A derivative is just a speedometer
Your car's odometer tells you where you are. Your speedometer tells you how fast that is changing. Calculus is the machinery for going from the first to the second — and back again.
On a graph, "how fast it is changing" means the steepness of the curve at one exact point. That is what a derivative measures. Steep upward curve → big positive derivative. Flat → zero. Sloping down → negative.
The problem with "slope at a point"
Slope needs two points: rise over run. But at a single point there is no run. So we cheat: take a second point a distance h away, compute the slope of the line joining them (the secant), then slide h toward zero and see what number the slope settles on.
That settling number is the limit, and it is the derivative.
Pull the h slider down. Watch the dashed cyan secant swing until it lies exactly on the amber tangent — and watch the "gap" readout collapse to zero.
The rules you will actually use
Nobody computes limits by hand after the first week. You memorise these six and you are done for most of school calculus.
| Rule | If f(x) = | then f′(x) = |
|---|---|---|
| Power | xⁿ | n·xⁿ⁻¹ |
| Constant | c | 0 |
| Sum | u + v | u′ + v′ |
| Product | u · v | u′v + uv′ |
| Quotient | u / v | (u′v − uv′) / v² |
| Chain | f(g(x)) | f′(g(x)) · g′(x) |
The mistake everyone makes
The chain rule. Differentiating sin(3x) gives 3·cos(3x), not cos(3x). The inner function 3x has its own derivative of 3, and it must come along. If a function is wrapped inside another function, you always pay the inside derivative too.
Integration: slicing an area into strips
Now run it backwards. Instead of "how fast is this changing", ask "how much has accumulated". Distance from speed. Area under a curve.
The trick is the same trick: approximate with something crude, then take a limit. Chop the area into rectangles, add them up, and let the rectangles get infinitely thin.
Push n from 6 to 80. The error readout falls toward zero. That limit is the integral.
The Fundamental Theorem
Differentiation and integration are inverse operations. This is the single most useful fact in calculus, because it means you never actually have to add up infinite rectangles — you just find an antiderivative and subtract.
Worked example
∫₀² x² dx. The antiderivative of x² is x³/3. So the answer is (2³/3) − (0³/3) = 8/3 ≈ 2.667. Set the visualization above to x² with n = 80 and compare the readout.
Differentiation from first principles
The rules are for speed; the limit definition is what they are built on, and the paper asks for a first-principles derivation at least once. The routine is fixed, so it is worth practising until it takes two minutes.
Write f(x + h), subtract f(x), divide by h, simplify until the h in the denominator cancels, then let h → 0.
Differentiate f(x) = x² + 3x from first principles.
f(x + h) = (x + h)² + 3(x + h) = x² + 2xh + h² + 3x + 3h.Expand fully. Writing (x+h)² as x² + h² is the error that ruins the whole derivation.f(x + h) − f(x) = 2xh + h² + 3h.Every term without an h cancels — which is the check that the expansion was right.- Divide by h:
(2xh + h² + 3h)/h = 2x + h + 3.Legitimate because h is approaching zero, not equal to it. - Let
h → 0:f′(x) = 2x + 3.Confirm with the power rule: d/dx(x²) = 2x and d/dx(3x) = 3 ✓
f′(x) = 2x + 3
What derivatives are actually used for
Three applications account for nearly all the marks in the applied half of the chapter, and all three come from the same fact: the derivative is a gradient.
- Tangents and normals. The tangent at
x = ahas gradientf′(a); the normal is perpendicular to it, so its gradient is−1/f′(a). Use point-slope form with the point(a, f(a)). - Maxima and minima. A stationary point has
f′(x) = 0. Classify it with the second derivative:f″ > 0is a minimum,f″ < 0a maximum, andf″ = 0is inconclusive, so test the sign of f′ either side. - Rates of change.
dy/dtis how fast y changes with time. Related rates chain them together:dV/dt = (dV/dr)(dr/dt).
A closed cylindrical can is to hold 500 cm³. Find the radius that minimises its total surface area.
- Surface area
A = 2πr² + 2πrhand volumeV = πr²h = 500.Two variables, one constraint — the standard shape of an optimisation problem. - From the constraint,
h = 500/(πr²). Substitute:A = 2πr² + 1000/r.Reduce to one variable before differentiating. This substitution step is where the marks are. dA/dr = 4πr − 1000/r². Set it to zero:4πr³ = 1000.Multiply through by r² to clear the fraction.r³ = 250/π, sor = 4.30cm.d²A/dr² = 4π + 2000/r³ > 0, so it is a minimum.The second-derivative test is a mark of its own, and "it is obviously a minimum" earns nothing.
r ≈ 4.30 cm, confirmed as a minimum by the second derivative
The three rules that handle everything else
Differentiating xⁿ covers polynomials and nothing more. Almost every function worth differentiating is built by combining simpler ones, so three rules are needed: one for a function inside another, one for a product, and one for a quotient.
Recognising which structure you are looking at matters more than the rules themselves. Ask what the last operation performed is: if something is being raised to a power or fed into a function, it is a chain; if two functions are multiplied, it is a product; if one is divided by another, it is a quotient.
| Function | Derivative | Note |
|---|---|---|
| eˣ | eˣ | unchanged — the defining property of e |
| e^(f(x)) | f′(x)·e^(f(x)) | chain rule; the inner derivative comes out front |
| ln x | 1/x | only for x > 0 |
| ln(f(x)) | f′(x)/f(x) | derivative over the original |
| sin x | cos x | x must be in radians |
| cos x | −sin x | the minus sign is examined constantly |
| tan x | sec²x | |
| tan⁻¹x | 1/(1 + x²) | in the formula book, but worth knowing |
- u, v
- the two component functionsname them explicitly before starting
- v²
- the denominator squarednot differentiated — just squared
- du/dx
- derivative of the inner functionthe chain rule is hiding in every one of these
These derivatives assume radians
The result d(sin x)/dx = cos x is only true when x is measured in radians. In degrees the derivative picks up a factor of π/180. This is the deeper reason radians are used throughout A Level analysis — the calculus is simply wrong otherwise.
When y is not given explicitly
Sometimes a curve is defined by an equation that cannot be rearranged into y = f(x), such as x² + y² = xy + 7. Differentiating implicitly handles this: differentiate every term with respect to x, and each time a y is differentiated, the chain rule contributes a dy/dx.
Other curves are given through a third variable: x = t − e^(2t), y = t + e^(2t). These are parametric, and the gradient is found by differentiating each with respect to the parameter and dividing.
- dy/dx
- the gradient being soughtcollect all such terms on one side, then factorise
- t
- the parameterit cancels in the division, leaving a gradient in t
Find dy/dx for the curve x² + y² = xy + 7.
- Differentiate term by term: 2x + 2y(dy/dx) = x(dy/dx) + y + 0.The y² gives 2y·dy/dx by the chain rule, and xy needs the product rule.
- Gather the dy/dx terms: 2y(dy/dx) − x(dy/dx) = y − 2x.All the unknowns to one side, everything else to the other — as with any linear equation.
- Factorise: (2y − x)(dy/dx) = y − 2x.dy/dx is a common factor, which is what makes it solvable.
- dy/dx = (y − 2x)/(2y − x).The answer involves both x and y, which is normal for an implicit curve and not a sign of error.
dy/dx = (y − 2x)/(2y − x)