Differentiation answers one question: how fast is something changing right now? The speed of a car at this instant. The rate a capacitor is charging. Whether a beam is getting stronger or weaker at this point. It is the most powerful tool in engineering maths.
dy/dx = instantaneous rate of changeGradient of curve at a pointSlope of tangent lineVelocity = ds/dtAcceleration = dv/dt
ℹ️ How to use: Drag the slider left and right to move the point along the curve y = x³ − 3x + 1. The gold tangent line shows the gradient at that exact point. Notice the gradient is 0 at the peaks and troughs — that is where dy/dx = 0.
🎯 Drag to move the tangent line — watch the gradient change
0.0
x = 0.0 | y = f(0) = 1.000 | dy/dx = f'(0) = -3.000 | tangent falling steeply
💡 Notice how the gradient (shown in gold) changes continuously as x moves. At the peaks and troughs the gradient = 0. This is the heart of differentiation.
What dy/dx actually means
Think of a car journey. The distance s changes with time t. The speed at any instant is how fast distance is changing — written ds/dt.
Average speed = Δs/Δt (over an interval)
Instantaneous speed = ds/dt (at one moment)
dy/dx = gradient of the curve at point x
💡 dy/dx is not a fraction — it is the limit of Δy/Δx as Δx→0. We call it the derivative of y with respect to x.
Why engineers need it
🚗 Velocity= rate of change of displacement
⚡ Current= rate of change of charge: i = dq/dt
🌡️ Cooling rate= dT/dt (how fast temp drops)
⚙️ Stress rate= dσ/dε (stiffness)
📐 Optimisation— find the maximum or minimum
🔊 Signal processing— rate of change of voltage
❓ What does f(x) mean?
In maths you will constantly see f(x), g(x), f′(x). These are not special — they are just shorthand names meaning “some function of x”.
f(x) = “a function of x” — any rule that turns x into a number
f(3) means “put x = 3 into f” — evaluate the function at x = 3
f′(x) means “the derivative of f” — the prime symbol ′ means differentiated
g(x) is just a second function, used when you need two at once
💡 Example: if f(x) = x², then f(3) = 9 and f′(x) = 2x. If g(x) = sin x, then g′(x) = cos x. That’s all there is to it.
The notation dy/dx explained
There are two ways to write the derivative — they mean exactly the same thing:
dy/dx — read “dee y by dee x”. Used when y is written as a formula in x (e.g. y = x²)
f′(x) — read “f prime of x”. Used when the function has been given a name like f
dy/dx = f′(x) — both mean “the gradient of the curve at x”
d²y/dx² = f″(x) — the second derivative (differentiate twice)
💡 In engineering you also see ds/dt (rate of change of displacement s with time t), dv/dt (acceleration), dI/dt (rate of change of current). The pattern is always: d(output)/d(input).
📐
Differentiation Rules
Three rules cover 90% of T Level problems. Learn the pattern, not the individual cases.
Power rule — worked examples
d/dx[x⁵] = 5x⁴ (bring 5 down, reduce to x⁴)
d/dx[x²] = 2x (n=2 → 2x¹ = 2x)
d/dx[x] = 1 (n=1 → 1x⁰ = 1)
d/dx[c] = 0 (constant — no change)
d/dx[x⁻¹] = −x⁻² = −1/x²
d/dx[x½] = ½x-½ = 1/(2√x)
💡 The pattern: bring the power down as a multiplier, then subtract 1 from the power. Works for any real n — positive, negative, fraction.
Full polynomials
Differentiate y = 3x⁴ − 2x³ + 5x − 7
3x⁴ → 3×4·x³ = 12x³
−2x³ → −2×3·x² = −6x²
5x → 5×1 = 5
−7 → 0 (constant)
dy/dx =12x³ − 6x² + 5 ✅
Differentiate y = 4√x + 3/x
Rewrite: 4x½ + 3x⁻¹
4x½ → 4·½·x-½ = 2/√x
3x⁻¹ →−3x⁻² = −3/x² ✅
ℹ️ How to use: Use the dropdown to choose a function. The navy curve is y (the original function). The red curve is dy/dx (the derivative). Notice the derivative is always one degree lower — x³ becomes 3x², x² becomes 2x, and so on.
