Differentiation

— Pure Mathematics 1 · AS & A Level (9709) · Chapter 7 —

★ Big picture: Calculus = maths of change. Differentiation is the 1st tool (integration is the 2nd). This chapter gives the rules + how to use them for gradients, tangents, normals, 2nd derivatives. Ch 8 applies them to practical problems.

Skills needed first: ① indices: write as axⁿ e.g. 3x√x = 3x^3/2, 12x = ½x⁻¹
② k(ax+b)ⁿ = k(ax+b)⁻ⁿ ③ ⊥ gradient = −1/m ④ line: y − y₁ = m(x − x₁)

1. Gradient of a Curve — Chords → Tangent

Old way (): draw a tangent by eye → only approximate, slow. New way: differentiation gives the exact gradient with no graph.

Tip: δ (delta) = a very small change in a quantity. δx = tiny step in x, δy = matching step in y.

Idea: take P(x, y) on the curve and a nearby point A. Slide A closer and closer to P. The chord PA gradient settles down to a limiting value = gradient of the curve at P.

A → P : chord PA → tangent at P

Explore 7.1numerical evidence · y = x²

P	A	B	C	chord gradients →
(2, 4)	(2.2, 4.84) → 4.2	(2.1, 4.41) → 4.1	(2.01, 4.0401) → 4.01	→ 4
(3, 9)	(3.2, 10.24) → 6.2	(3.1, 9.61) → 6.1	(3.01, 9.0601) → 6.01	→ 6

gradient at x = 2 is 4, at x = 3 is 6 ⇒ pattern: gradient at (a, a²) = 2a, so at (x, x²) it is 2x.

Proof for y = x² ("from first principles")

P = (x, x²), A = (x + δx, (x + δx)²)

gradient PA = y₂ − y₁x₂ − x₁ = (x+δx)² − x²(x+δx) − x = x² + 2xδx + (δx)² − x²δx = 2xδx + (δx)²δx = 2x + δx δx → 0 : gradient → 2x

Syllabus note: you do not need to differentiate from first principles, but you must understand that the gradient of a curve at a point is the limit of the gradients of a suitable sequence of chords.

Explore 7.2y = x³

Do the same with chords on y = x³ ⇒ pattern: gradient at (x, x³) is 3x² (confirm by first principles: (x+δx)³ − x³ = 3x²δx + 3x(δx)² + (δx)³ ÷ δx → 3x²).

2. Notation — three ways to say the same thing

Style	1st derivative	2nd derivative	Named after
y = x²	dydx = 2x	d²ydx²	Leibniz
f(x) = x²	f′(x) = 2x	f″(x)	Lagrange
operator	ddx(x²) = 2x	—	(Newton wrote ẏ)

dydx = the derivative of y w.r.t. x = the gradient function of y = f(x).
ddx(x²) = 2x reads: "differentiate x² with respect to x, result is 2x".

3. Differentiating Power Functions

Investigating y = x⁴, x⁵, x⁶ gives 4x³, 5x⁴, 6x⁵ ⇒ general rule:

★ Power ruleddx(xⁿ) = n xⁿ⁻¹ (true for any real n, not just positive integers)

Memory: "Multiply by the power, then take 1 off the power."

y = x² ⇒ dy/dx = 2 × x²⁻¹ = 2x.

x⁷7x⁷⁻¹ = 7x⁶

1/x²= x⁻² (rewrite first!)−2x⁻²⁻¹ = −2x⁻³= −2/x³

f(x) = √x= x^1/2f′(x) = ½ x^{½ − 1} = ½x^−½= 12√x

y = 2 (a constant)y = 2x⁰dy/dx = 0·x⁻¹= 0

Remember: derivative of any constant = 0 (a horizontal line has gradient 0).

Two more rules you must use

Scalar multiple rule (k = constant)ddx[k f(x)] = k ddx[f(x)]

Addition / subtraction ruleddx[f(x) ± g(x)] = ddx[f(x)] ± ddx[g(x)]

So: differentiate term by term, keep the constants multiplying, constants on their own vanish.

