for a product of two functions of x.
👁️ Visual aid: the tangent line just touches the curve at one point — its slope is exactly the value of dy/dx there.
✏️Worked Example 1 — Power rule on a polynomialType: Power rule
Question Differentiate y = 4x³ − 2x² + 5x − 7.
Apply term by term
d/dx(4x³)=12x², d/dx(−2x²)=−4x, d/dx(5x)=5, d/dx(−7)=0Answer dy/dx = 12x² − 4x + 5
🔗Worked Example 2 — The chain ruleType: Chain rule
Question Differentiate y = (3x+1)⁴.
Set up
u=3x+1, y=u⁴; du/dx=3, dy/du=4u³Multiply
dy/dx = 4u³ × 3 = 12u³Answer dy/dx = 12(3x+1)³
✖️Worked Example 3 — The product ruleType: Product rule
Question Differentiate y = x²sinx.
Set up
u=x², v=sinx; du/dx=2x, dv/dx=cosxApply formula
dy/dx = u(dv/dx)+v(du/dx) = x²cosx + 2xsinxAnswer dy/dx = x²cosx + 2xsinx
⚙️Worked Example 4 — Engineering context (velocity from displacement)Type: Engineering
Question A particle's displacement is s = 2t³ − 6t² + 4t (m). Find the velocity function.
Recall Velocity v = ds/dt (the derivative of displacement)
Differentiate
v = 6t² − 12t + 4Answer v = 6t² − 12t + 4 (m/s) — an equation the engineer can now use to find velocity at any specific time.
🧠 Common slip: forgetting to multiply by the inner derivative when using the chain rule — differentiating (3x+1)⁴ as just 4(3x+1)³, missing the ×3 from the inner function, is a very common error.
📈
Turning Points & Rates of Change
Where a curve is momentarily flat — and how the second derivative reveals whether that's a peak or a trough.
🔧Turning Point ClassifierEnter a quadratic ax²+bx+c for y' (the first derivative)
Enter a, b, c for y' = ax²+bx+c, then press Find Turning Points to solve y'=0 and classify each using y''…
📖 Key Vocabulary
Turning point (stationary point) — a point where the gradient dy/dx equals zero.
Local maximum — a turning point where the curve rises then falls; d²y/dx² < 0.
Local minimum — a turning point where the curve falls then rises; d²y/dx² > 0.
Point of inflection — where d²y/dx²=0 and concavity changes; not a max or min.
👁️ Visual aid: the tangent is horizontal (gradient zero) at both labelled points — the second derivative's sign tells them apart.
✏️Worked Example 1 — Finding and classifying a turning pointType: Full classification
Question Find and classify the turning points of y = x³ − 3x².
First derivative
dy/dx = 3x² − 6x; set to 0: 3x(x−2)=0 → x=0 or x=2Second derivative
d²y/dx² = 6x−6Classify At x=0: y''=−6 (max). At x=2: y''=6 (min)
Answer Local max at x=0 (y=0), local min at x=2 (y=−4)
🚗Worked Example 2 — Rate of change at a specific instantType: Instantaneous rate
Question For s = 5t² − 3t, find the velocity at t=2s.
Differentiate
v = ds/dt = 10t − 3Substitute t=2
v = 10(2) − 3Answer v = 17 m/s at t=2s
🎯Worked Example 3 — Optimisation (maximum area)Type: Optimisation
Question A rectangular enclosure has perimeter 40m. Find the dimensions that maximise the area.
Set up
2x+2y=40 → y=20−x; Area A=xy=x(20−x)=20x−x²Differentiate & solve
dA/dx=20−2x=0 → x=10Check
d²A/dx²=−2<0 → maximum confirmedAnswer x=10m, y=10m — a square gives the maximum area (100m²) for this fixed perimeter.
⚙️Worked Example 4 — Engineering context (acceleration from velocity)Type: Engineering
Question A lift's velocity is v = 3t² − 12t + 9 (m/s). Find the acceleration function and the time(s) when acceleration is zero.
Differentiate
a = dv/dt = 6t − 12Solve a=0
6t−12=0 → t=2Answer a = 6t−12 (m/s²); acceleration is zero at t=2s — the instant the lift's velocity is momentarily at a maximum or minimum.
🧠 Common slip: finding where dy/dx=0 but forgetting to check the second derivative — without that check, you can't tell a maximum from a minimum (or an inflection point).
⚙️
Engineering Applications
Four live scenarios — each with its own calculator. Enter your own numbers.
🚗 Velocity & Acceleration from Displacement
Differentiating displacement gives velocity; differentiating again gives acceleration — the foundation of kinematics.
v = ds/dt, a = dv/dt
📟 Try it — Kinematics Differentiator
Enter A, B, C and a time t, then press Calculate for velocity and acceleration…
📦 Optimising a Container's Dimensions
Differentiation finds the dimensions that minimise material use or maximise volume for a given constraint.
