Lesson 3: Differentiation & Integration (P1) | A Level Mathematics

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1. Differentiation — Standard Results P1

Standard Derivatives — P1 Core

d/dx (xⁿ) = nxⁿ⁻¹

d/dx (eˣ) = eˣ d/dx (eᵃˣ) = aeᵃˣ

d/dx (ln x) = 1/x d/dx (ln ax) = 1/x

d/dx (sin x) = cos x d/dx (sin ax) = a cos ax

d/dx (cos x) = −sin x d/dx (cos ax) = −a sin ax

d/dx (tan x) = sec²x d/dx (tan ax) = a sec²ax

Chain Rule: dy/dx = (dy/du)(du/dx)

Product Rule: d/dx(uv) = u'v + uv'

Quotient Rule: d/dx(u/v) = (u'v − uv')/v²

📐 Worked Example 1 — Differentiating Mixed Functions

Differentiate: (a) y = x³ ln x (b) y = sin²(3x) (c) y = e^(2x)/(x²+1)

(a) Product rule: u=x³, v=ln x
dy/dx = 3x²·ln x + x³·(1/x) = 3x² ln x + x²= x²(3ln x + 1)

(b) y=(sin 3x)². Chain rule: outer (...)² → inner sin 3x → 3cos 3x
dy/dx = 2sin 3x · 3cos 3x = 6 sin 3x cos 3x = 3 sin 6x

(c) Quotient rule: u=e^(2x), v=x²+1
dy/dx = [2e^(2x)(x²+1) − e^(2x)·2x]/(x²+1)²
= 2e^(2x)(x²−x+1)/(x²+1)²

2. Applications — Tangents, Normals & Stationary Points P1

Tangent, Normal, and Stationary Point Procedure:
Tangent at (x₁,y₁): gradient m = dy/dx|_{x=x₁}; equation: y−y₁ = m(x−x₁)
Normal at (x₁,y₁): gradient = −1/m; equation: y−y₁ = (−1/m)(x−x₁)
Stationary points: Set dy/dx = 0; classify using d²y/dx² or sign change.
d²y/dx² > 0 → minimum | d²y/dx² < 0 → maximum | d²y/dx² = 0 → use sign test.

📐 Worked Example 2 — Stationary Points

Find and classify all stationary points of y = x⁴ − 8x² + 3.

dy/dx = 4x³ − 16x = 4x(x²−4) = 4x(x−2)(x+2)
Stationary: x = 0, x = 2, x = −2

y values: y(0)=3; y(2)=16−32+3=−13; y(−2)=−13
Points: (0,3), (2,−13), (−2,−13)

d²y/dx² = 12x²−16
x=0: d²y/dx²=−16 < 0 → local maximum at (0, 3)
x=±2: d²y/dx²=32 > 0 → local minima at (±2, −13)

📐 Worked Example 3 — Tangent and Normal

The curve y = 4/x + √x has a tangent at the point where x = 4. Find the equations of the tangent and normal.

At x=4: y = 4/4 + √4 = 1 + 2 = 3. Point: (4, 3).

dy/dx = −4/x² + 1/(2√x). At x=4: dy/dx = −4/16 + 1/4 = −1/4 + 1/4 = 0
Gradient of tangent = 0 → horizontal tangent.

Tangent: y = 3 (horizontal line)
Normal: x = 4 (vertical line, perpendicular to horizontal)

Increasing and Decreasing Functions

f is increasing on an interval when f'(x) > 0 for all x in that interval.
f is decreasing on an interval when f'(x) < 0 for all x in that interval.
The function changes from increasing to decreasing (or vice versa) at stationary points.

Optimisation

📐 Worked Example 4 — Optimisation

A closed cylinder has total surface area 600π cm². Find the radius that maximises the volume.

SA = 2πr² + 2πrh = 600π → r² + rh = 300 → h = (300−r²)/r

V = πr²h = πr²·(300−r²)/r = π r(300−r²) = π(300r − r³)

dV/dr = π(300 − 3r²) = 0 → r² = 100 → r = 10 cm

d²V/dr² = −6πr. At r=10: −60π < 0 → maximum ✓
h = (300−100)/10 = 20 cm. V_max = π×10×(300−100) = 2000π cm³

3. Connected Rates of Change P1

Chain Rule for Rates of Change

dy/dt = (dy/dx) · (dx/dt)

Parametric differentiation: dy/dx = (dy/dt)/(dx/dt)

📐 Worked Example 5 — Connected Rates

The volume of a sphere is increasing at 8π cm³/s. Find the rate of increase of the surface area when r = 3 cm.