⚡ Power Rule Visualiser — pick a function, see curve + derivative
y = x² (navy) | dy/dx = 2x (red) — derivative is always one degree lower
📍
Gradient at a Point — Tangent Lines
Once you have dy/dx, substitute any x-value to get the exact gradient of the curve at that point. The tangent is the straight line that just touches the curve at that point with that gradient.
Step 1: Differentiate to get dy/dx Step 2: Substitute x to get the gradient Step 3: Use y−y₁=m(x−x₁) for the tangent equation
Step-by-step method
Find tangent to y=x³−3x+1 at x=2
Differentiate: dy/dx = 3x²−3
Gradient at x=2: m = 3(4)−3 = 9
Point (x=2): y = 8−6+1 = 3 → P=(2,3)
Tangent eq: y−3 = 9(x−2)
Simplify:y = 9x − 15 ✅
Normal line
The normal is perpendicular to the tangent at the same point. If tangent gradient = m, normal gradient = −1/m.
Tangent gradient m Normal gradient = −1/m
Normal to y=x³−3x+1 at x=2
Tangent m=9: Normal m=−1/9
Point (2,3): y−3=−(1/9)(x−2)
Normal:y = −x/9 + 29/9 ✅
📚 Need a hint for finding tangents? Click each step:
Differentiate the function to get dy/dx
Substitute the given x-value into dy/dx to get the gradient m
Use y − y₁ = m(x − x₁) with the point (x₁, y₁) on the curve
📍 Draw the tangent — pick a curve and drag to any point
ℹ️ How to use: Choose a curve from the dropdown, then drag the x-slider. The tangent line draws at that point showing the gradient m. Watch how a steep curve has a large |m| and a flat section has m near 0.
1.0
Gradient m = 2.0 | Tangent: y = 2x − 1 | Rising steeply
🏔️
Stationary Points — Maxima, Minima, Inflection
Where the gradient is zero, the curve is momentarily flat — a peak, a trough, or a point of inflection. Finding these is how engineers optimise designs.
Find stationary points: set dy/dx = 0 and solve for x Classify: use d²y/dx² (+ve → min, −ve → max)
Full method — step by step
Find and classify stationary points of y=x³−3x+1
Step 1: dy/dx = 3x²−3
Set to zero: 3x²−3=0 → x²=1 → x=±1
y at x=−1: (−1)³−3(−1)+1=−1+3+1=3
y at x=1: 1−3+1=−1
d²y/dx²: 6x
At x=−1: 6(−1)=−6<0 → LOCAL MAXIMUM (−1, 3)
At x=+1: 6(1)=+6>0 → LOCAL MINIMUM (1, −1)
Table of signs method
If d²y/dx²=0, test signs of dy/dx either side of the point:
Sign pattern of dy/dx
Type
+ → 0 → −
Local maximum ↗↘
− → 0 → +
Local minimum ↘↗
+ → 0 → + or − → 0 → −
Inflection point
💡 Engineering context: The maximum material in a beam cross-section, minimum cost of a container, maximum power transfer — all found by setting dy/dx=0.
🏔 Find stationary points — watch dy/dx cross zero
ℹ️ How to use: Choose a curve. The navy curve is y, the red dashed curve is dy/dx. Where the red curve crosses zero (the x-axis) is where a stationary point exists. The slider moves a point along the curve — watch the gradient readout change sign at the peaks and troughs.
x = -3.0
dy/dx = 12 | Curve is rising here
🔁
The Second Derivative — d²y/dx²
Differentiate again to get the rate of change of the gradient. It tells you about concavity (is the curve bending up or down?) and is essential for classifying stationary points.
d²y/dx² = derivative of dy/dx Positive → concave up (holds water 🥣) Negative → concave down (sheds water ⌒)
Calculating second derivatives
y = x⁴ − 6x² + 5
First deriv: dy/dx = 4x³ − 12x
Second deriv: d²y/dx² = 12x² − 12
At x=0: d²y/dx²=−12 < 0 → concave down
At x=2: d²y/dx²=48−12=36 > 0 → concave up
What the second derivative shows
d²y/dx² > 0 → concave up, local MINIMUM
d²y/dx² < 0 → concave down, local MAXIMUM
d²y/dx² = 0 → possible inflection (inconclusive)
In kinematics: d²s/dt² = acceleration
Positive acceleration → speeding up
💡 Think of d²y/dx² as the curvature of the road. Positive = valley shape. Negative = hill shape.
