First principles, differentiating x^n, gradient functions, rates of change.
Answer at least 3 of 3 correctly to complete this section.
How fast is a car going at a particular instant? How quickly is a population growing right now? At what rate is temperature changing this second?
All of these are rates of change — and differentiation is the tool that answers them. In geometry, the derivative gives you the gradient of a curve at any point. In the real world, it gives you the instantaneous rate of change of anything that varies.
Differentiation is the single most-used technique in A-Level Maths. It appears in curve sketching, optimisation, kinematics, and practically every applied topic you will meet. Master the power rule here, and the rest of calculus builds on it.
Consider the curve y=x2. To find the gradient at a point, you could draw a secant line through two nearby points and calculate its slope.
If the two points are (x,x2) and (x+h,(x+h)2), the gradient of the secant is:
h(x+h)2−x2=hx2+2xh+h2−x2=h2xh+h2=2x+h
As h→0, the secant becomes a tangent, and the gradient becomes exactly 2x.
This is differentiation from first principles — the limit definition:
f′(x)=limh→0hf(x+h)−f(x)
You need to know this definition exists, but for polynomials you will always use the power rule shortcut instead.
When you differentiate y=x2, you get dxdy=2x. This is a new function that tells you the gradient at every point on the curve:
Notation: dxdy, f′(x), and y˙ all mean “the derivative”. At A-Level you will mainly use the first two. dxdy is not a fraction — it is a single symbol meaning “the rate of change of y with respect to x”.
Drag the point along the curve below. Watch how the gradient changes. Can you predict where the gradient is zero? Where is it steepest?
The gradient is approximately zero here — this could be a turning point!
Answer at least 3 of 4 correctly to complete this section.
For any real number n:
dxd(xn)=nxn−1
In words: bring down the power as a multiplier, then reduce the power by one.
With a coefficient:
dxd(axn)=anxn−1
The constant a just multiplies through.
| y | dxdy | Note |
|---|---|---|
| x5 | 5x4 | Bring down 5, reduce power to 4 |
| 3x4 | 12x3 | 3×4=12, power drops to 3 |
| 7x | 7 | Think of 7x1: power becomes x0=1 |
| 5 | 0 | Constants differentiate to zero |
| −2x3 | −6x2 | The negative carries through |
Differentiate f(x)=3x4−2x3+7x−5
Differentiate each term separately:
f′(x)=12x3−6x2+7−0=12x3−6x2+7
Watch out: The derivative of 7x is 7, not 7x. And the derivative of −5 is 0, not −5.
Answer at least 3 of 4 correctly to complete this section.
The power rule only works when the expression is written as a sum of terms of the form axn. Before you can differentiate, you often need to rewrite the expression first.
x=x1/2,3x=x1/3,xx=x3/2
x1=x−1,x21=x−2,x43=3x−4
x2(3x−1)=3x3−x2
Watch out: You can only use the power rule on sums of axn terms. If you see a product like x2(3x−1), you must expand it first. (The product rule comes later in the course.)
Differentiate y=x24+3x
Step 1: Rewrite as powers of x:
y=4x−2+3x1/2
Step 2: Differentiate each term:
dxdy=4×(−2)x−3+3×21x−1/2=−8x−3+23x−1/2
Step 3: Rewrite in the original form (if the question asks):
dxdy=−x38+2x3
Differentiate y=x2(3x−1)
Expand: y=3x3−x2
dxdy=9x2−2x
| Original form | Rewrite as | Derivative |
|---|---|---|
| x | x1/2 | 21x−1/2 |
| x1 | x−1 | −x−2 |
| x35 | 5x−3 | −15x−4 |
| xx | x3/2 | 23x1/2 |
| x2 | 2x−1/2 | −x−3/2 |
Answer at least 3 of 4 correctly to complete this section.
Once you have the derivative, you can find the gradient at any specific point by substituting the x-value.
Find the gradient of y=2x3−x+4 at the point where x=2
Step 1: Differentiate:
dxdy=6x2−1
Step 2: Substitute x=2:
dxdy=6(2)2−1=6(4)−1=24−1=23
The gradient at x=2 is 23.
Find the x-coordinates where the gradient of y=x3−6x2+9x+1 is zero.
Step 1: Differentiate:
dxdy=3x2−12x+9
Step 2: Set equal to zero and solve:
3x2−12x+9=0
x2−4x+3=0
(x−1)(x−3)=0
x=1 or x=3
These are the stationary points of the curve — where it changes direction.
Exam tip: “Find the stationary points” means “set dxdy=0 and solve”. This is one of the most common exam questions on differentiation.
Answer at least 3 of 4 correctly to complete this section.
The curve C has equation y=2x3−5x2+4
(a) Find dxdy
(b) Find the gradient of C at the point (2,0)
(c) Find the coordinates of the stationary points of C
(a) dxdy=6x2−10x
(b) At x=2: dxdy=6(4)−10(2)=24−20=4
(c) Set dxdy=0:
6x2−10x=0 2x(3x−5)=0 x=0 or x=35
At x=0: y=0−0+4=4, so (0,4)
At x=35: y=2(27125)−5(925)+4=27250−9125+4=27250−375+108=−2717
Stationary points: (0,4) and (35,−2717)
Differentiate y=xx3+2x, and find the gradient at x=3
Simplify first:
y=xx3+x2x=x2+2
dxdy=2x
At x=3: gradient =6
Answer at least 3 of 4 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
Simplifying surds, laws of indices, rationalising denominators.
Completing the square, discriminant, solving quadratic equations and inequalities.
Finding equations of tangents and normals to curves at given points.
Finding and classifying stationary points. Second derivative test. Optimisation problems.
Differentiating composite functions using the chain rule.
Differentiating products of two functions using the product rule.
Integration as the reverse of differentiation, integrating x^n, finding constants of integration.
Using change of sign to locate roots of equations, interval bisection.
Using the Newton-Raphson formula to find approximate roots, graphical interpretation.
Using differentiation and integration for variable acceleration, displacement-velocity-acceleration relationships.