Step-by-step worked examples and graded practice questions on differentiation — from first principles through the power rule to tangents, normals and stationary points. Written to the Edexcel Pure Year 1 specification and equally suitable for AQA and OCR A.
📚 Pure Year 1 (AS)✅ 15 Practice Questions🔍 7 Worked Examples⚠️ Common Mistakes
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Differentiation is the process of finding the gradient function of a curve — a formula that gives you the gradient at any point on the curve, rather than the single gradient a straight line has. At GCSE you found the gradient of a straight line using rise over run; a curve's gradient changes continuously as x changes, so differentiation produces a new function that calculates it for any x-value.
If y = f(x), the derivative (gradient function) is written:
dy/dx — "dee y by dee x", when the curve is written as y = ...
f′(x) — "f dash x", when the curve is written as a function f(x)
d²y/dx² or f″(x) — the second derivative, found by differentiating dy/dx again
Differentiation is one of the two core skills of A-Level calculus, alongside integration — its reverse process.
Differentiation from first principles
Every differentiation rule ultimately comes from one definition. Take two points on a curve, P and a nearby point Q, a small horizontal distance h further along. As Q slides towards P (as h → 0), the gradient of the chord PQ gets closer and closer to the true gradient of the curve at P — the gradient of the tangent.
This gives the formal definition of the derivative:
Definitionf′(x) = lim (h → 0) f(x + h) − f(x)h
As Q slides along y = x² towards P(1, 1), the gradient of chord PQ approaches the gradient of the tangent at P. With h = 1, chord PQ has gradient 3; the true tangent gradient at P is 2.
Divide by h: [f(x + h) − f(x)] / h = 3x² + 3xh + h²
4
Let h → 0: 3x² + 3xh + h² → 3x²
Answerf′(x) = 3x²
Notice the pattern: x² → 2x, and x³ → 3x². In both cases the power comes down as a multiplier and the power itself drops by 1. This pattern is the power rule.
The power rule
For any real number n:
Power ruley = xⁿ ⟹ dy/dx = nxⁿ⁻¹
Two further rules let you differentiate almost any polynomial term by term:
Constant multiple rule: d/dx[a·f(x)] = a·f′(x) — a constant in front of a term stays put
Sum/difference rule: d/dx[f(x) ± g(x)] = f′(x) ± g′(x) — differentiate each term separately
Constants differentiate to zero: a constant term (no x) has zero gradient, so it disappears entirely
Worked Example 3
Differentiate y = 3x⁴ − 2x² + 5x − 7.
1
Differentiate each term using nxⁿ⁻¹: 3x⁴ → 3 × 4x³ = 12x³
2
−2x² → −2 × 2x¹ = −4x
3
5x → 5x⁰ × 5 = 5 (since x¹ → 1x⁰ = 1)
4
−7 → 0 (a constant term always differentiates to zero)
Answerdy/dx = 12x³ − 4x + 5
Negative and fractional powers
The power rule works for any real power — including negative and fractional ones — but you can only apply it once an expression is written as a power of x. Roots and fractions must be rewritten first.
Expression
Index form
√x
x^(1/2)
∛x
x^(1/3)
1/x
x⁻¹
1/x²
x⁻²
1/√x
x^(−1/2)
Worked Example 4
Differentiate y = 4√x + 3/x², simplifying your answer.
1
Rewrite in index form: y = 4x^(1/2) + 3x⁻²
2
Differentiate term by term: dy/dx = 4 × (1/2)x^(−1/2) + 3 × (−2)x⁻³
3
Simplify the coefficients: dy/dx = 2x^(−1/2) − 6x⁻³
4
Rewrite back in surd/fraction form: dy/dx = 2/√x − 6/x³
Answerdy/dx = 2/√x − 6/x³
Tangents and normals
The derivative gives you the gradient of the curve at a single point — which means it also gives you the gradient of the tangent (the line that just touches the curve there). The normal is the line perpendicular to the tangent at that same point, so its gradient is the negative reciprocal of the tangent's gradient.