🔄 Concavity visualiser — drag to see d²y/dx² in action
ℹ️ How to use: Drag the slider along the curve. The green arc below the point shows concavity — opens upward (∪) when d²y/dx² > 0, downward (∩) when < 0. Watch it flip sign as you pass through a stationary point.
x = -2.0
d²y/dx² = 12 | Concave UP (∪) — minimum region
🚗
Motion — Velocity and Acceleration
If you know position s as a function of time t, differentiation gives you velocity and acceleration instantly. This is how engineers analyse engine performance, robot motion, and vehicle dynamics.
v = ds/dta = dv/dt = d²s/dt² Positive v = moving forward. Negative v = moving backward.
🚗 Live Motion Simulator — s = t³ − 6t² + 9t
Drag the slider to drive the car
ℹ️ How to use: Drag the time slider to move through the journey. Watch the car move on the road, the graphs update, and the readouts show displacement (s), velocity (v) and acceleration (a) at that exact moment. Press Play to animate automatically.
t = 0.00 s
Displacement s
0.000 m
s = t³ − 6t² + 9t
Velocity v = ds/dt
9.000 m/s
v = 3t² − 12t + 9
Acceleration a = dv/dt
-12.00 m/s²
a = 6t − 12
t = 0 | Moving forward and decelerating
▬ s (displacement)▬ v (velocity)▬ a (acceleration)
Full analysis
s = t³ − 6t² + 9t (metres, seconds)
Velocity: v = 3t² − 12t + 9 m/s
Acceleration: a = 6t − 12 m/s²
v=0 when: 3t²−12t+9=0 → t=1 or t=3
t=1: s=4m (momentarily stopped, turns back)
t=3:s=0m (back at start, turns forward again)
Interpreting the signs
t range
v sign
a sign
Motion
0 to 1
+
−
Forward, slowing
1 to 2
−
−
Backward, speeding
2 to 3
−
+
Backward, slowing
3 to 4
+
+
Forward, speeding
💡 Speed ≠ velocity. Speed = |v|. You slow down when v and a have opposite signs.
⚙️
Engineering Optimisation
Set dy/dx=0 to find maximum or minimum values. This is the most applied use of differentiation in T Level / Level 3 assessments — maximising volume, minimising material, maximising power.
Method: 1. Express quantity as function of one variable 2. Differentiate 3. Set = 0 & solve 4. Verify with d²y/dx²
🥫 Maximum volume cylinder
A cylinder has a fixed total surface area of 600π cm². Find the radius that maximises the volume.
SA = 2πr² + 2πrh = 600π → h = (300−r²)/r
Volume: V = πr²h = πr²·(300−r²)/r = π(300r−r³)
dV/dr: π(300 − 3r²)
Set=0: 300−3r²=0 → r²=100 → r=10
d²V/dr²: −6πr = −60π < 0 ✓ maximum
h: (300−100)/10=20cm r=h/2 (classic result!)
V_max:π×100×20 = 2000π = 6283 cm³ ✅
📦 Minimum surface area box
An open box (no lid) is made from a square sheet. Squares of side x are cut from each corner. Sheet is 30cm × 30cm. Find x for maximum volume.
⚙️ Live optimisation — find the maximum or minimum
ℹ️ How to use: Choose an engineering optimisation scenario. The graph shows the quantity you want to maximise or minimise. Drag the slider to see how it changes. The solver finds the exact optimum using dy/dx = 0 and shows every step.
x = 2.0
Optimal x = 2 | Maximum volume = 128 cm³
∑
More Differentiation Rules
Trig, exponential and log derivatives. These appear in AC circuit analysis, cooling problems, and wave equations.
∫ Standard Derivatives ExplorerTrig, exponential, log & chain rule
ℹ️ How to use: Use the dropdown to choose a function, and type the constant (a) where relevant. The solver shows the standard result and, for compound functions like sin(ax) or eax, explains the chain rule step by step — just like the integral explorer.
Select a function above to see the working.
💡The chain rule pattern: when differentiating f(ax+b), differentiate the outside function as normal, then multiply by a — the derivative of the bracket. This is the opposite of the “divide by a” rule used in integration.