V = (4/3)πr³ → dV/dr = 4πr²
SA = 4πr² → dSA/dr = 8πr

dV/dt = 8π. Chain rule: dV/dt = (dV/dr)(dr/dt)
8π = 4πr² · dr/dt → at r=3: dr/dt = 8π/(4π×9) = 2/9 cm/s

dSA/dt = (dSA/dr)(dr/dt) = 8πr × 2/9 = 16πr/9
At r=3: dSA/dt = 16π/3 cm²/s

📐 Worked Example 6 — Parametric Differentiation

A curve is defined by x = t² + 2t, y = t³ − 3t. Find dy/dx in terms of t, and find the gradient when t = 2.

dx/dt = 2t+2 dy/dt = 3t²−3

dy/dx = (3t²−3)/(2t+2) = 3(t²−1)/(2(t+1)) = 3(t−1)/2

At t=2: dy/dx = 3(2−1)/2 = 3/2

4. Integration — Standard Results and Techniques P1

Standard Integrals — P1 Core

∫xⁿ dx = xⁿ⁺¹/(n+1) + c (n ≠ −1)

∫1/x dx = ln|x| + c

∫eˣ dx = eˣ + c ∫eᵃˣ dx = (1/a)eᵃˣ + c

∫sin x dx = −cos x + c ∫sin ax dx = −(1/a)cos ax + c

∫cos x dx = sin x + c ∫cos ax dx = (1/a)sin ax + c

∫sec²x dx = tan x + c ∫sec²ax dx = (1/a)tan ax + c

∫(ax+b)ⁿ dx = (ax+b)ⁿ⁺¹/[a(n+1)] + c (n ≠ −1)

∫1/(ax+b) dx = (1/a)ln|ax+b| + c

Integration by substitution: For ∫f(g(x))·g'(x) dx, let u = g(x), then du = g'(x)dx and the integral becomes ∫f(u) du. This is the reverse of the chain rule.

📐 Worked Example 7 — Integration by Substitution

Find: (a) ∫x(x²+3)⁵ dx (b) ∫sin³x cos x dx (c) ∫(2x+1)e^(x²+x) dx

(a) Let u=x²+3, du=2x dx → x dx = du/2
∫u⁵(du/2) = u⁶/12 + c = (x²+3)⁶/12 + c

(b) Let u=sin x, du=cos x dx
∫u³ du = u⁴/4 + c = sin⁴x/4 + c

(c) Let u=x²+x, du=(2x+1)dx
∫eᵘ du = eᵘ + c = e^(x²+x) + c

Definite Integrals

📐 Worked Example 8 — Definite Integral

Evaluate: (a) ∫₁³ (2x + 1/x²) dx (b) ∫₀^(π/3) 2sin(2x) dx

(a) ∫(2x + x⁻²) dx = [x² − x⁻¹]₁³
= (9 − 1/3) − (1 − 1) = (26/3) − 0 = 26/3

(b) [−cos 2x]₀^(π/3)
= −cos(2π/3) − (−cos 0) = −(−½) + 1 = ½ + 1 = 3/2

5. Areas and Volumes of Revolution P1

Area and Volume Formulae

Area between curve and x-axis: A = ∫ₐᵇ y dx (take |value| if below axis)

Area between two curves: A = ∫ₐᵇ (y_upper − y_lower) dx

Area between curve and y-axis: A = ∫_c^d x dy

Volume of revolution about x-axis: V = π ∫ₐᵇ y² dx

Volume of revolution about y-axis: V = π ∫_c^d x² dy

📐 Worked Example 9 — Area Between Two Curves

Find the area enclosed between y = x² and y = 4x − x².

Intersections: x² = 4x−x² → 2x²−4x = 0 → 2x(x−2) = 0
x = 0 and x = 2.

Check which is upper: at x=1: y₁=1, y₂=3. Upper = 4x−x².
A = ∫₀² [(4x−x²) − x²] dx = ∫₀² (4x−2x²) dx

= [2x² − (2/3)x³]₀² = 8 − 16/3 = (24−16)/3 = 8/3 sq units

📐 Worked Example 10 — Volume of Revolution

The region bounded by y = √(x+1), the x-axis, x = 0 and x = 3 is rotated 360° about the x-axis. Find the volume generated.

V = π∫₀³ y² dx = π∫₀³ (x+1) dx

= π[x²/2 + x]₀³ = π[(9/2+3)−0] = π × 15/2 = 15π/2 cubic units

⚠ Common Integration Mistakes:
1. Forgetting +c in indefinite integrals — always required.
2. Area below x-axis gives a negative integral — take the absolute value for the actual area. Split the integral at the x-intercepts.
3. In volume of revolution: V = π∫y² dx, NOT π∫y dx. Square y first, then integrate.
4. When curves intersect, check which is upper — don't assume from the equation.