Both lines are straight lines, so once you have a point and a gradient you find the equation using y − y₁ = m(x − x₁) from GCSE:
Tangent gradient at x = a: m = f′(a)
Normal gradient at x = a: m = −1 / f′(a)
The tangent to y = x² at the point P(2, 4) has gradient 4, found by differentiating dy/dx = 2x and substituting x = 2.
Worked Example 5
Find the equations of the tangent and the normal to the curve y = x² at the point where x = 2.
1
Find the y-coordinate: y = 2² = 4, so the point is (2, 4)
2
Differentiate: dy/dx = 2x
3
Gradient of the tangent at x = 2: m = 2(2) = 4
4
Tangent equation: y − 4 = 4(x − 2) ⟹ y = 4x − 4
5
Gradient of the normal: m = −1/4
6
Normal equation: y − 4 = −¼(x − 2) ⟹ y = −¼x + 4.5
AnswerTangent: y = 4x − 4 Normal: y = −¼x + 4.5
Increasing and decreasing functions
A function is increasing on an interval if f′(x) > 0 for every x in that interval, and decreasing if f′(x) < 0 throughout. To prove a function is increasing or decreasing over a range, differentiate it and show the sign of f′(x) doesn't change within that range.
Worked Example 6
Show that f(x) = x³ − 3x² + 8 is decreasing for 0 < x < 2.
1
Differentiate: f′(x) = 3x² − 6x
2
Factorise: f′(x) = 3x(x − 2)
3
For 0 < x < 2: x is positive, and (x − 2) is negative
4
So 3x(x − 2) is positive × negative = negative throughout the interval
Answerf′(x) < 0 for 0 < x < 2, so f(x) is decreasing there
Stationary points and the second derivative test
A stationary point is a point on a curve where the gradient is momentarily zero — dy/dx = 0. To find the nature of a stationary point, differentiate again to get the second derivative, d²y/dx², and substitute in the x-value:
Second derivative
Nature of stationary point
d²y/dx² > 0
Local minimum
d²y/dx² < 0
Local maximum
d²y/dx² = 0
Inconclusive — check the gradient either side instead
Worked Example 7
Find and classify the stationary points of y = x³ − 3x² − 9x + 5.
So x = 3 or x = −1. Find y: at x = −1, y = −1 − 3 + 9 + 5 = 10; at x = 3, y = 27 − 27 − 27 + 5 = −22
4
Second derivative: d²y/dx² = 6x − 6
5
At x = −1: d²y/dx² = −12 < 0 ⟹ local maximum at (−1, 10)
6
At x = 3: d²y/dx² = 12 > 0 ⟹ local minimum at (3, −22)
AnswerLocal max (−1, 10); local min (3, −22)
Exam tips
💡 Tip 1
Always rewrite before you differentiate
You cannot apply the power rule to √x or 1/x² directly — rewrite them as x^(1/2) and x⁻² first. This is the single most common reason marks are dropped on otherwise correct working.
💡 Tip 2
State "dy/dx = 0" explicitly for stationary points
Examiners award a method mark just for writing the line "at a stationary point, dy/dx = 0" before you solve — don't skip straight to the answer.
💡 Tip 3
Show the limit step in first-principles questions
"Prove from first principles" questions carry marks for showing [f(x+h) − f(x)]/h explicitly and then stating "as h → 0" — a correct final answer with no working earns very few marks.
💡 Tip 4
Substitute into the second derivative, not the first
To classify a stationary point you substitute its x-value into d²y/dx², never back into dy/dx (which is always zero there by definition) or into y itself.
💡 Tip 5
Leave surd and fraction answers unless told otherwise
Edexcel mark schemes usually accept 2/√x or 2x^(−1/2) equally, but check whether the question asks for a specific form ("give your answer as a single fraction", etc.) before you simplify further.
Common mistakes
Common Mistake 1
Differentiating 1/x² without rewriting first
Students often try to differentiate 1/x² directly and get confused. Rewrite it as x⁻² first: d/dx(x⁻²) = −2x⁻³ = −2/x³. Always convert roots and fractions to index form before applying the power rule.