Chain rule
d/dx[f(g(x))] = f'(g(x)) × g'(x)
d/dx[(3x+1)⁵] = 5(3x+1)⁴ × 3 = 15(3x+1)⁴
d/dx[sin(2x)] = cos(2x) × 2 = 2cos(2x)
d/dx[e3x] = e3x × 3 = 3e3x
💡 The chain rule: differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
Engineering applications
AC voltage: v=325sin(314t). Find dv/dt at t=0.
dv/dt: 325×314×cos(314t)
At t=0: 325×314×cos(0)=325×314×1
Answer:102,050 V/s ✅
Cooling: T=20+80e−0.1t. Find dT/dt at t=5.
dT/dt: 80×(−0.1)×e−0.1t = −8e−0.1t
At t=5: −8×e−0.5 = −8×0.6065
Answer:−4.85 °C/s ✅
❓ What does f(x) and g(x) mean?
You will often see rules written using f(x) and g(x). These are just shorthand names for any function of x — a placeholder, like a variable name in programming.
f(x) means...
“Some function of x” — could be any expression involving x.
Examples:
• f(x) = x² + 3x
• f(x) = sin x
• f(x) = e2x
f′(x) means the derivative of f(x).
g(x) means...
A second function, different from f(x), also involving x.
Examples:
• g(x) = x³ − 1
• g(x) = cos x
• g(x) = ln x
g′(x) means the derivative of g(x).
Why use f(x) and g(x)? A real example:
The chain rule is written as:
d/dx[f(g(x))] = f′(g(x)) × g′(x)
In plain English: “differentiate the outside function, leave the inside unchanged, then multiply by the derivative of the inside.”
Concrete example — y = sin(3x²):
• The outside is sin(…) — so f(x) = sin x, and f′(x) = cos x
• The inside is 3x² — so g(x) = 3x², and g′(x) = 6x
• Chain rule: cos(3x²) × 6x = 6x cos(3x²)
💡 Memory trick
Think of f(x) and g(x) like boxes. f is the outer box, g is the inner box. The chain rule says: differentiate the outer box, keep what’s inside it, multiply by the derivative of the inner box.
✅ Product & Quotient Rule preview
Product rule (two things multiplied):
d/dx[f⋅g] = f′g + fg′
📐 Standard derivatives explorer — see the curve and its derivative
ℹ️ How to use: Choose a trig, exponential or log function from the dropdown. The navy curve is the original function; the red dashed curve is its derivative. Drag the slider to move along both curves simultaneously and read off exact values.
x = 0
y = sin(0) = 0 | dy/dx = cos(0) = 1.000
🧮
Differentiation Solver
From basic power rule to chain rule — every step explained in plain English. Start simple, build up to Level 3 complexity.
🟢 Level 1 — Power Rule (start here)
📐 Differentiate any Power or Polynomiald/dx[axⁿ] = naxⁿ⁻¹
ℹ️ How to use: Use the dropdown to choose a function. The solver shows each term being differentiated one at a time, with a plain-English explanation of every step. Start with simple ones and build up.
Select a function above to see every step explained.
💡The Power Rule in words: “Bring the power down in front as a multiplier, then reduce the power by 1.”
— x² → 2x¹ = 2x — x³ → 3x²
🟡 Level 2 — Gradient & Tangent at a Point
📍 Find Gradient & Tangent LineSubstitute x into dy/dx
ℹ️ How to use: Choose a curve from the dropdown. Enter an x-value. The solver finds dy/dx, evaluates the gradient at that point, finds the y-coordinate, then writes the full equation of the tangent line — step by step in plain English.
Select a curve and enter an x-value above.
💡Tangent line formula: y − y₁ = m(x − x₁) — where m = dy/dx at that point, and (x₁, y₁) is the point on the curve.
ℹ️ How to use: Choose a polynomial. The solver sets dy/dx = 0, solves for x, finds the y-coordinates, then uses the second derivative test to classify each point as a maximum, minimum or point of inflection — with full explanation of what each classification means.
Select a curve above to find all stationary points with full working.
💡Second derivative test: If d²y/dx² > 0 at a stationary point → minimum (curve curves upward like a ∪). If d²y/dx² < 0 → maximum (curves downward like a ∩). If = 0, the test is inconclusive.