Worked example — differentiate 3x⁴ − 12x² + 4√x + 5

rewrite: 3x⁴ − ½x⁻² + 4x^−½ + 5x⁰ 3(4x³) − ½(−2x⁻³) + 4(−½ x^−3/2) + 0 = 12x³ + x⁻³ − 2x^−3/2 = 12x³ + 1x³ − 2√(x³)

Common mistake: the power rule works on xⁿ only. You can't "do the power rule" on 1x²+1 or on a product like x(2x−1)(x+3) directly → expand / rewrite first, or use the chain rule (§5).

Gradient at a point — y = x(2x − 1)(x + 3) at (1, 4)

expand: y = 2x³ + 5x² − 3x dy/dx = 6x² + 10x − 3 x = 1: 6(1)² + 10(1) − 3 gradient = 13

Finding unknown constants — y = ax⁴ + bx² + x has gradient 3 at x = 1 and −51 at x = −2

dy/dx = 4ax³ + 2bx + 1 x = 1: 4a + 2b + 1 = 3 ⇒ 2a + b = 1 … (1) x = −2: −32a − 4b + 1 = −51 ⇒ 8a + b = 13 … (2) (2) − (1): 6a = 12 ⇒ a = 2 sub in (1): 4 + b = 1 ⇒ b = −3

Method: differentiate → plug in the x-value → set equal to the given gradient → solve the simultaneous equations.

4. The Chain Rule (composite functions)

To differentiate y = (3x − 2)⁷ don't expand (very long!). Let u = 3x − 2 ⇒ y = u⁷: y is now a function of u, and u a function of x.

★ Chain ruledydx = dydu × dudx

rate of y wrt x = (rate of y wrt u) × (rate of u wrt x)

y = (3x − 2)⁷ ; u = 3x − 2, y = u⁷ du/dx = 3 and dy/du = 7u⁶ dy/dx = 7u⁶ × 3 = 7(3x − 2)⁶ × 3 = 21(3x − 2)⁶

Mental method ("outside–inside"): inside of (3x−2)⁷ is 3x − 2. ① differentiate the outside (keep inside): 7(3x−2)⁶ ② differentiate the inside: 3 ③ multiply: 21(3x−2)⁶.

y = 2(3x² + 1)⁵

u = 3x² + 1, y = 2u⁻⁵ du/dx = 6x dy/du = −10u⁻⁶ dy/dx = −10(3x²+1)⁻⁶ × 6x = −60x(3x²+1)⁶

Mentally: write as 2(3x²+1)⁻⁵ → outside −10(3x²+1)⁻⁶, inside 6x, multiply.

Curve y = √(ax+b) passes through (12, 4) with gradient ¼ there — find a, b

(12, 4): 4 = √(12a + b) … (1) y = (ax+b)^½ ⇒ dy/dx = ½(ax+b)^−½ × a = a2√(ax+b) x = 12: 14 = a2√(12a+b) ⇒ 2a = √(12a+b) … (2) (1) & (2): 2a = 4 ⇒ a = 2 (1): 4 = √(24 + b) ⇒ 16 = 24 + b ⇒ b = −8

Common mistake: forgetting the × derivative of the inside (the "3" in 21(3x−2)⁶, the "6x", the "a"). Also, roots → powers first: √(ax+b) = (ax+b)^½.

5. Tangents and Normals

Tip: in the formula use the numerical value of m (gradient at that point), not the dy/dx expression.

Tangent: touches the curve at A(x₁, y₁), gradient = m = dy/dx at A. Normal: the line perpendicular to the tangent at A.

normal ⟂ tangent at the same point

★ Tangenty − y₁ = m(x − x₁)

★ Normaly − y₁ = −1m(x − x₁)

Special case m = 0: tangent is horizontal (y = y₁) ⇒ normal is vertical, equation x = x₁ (the −1/m formula fails).