Common Mistake 2
Losing the negative sign with negative powers
For y = x⁻², students often write dy/dx = 2x⁻³ instead of −2x⁻³. Multiply by the original power (−2), including its sign, and then reduce the power by 1.
Common Mistake 3
Using the tangent gradient for the normal
The normal's gradient is the negative reciprocal of the tangent's gradient, not the same value. If the tangent gradient is 4, the normal gradient is −1/4, not 4 and not −4.
Common Mistake 4
Forgetting a constant term differentiates to zero
For y = x² + 7, students sometimes carry the +7 through into dy/dx. A constant has zero gradient everywhere, so it simply disappears: dy/dx = 2x, not 2x + 7.
Common Mistake 5
Assuming every stationary point is a maximum
Finding dy/dx = 0 only locates a stationary point — it doesn't tell you its type. Always check the sign of d²y/dx² (or the gradient either side) before calling a point a maximum or minimum.
Practice questions
Work through each question before checking the worked solution — differentiation is a skill you build by doing it, not by reading about it.
Core Skills
Q1Differentiate y = x⁵.Core Skills
Q2Differentiate y = 4x³ − 2x + 9.Core Skills
Q3Find dy/dx for y = 7x² − 3x + 1.Core Skills
Q4Differentiate f(x) = 6x − x² + 10.Core Skills
Q5Find f′(x) for f(x) = 2x⁴ + x³ − 5.Core Skills
Exam-Style
Q6Differentiate y = 5√x − 2/x, simplifying your answer.Exam-Style
Q7Find the gradient of the curve y = 2x³ − 5x + 1 at the point where x = −1.Exam-Style
Q8Find the equation of the tangent to y = x² − 4x + 1 at the point where x = 3.Exam-Style
Q9Find the equation of the normal to the curve y = x³ at the point (1, 1).Exam-Style
Q10Show that f(x) = x² − 6x + 10 is decreasing for x < 3.Exam-Style
A* Challenge
Q11Find and classify the stationary points of y = x³ − 3x² − 9x + 2.A* Challenge
Q12Differentiate y = 5x^(3/2) − 4x^(−1/2) + 2, simplifying your answer.A* Challenge
Q13Find the equation of the normal to y = x³ − 2x² + 1 at the point where x = 2, in the form ax + by + c = 0.A* Challenge
Q14Find the range of values of x for which y = x³ − 6x² + 9x + 1 is increasing.A* Challenge
Q15Prove from first principles that the derivative of x³ is 3x².A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — dy/dx = 5x⁴
1
Apply the power rule to x⁵: multiply by the power (5) and reduce the power by 1.
Increasing where dy/dx > 0: (x − 1)(x − 3) > 0, which holds when x < 1 or x > 3
Increasing for x < 1 or x > 3
Q15 — f′(x) = 3x² (full proof)
1
f(x + h) = (x + h)³ = x³ + 3x²h + 3xh² + h³
2
f(x + h) − f(x) = 3x²h + 3xh² + h³
3
[f(x + h) − f(x)] / h = 3x² + 3xh + h²
4
As h → 0: 3x² + 3xh + h² → 3x²
f′(x) = 3x² ∎
Summary
The derivative is defined as f′(x) = lim (h → 0) [f(x + h) − f(x)] / h — the limiting gradient of a chord as it shrinks to a point.
Power rule: xⁿ → nxⁿ⁻¹, applied term by term; constants always differentiate to zero.
Rewrite roots and fractions as powers of x before differentiating.
Tangent gradient at x = a is f′(a); normal gradient is −1/f′(a).
f′(x) > 0 means increasing; f′(x) < 0 means decreasing.
Stationary points occur where f′(x) = 0; classify them with f″(x) — positive means minimum, negative means maximum.
Next in the Pure Maths calculus strand: Further Differentiation (Year 2), which extends these rules to trigonometric, exponential and logarithmic functions, the chain, product and quotient rules, and implicit differentiation.
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