Set dA/dx = 0, check d²A/dx² for max/min
📟 Try it — Optimisation Helper (quadratic area A=px²+qx)
Enter p and q for A=px²+qx, then press Calculate for the optimum x…
🌡️ Rate of Temperature Change
Differentiating a temperature-time model gives the instantaneous heating or cooling rate at any moment.
dT/dt at a specific time t
📟 Try it — Rate of Change Calculator (T=At²+Bt+C)
Enter A, B and t, then press Calculate for dT/dt…
🔩 Stress Gradient Along a Beam
Differentiating a stress distribution function locates the point of maximum stress along a structural member.
Maximum stress where dσ/dx = 0
📟 Try it — Stress Turning Point (σ'=ax+b)
Enter a and b for σ'=ax+b, then press Calculate for the position of zero stress-gradient…
🎮
Quick-Fire Differentiation Game
12 questions on rules and turning points. How many can you get?
⚡ Differentiation Challenge
Score: 0Question: 1/12Streak: 0🔥
Press Start to begin!
📝
Practice Questions
Graded from warm-up to challenge. Attempt fully before revealing the working.
Question 1 🥉2 marks
Differentiate y = 5x⁴.
dy/dx = 20x³
Question 2 🥉2 marks
Differentiate y = 7x² − 3x + 2.
dy/dx = 14x − 3
Question 3 🥉2 marks
Differentiate y = sin(2x).
dy/dx = 2cos(2x)
Question 4 🥉2 marks
For s=4t²−3t, find ds/dt.
8t − 3
Question 5 🥈3 marks
Differentiate y = (2x+5)³ using the chain rule.
u=2x+5, dy/du=3u², du/dx=2
dy/dx = 6(2x+5)²
dy/dx = 6(2x+5)²
Question 6 🥈4 marks
Find the turning point of y=x²−6x+5 and classify it.
dy/dx=2x−6=0 → x=3
d²y/dx²=2>0
Minimum at x=3 (y=−4)
d²y/dx²=2>0
Minimum at x=3 (y=−4)
Question 7 🥈3 marks
Differentiate y = x·cosx using the product rule.
u=x,v=cosx; du/dx=1, dv/dx=−sinx
dy/dx = −xsinx + cosx
dy/dx = −xsinx + cosx
Question 8 🥈4 marks
For v=4t²−8t+3 (m/s), find the acceleration at t=3s.
a=dv/dt=8t−8
At t=3: a=8(3)−8=16 m/s²
At t=3: a=8(3)−8=16 m/s²
Question 9 🥇4 marks
Find and classify both turning points of y=2x³−9x²+12x.
dy/dx=6x²−18x+12=6(x²−3x+2)=6(x−1)(x−2)=0 → x=1,2
d²y/dx²=12x−18
At x=1: y''=−6 (max); at x=2: y''=6 (min)
Max at x=1 (y=5), min at x=2 (y=4)
d²y/dx²=12x−18
At x=1: y''=−6 (max); at x=2: y''=6 (min)
Max at x=1 (y=5), min at x=2 (y=4)
Question 10 🥇5 marks
A rectangular field with one side against a wall (no fence needed there) has 60m of fencing for the other 3 sides. Find the dimensions for maximum area.
Let width=x (2 sides), length=y (1 side): 2x+y=60 → y=60−2x
A=xy=x(60−2x)=60x−2x²
dA/dx=60−4x=0 → x=15
y=60−30=30
d²A/dx²=−4<0 (max)
x=15m, y=30m, giving max area 450m²
A=xy=x(60−2x)=60x−2x²
dA/dx=60−4x=0 → x=15
y=60−30=30
d²A/dx²=−4<0 (max)
x=15m, y=30m, giving max area 450m²
Question 11 🥇5 marks
Differentiate y = sin(3x)·e^(2x) using the product rule.
u=sin(3x), v=e^(2x); du/dx=3cos(3x), dv/dx=2e^(2x)
dy/dx = 3cos(3x)e^(2x) + 2sin(3x)e^(2x) = e^(2x)[3cos(3x)+2sin(3x)]
dy/dx = 3cos(3x)e^(2x) + 2sin(3x)e^(2x) = e^(2x)[3cos(3x)+2sin(3x)]
Question 12 🥇5 marks
A projectile's height is h=20t−5t² (m). Find the maximum height reached and the time it occurs.
dh/dt=20−10t=0 → t=2s
d²h/dt²=−10<0 (max confirmed)
h=20(2)−5(4)=40−20
Max height = 20m at t=2s
d²h/dt²=−10<0 (max confirmed)
h=20(2)−5(4)=40−20
Max height = 20m at t=2s