ℹ️ How to use: The chain rule is used when you have a function inside another function, like (3x+2)⁴ or √(2x−1). Choose a composite function from the dropdown. The solver identifies the inner and outer functions, differentiates each separately, then multiplies them together — showing every reasoning step.
Select a composite function above to see the chain rule applied step by step.
💡When to use the chain rule: whenever you can spot a “function inside a function”. Ask yourself: can I write this as y = [something]ⁿ? If yes, use the chain rule. Let u = the inside part, then y = uⁿ.
🎮
🎮 Diff Dash! — Differentiation Game
Race against yourself to differentiate functions correctly. Choose the right derivative from 4 options. Beat your high score!
⚡ DIFF DASH — Power Rule Speed Round
ℹ️ How to play: A differentiation question appears. Click the correct answer from the options. Score and streak update after each — click Next to continue.
Score: 0Streak: 0🔥High Score: 0Q: 1/10
Differentiate:
y = x³
Game Over!
Final score: 0/10
🏔️ PEAK FINDER — Stationary Points Game
Score: 0Q: 1/8
Given dy/dx = 0 at a point, and the value of d²y/dx² below — what type of stationary point?
d²y/dx² = +8
Done!
Score: 0/8
🔗 MATCH UP — Function to Derivative
Drag or tap to match each function with its derivative
⚙️
Applications
Where this topic is used in engineering, manufacturing, maintenance and daily life. Use the calculator below to try real numbers for each scenario.
🚗Mechanical:v=ds/dt, a=dv/dt. Velocity & acceleration from displacement.
⚡Electrical:i=C·dV/dt — capacitor current from changing voltage.
🔧Optimisation:Minimise material: dA/dx=0. Optimal can dimensions.
🌡️Thermal:dT/dt=−k(T−T_env). Cooling rate of a machine part.
ℹ️ How to use: Pick a real engineering scenario from the dropdown. Type the values for that scenario and the solver differentiates the formula and evaluates it at your chosen point — showing every step in plain English, just like the examples above.
Select a scenario above to see the working.
👆
Try: Car at t=2s ·
Capacitor C=1μF at t=0.5s ·
Cooling at t=5s
✏️
Practice Questions
Show all steps. Use correct notation: dy/dx, d²y/dx², f'(x).
🟢 Tier 1 — Basic Differentiation
ℹ️ How to use: Work through each question and write down your answer. When ready, click Show Answer for the full step-by-step solution.
Find the equation of the tangent to y=x²+3x at x=2.
[4]
dy/dx=2x+3. At x=2: m=7. y=(4+6)=10. Point (2,10). y−10=7(x−2) → y=7x−4
Q4
Find the stationary points of y=2x³−9x²+12x−4 and classify them.
[6]
Set dy/dx=0, solve, then use d²y/dx² to classify
dy/dx=6x²−18x+12=6(x²−3x+2)=6(x−1)(x−2)=0 → x=1 or x=2 At x=1: y=2−9+12−4=1. d²y/dx²=12x−18=−6<0 → LOCAL MAX (1,1) At x=2: y=16−36+24−4=0. d²y/dx²=6>0 → LOCAL MIN (2,0)
🟡 Tier 2 — Applied Problems
Q5
A particle moves so that s = 3t³ − 18t² + 24t. Find (a) velocity at t=2, (b) when it is stationary, (c) acceleration at t=3.
[6]
v=ds/dt=9t²−36t+24 (a) v(2)=36−72+24=−12 m/s (moving backward) (b) 9t²−36t+24=0 → 3t²−12t+8=0 → t=(12±√(144−96))/6=(12±√48)/6 → t≈0.85s and t≈3.15s (c) a=18t−36. a(3)=54−36=18 m/s²
Q6
A rectangular storage area is enclosed on three sides by fencing (total 120m). The fourth side is a wall. Find the dimensions that maximise the enclosed area.
[5]
Let width=x. Then 2x+length=120 → length=120−2x A=x(120−2x)=120x−2x² dA/dx=120−4x=0 → x=30m Length=60m d²A/dx²=−4<0 ✓ maximum 30m × 60m = 1800m² ✅
🔴 Tier 3 — T Level Challenge
Q7
An open cylindrical can must hold 500π cm³. Find the radius that minimises the total surface area (base + curved side).
[6]
Express SA in terms of r only using V=πr²h=500π to eliminate h