Pearson Edexcel International Advanced Level Mathematics: Differentiating and integrating powers of x: gradients, tangents, normals and indefinite integrals (YMA01)

This guide is aligned to the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (Issue 3, April 2019). It covers Unit P1: Pure Mathematics 1, section 4 (Differentiation, 4.1 to 4.3) and section 5 (Integration, 5.1 and 5.2). As the specification states, P1 is a compulsory unit for IAS Mathematics and Pure Mathematics and for IAL Mathematics and Pure Mathematics. There are no tiers or optional strands inside this content.

Use it with the revision notes and the practice questions. The wider course sits on the Edexcel A Level Mathematics hub, the printable checklist lists every topic, and the free 10-minute diagnostics show where to start. The older Pure Mathematics 1 overview gives the unit as a whole.

What this unit covers

Spec	What you must be able to do
4.1	Know the derivative as the gradient of the tangent, the limit definition, the rate-of-change meaning, and second derivatives; use f′(x), f″(x) and dy/dx
4.2	Differentiate xⁿ and sums, differences and constant multiples, including products you must expand and fractions you must split
4.3	Work out gradients at given points, and write down tangent and normal equations
5.1	Know that integration reverses differentiation and that a constant of integration is needed
5.2	Integrate xⁿ (n ≠ −1) and related sums, differences and multiples; given f′(x) and a point on the curve, find y = f(x)

The specification states that knowledge of the chain rule is not required for 4.1. Integrating 1/x, which is the case n = −1, is excluded here.

1. The derivative as a gradient (4.1)

A curve has a different gradient at every point. The gradient at a point means the gradient of the tangent there. Take a point A at x and a second point B at x + h on y = f(x). The chord AB has gradient

[f(x + h) − f(x)] / h

Let h shrink and B closes in on A, so the chord approaches the tangent. The derivative is the value the chord gradient settles at:

f′(x) = lim (h → 0) [f(x + h) − f(x)] / h

Worked example 1. Find the derivative of f(x) = x³ − 2x from the definition, then the gradient at x = 2.

f(x + h) = (x + h)³ − 2(x + h) = x³ + 3x²h + 3xh² + h³ − 2x − 2h
f(x + h) − f(x) = 3x²h + 3xh² + h³ − 2h
divide by h:       3x² + 3xh + h² − 2
let h → 0:         f′(x) = 3x² − 2
f′(2) = 3(4) − 2 = 10

The gradient at x = 2 is 10. The steps matter: expand, subtract, divide by h, and only then let h tend to 0. Setting h = 0 earlier gives 0/0.

Rate of change. The derivative dy/dx measures how quickly y changes as x changes. If the depth h cm of water in a tank is h = 4t + t²/2 after t minutes, then dh/dt = 4 + t. At t = 6 the depth is increasing at 10 cm per minute. Always give the units of the rate: units of y per unit of x.

Second derivatives. Differentiating f′(x) again gives f″(x), also written d²y/dx². It tells you how quickly the gradient itself is changing.

Worked example 2. For y = x⁴ − 6x² + 8/x, find dy/dx and d²y/dx², and the value of d²y/dx² at x = 2.

y = x⁴ − 6x² + 8x⁻¹
dy/dx = 4x³ − 12x − 8x⁻²
d²y/dx² = 12x² − 12 + 16x⁻³
at x = 2: 12(4) − 12 + 16/8 = 48 − 12 + 2 = 38

2. Differentiating xⁿ (4.2)

The rule is: if f(x) = xⁿ then f′(x) = nxⁿ⁻¹. Multiply by the power, then reduce the power by 1. It works for any rational n, including negatives and fractions. Constants differentiate to 0, and a term kxⁿ becomes knxⁿ⁻¹. Sums and differences are differentiated term by term.

The skill is rewriting first. Put every term into the form axⁿ before you differentiate:

Written as	Rewrite as
√x	x^(1/2)
1/x³	x⁻³
1/√x	x^(−1/2)
5/(2x²)	(5/2)x⁻²
³√(x²)	x^(2/3)

There is no product rule or quotient rule in this unit. You get round products by expanding brackets and round fractions by dividing each term by the denominator.

Worked example 3 (a product). Differentiate y = (3x + 2)(2x − 5).

y = 6x² − 15x + 4x − 10 = 6x² − 11x − 10
dy/dx = 12x − 11

Worked example 4 (a fraction). Find g′(x) for g(x) = (x³ + 4x − 8)/(2√x), and evaluate g′(4).

g(x) = (1/2)x³·x^(−1/2) + 2x·x^(−1/2) − 4x^(−1/2)
     = (1/2)x^(5/2) + 2x^(1/2) − 4x^(−1/2)
g′(x) = (5/4)x^(3/2) + x^(−1/2) + 2x^(−3/2)
g′(4) = (5/4)(8) + 1/2 + 2/8 = 10 + 0.5 + 0.25 = 43/4

Note that each term of the numerator is divided by the denominator, and the powers are added as indices: x³ ÷ x^(1/2) = x^(5/2).