Tangent & normal to y = 2x² + 8x² − 9 at x = 2

y = 2x² + 8x⁻² − 9 ; dy/dx = 4x − 16x⁻³ x = 2: y = 8 + 2 − 9 = 1 ⇒ point (2, 1) m = 4(2) − 16(2)⁻³ = 8 − 2 = 6 tangent: y − 1 = 6(x − 2) ⇒ y = 6x − 11 normal: y − 1 = −16(x − 2) 6y − 6 = −x + 2 ⇒ x + 6y = 8

y = (4 − √x)³ : normals at P(4, 8) and Q(9, 1) meet at R. (a) find R (b) area of triangle PQR

dy/dx = 3(4 − √x)² × (−½x^−½) = −3(4 − √x)²2√x x = 4: m = −3(2)²4 = −3 ⇒ normal gradient ⅓ x = 9: m = −3(1)²6 = −½ ⇒ normal gradient 2 normal at P: y − 8 = ⅓(x − 4) ⇒ 3y = x + 20 … (1) normal at Q: y − 1 = 2(x − 9) ⇒ y = 2x − 17 … (2) 3(2x − 17) = x + 20 ⇒ 5x = 71 ⇒ x = 14.2 y = 2(14.2) − 17 = 11.4 ⇒ R = (14.2, 11.4)

area = surrounding rectangle − 3 corner right-angled triangles

rectangle: 10.2 × 10.4 = 106.08 triangles: ½(5)(7) + ½(5.2)(10.4) + ½(10.2)(3.4) = 17.5 + 27.04 + 17.34 = 61.88 area = 106.08 − 61.88 = 44.2 units²

Check trick: coordinates known? shoelace gives ½|4(1−11.4) + 9(11.4−8) + 14.2(8−1)| = ½(88.4) = 44.2 ✓

6. Second Derivatives

Differentiate y → dy/dx (1st derivative). Differentiate again → ddx(dydx) = d²ydx² (2nd derivative).

y = x³ + 5x² − 3x + 2	f(x) = x³ + 5x² − 3x + 2
dy/dx = 3x² + 10x − 3	f′(x) = 3x² + 10x − 3
d²y/dx² = 6x + 10	f″(x) = 6x + 10

y = 5(2x − 3)³ — find d²y/dx²

y = 5(2x − 3)⁻³ dy/dx = −15(2x − 3)⁻⁴ × 2 = −30(2x − 3)⁻⁴ (chain rule) d²y/dx² = 120(2x − 3)⁻⁵ × 2 (chain rule again) = 240(2x − 3)⁵

y = x³ + 3x² − 9x + 2 : (a) range of x where dy/dx and d²y/dx² are both negative (b) show d²y/dx² ≠ (dy/dx)²

dy/dx = 3x² + 6x − 9 < 0 x² + 2x − 3 < 0 ⇒ (x + 3)(x − 1) < 0 ⇒ −3 < x < 1 … (1) d²y/dx² = 6x + 6 < 0 ⇒ x < −1 … (2) both: −3 < x < −1

sketch the quadratic, read off where it is below the axis

combine the conditions on a number line

(b) d²y/dx² = 6x + 6 but (dy/dx)² = (3x² + 6x − 9)² ⇒ not equal.

Common mistake: d²ydx² means "differentiate twice", it is NOT (dy/dx)² (the square of the first derivative).

✓ Full Chapter Cheat-Sheet

Gradient of curve at a point = limit of chord gradients; dy/dx = f′(x) = gradient function. d²y/dx² = f″(x).
Power: d/dx(xⁿ) = nxⁿ⁻¹ (any n) | constant → 0 | rewrite roots & fractions as powers first.
Scalar: d/dx[kf(x)] = k·d/dx[f(x)] | Sum/diff: differentiate term by term.
Chain: dy/dx = dy/du × du/dx (outside × inside) — e.g. d/dx (ax+b)ⁿ = n·a·(ax+b)ⁿ⁻¹.
Tangent: y − y₁ = m(x − x₁) | Normal: y − y₁ = −(1/m)(x − x₁); if m = 0 normal is x = x₁.
Steps for tangent/normal: find y at given x → dy/dx → m numerically → build the line.
Unknown constants: differentiate, substitute gradient/point conditions, solve simultaneously.
Gradient < 0 / > 0 ranges: solve the inequality (sketch the quadratic); combine on a number line.