3. Gradients, tangents and normals (4.3)

To find the gradient at a point, differentiate and substitute the x-coordinate. The tangent at a point has that gradient m and passes through the point. The normal is perpendicular to the tangent, so its gradient is −1/m, provided m ≠ 0.

Use y − y₁ = m(x − x₁) for both lines. If a question asks for a particular form such as ax + by + c = 0 with integer coefficients, finish by rearranging.

Worked example 5. The curve C has equation y = 3x² − 4√x for x > 0. Find the equations of the tangent and the normal at the point where x = 4.

y = 3x² − 4x^(1/2)
dy/dx = 6x − 2x^(−1/2)
at x = 4: y = 3(16) − 4(2) = 40; dy/dx = 24 − 1 = 23
Tangent: y − 40 = 23(x − 4)  →  y = 23x − 52
Normal gradient = −1/23
Normal: y − 40 = −(1/23)(x − 4)
        23y − 920 = −x + 4
        x + 23y − 924 = 0

A common variation gives you the gradient and asks for the point. Set dy/dx equal to the required value, solve for x, then find y from the original equation. Lines are parallel when their gradients are equal.

4. Indefinite integration (5.1 and 5.2)

Integration reverses differentiation. If F′(x) = f(x), then ∫f(x) dx = F(x) + c. The constant c is the constant of integration: many curves have the same gradient function, and they differ by a vertical shift. An indefinite integral without + c is incomplete, and you should write it every time.

The rule for powers is, for n ≠ −1:

∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c

Add 1 to the power, then divide by the new power. Check by differentiating: the result must return the integrand. Constant multiples come outside and sums are integrated term by term. As with differentiation, rewrite roots and reciprocals as powers first, and expand brackets before integrating.

Worked example 6. Find ∫(x²/2 − 3/√x) dx.

x²/2 − 3x^(−1/2)
∫ = (1/2)(x³/3) − 3 · x^(1/2)/(1/2) + c
  = x³/6 − 6√x + c

Dividing by 1/2 means multiplying by 2, which is where the 6 comes from.

Worked example 7 (expanding first). Find ∫(3x + 1)(x − 2)/√x dx.

(3x + 1)(x − 2) = 3x² − 5x − 2
divide by x^(1/2): 3x^(3/2) − 5x^(1/2) − 2x^(−1/2)
∫ = 3 · x^(5/2)/(5/2) − 5 · x^(3/2)/(3/2) − 2 · x^(1/2)/(1/2) + c
  = (6/5)x^(5/2) − (10/3)x^(3/2) − 4x^(1/2) + c

5. Finding a curve from its gradient (5.2)

If you know f′(x) and one point on the curve, integrate, then use the point to find c.

Worked example 8. A curve has f′(x) = 6√x − 4/x² and passes through (4, 20). Find f(x).

f′(x) = 6x^(1/2) − 4x⁻²
f(x) = 6 · x^(3/2)/(3/2) − 4 · x⁻¹/(−1) + c
     = 4x^(3/2) + 4/x + c
at (4, 20): 4(8) + 4/4 + c = 20  →  33 + c = 20  →  c = −13
f(x) = 4x^(3/2) + 4/x − 13

Check by differentiating: 6x^(1/2) − 4x⁻², which matches. A single point fixes one constant. A question that asks for f(x) from f″(x) needs two integrations and two conditions.

Common errors

• Differentiating a product by multiplying the derivatives of the factors. Expand first.
• Leaving 1/x² or √x unconverted, then applying the rule to the wrong power.
• Writing the new power as n − 1 when integrating, or forgetting to divide by the new power.
• Dividing by n + 1 incorrectly when it is a fraction. Dividing by 3/2 means multiplying by 2/3.
• Forgetting + c, or finding c and then dropping it from the final equation.
• Using the tangent gradient as the normal gradient, or taking the reciprocal and forgetting the sign change.
• Substituting into the original function instead of dy/dx when the gradient is wanted (or the reverse for the y-coordinate).
• Letting h = 0 too soon in a first-principles argument.

Where to go next

Practise with the practice questions, then condense with the revision notes. For wider P1 recall use the Pure 1 revision notes and the Pure 1 practice set.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification Issue 3, April 2019, published by Pearson Education Limited. This page covers Unit P1 sections 4.1 to 4.3 and 5.1 to 5.2.