6. Kinematics Using Calculus P1

Kinematics Relationships

v = ds/dt a = dv/dt = d²s/dt²

s = ∫v dt v = ∫a dt

v · dv/ds = a (useful when a is a function of s)

📐 Worked Example 11 — Kinematics

A particle moves such that its velocity at time t is v = 3t² − 12t + 9 m/s (t ≥ 0). Find: (a) when the particle is at rest (b) the acceleration when t = 3 (c) the displacement from the start in the first 4 seconds (d) the total distance travelled in the first 4 seconds.

(a) v = 0: 3t²−12t+9 = 3(t−1)(t−3) = 0 → t=1 and t=3

(b) a = dv/dt = 6t−12. At t=3: a = 18−12 = 6 m/s²

(c) s = ∫₀⁴(3t²−12t+9) dt = [t³−6t²+9t]₀⁴
= (64−96+36)−0 = 4 m

(d) v changes sign at t=1 and t=3. Calculate s in each section:
s(0→1): [t³−6t²+9t]₀¹ = 1−6+9 = 4 m
s(1→3): [t³−6t²+9t]₁³ = (27−54+27)−(1−6+9) = 0−4 = −4 m (moved back 4 m)
s(3→4): [t³−6t²+9t]₃⁴ = (64−96+36)−0 = 4 m
Total distance = 4+4+4 = 12 m

7. Implicit Differentiation P1

Implicit differentiation: When y cannot be made the subject explicitly, differentiate both sides with respect to x, using the chain rule for any term in y: d/dx[f(y)] = f'(y) · dy/dx.

📐 Worked Example 12 — Implicit Differentiation

Find dy/dx for: (a) x² + y² = 25 (b) x²y + y³ = 8 (c) Find the gradient of x³ + y³ − 3xy = 5 at (2, 1).

(a) Differentiate implicitly: 2x + 2y(dy/dx) = 0
dy/dx = −x/y

(b) Product rule on x²y: 2xy + x²(dy/dx) + 3y²(dy/dx) = 0
dy/dx(x²+3y²) = −2xy
dy/dx = −2xy/(x²+3y²)

(c) 3x² + 3y²(dy/dx) − 3y − 3x(dy/dx) = 0
dy/dx(3y²−3x) = 3y−3x²
dy/dx = (y−x²)/(y²−x)
At (2,1): dy/dx = (1−4)/(1−2) = −3/−1 = 3

📝 Exam Practice Questions

Q1 [4 marks] — Differentiate with respect to x:
(a) y = (3x²−1)⁴ (b) y = x² e^(−2x) (c) y = ln(sin x) (d) y = cos³(2x)

(a) dy/dx = 4(3x²−1)³ × 6x = 24x(3x²−1)³

(b) Product: 2xe^(−2x) + x²×(−2)e^(−2x) = e^(−2x)(2x−2x²) = 2xe^(−2x)(1−x)

(c) Chain: (1/sin x)×cos x = cot x

(d) y=(cos 2x)³. dy/dx = 3cos²(2x)×(−sin 2x)×2 = −6cos²(2x)sin(2x)

Q2 [5 marks] — The curve y = 2x³ − 9x² + 12x − 3 has two stationary points. (a) Find the stationary points and classify each. (b) Find the range of values of x for which the curve is decreasing. (c) Find the equation of the tangent at x = 1.

(a) dy/dx = 6x²−18x+12 = 6(x−1)(x−2) = 0 → x=1, x=2
y(1)=2−9+12−3=2; y(2)=16−36+24−3=1
d²y/dx²=12x−18. At x=1: −6<0 → max at (1,2). At x=2: 6>0 → min at (2,1).

(b) Decreasing: dy/dx<0 → 6(x−1)(x−2)<0 → 1<x<2

(c) At x=1, y=2, dy/dx=0. Tangent: y=2

Q3 [4 marks] — Find the following integrals:
(a) ∫(x + 1/x)² dx (b) ∫₀¹ x/(2x²+1) dx

(a) Expand: (x+1/x)² = x²+2+1/x²
∫(x²+2+x⁻²) dx = x³/3 + 2x − 1/x + c

(b) Let u=2x²+1, du=4x dx → x dx=du/4. Limits: x=0→u=1; x=1→u=3
∫₁³ (1/u)(du/4) = [¼ln u]₁³ = ¼ln3−0 = ¼ ln 3

Q4 [5 marks] — The curve y = 4/x and the line y = 5 − x enclose a region R. (a) Find the x-coordinates of the intersection points. (b) Find the area of R. (c) Find the volume when R is rotated 360° about the x-axis.

(a) 4/x = 5−x → x(5−x)=4 → x²−5x+4=0 → (x−1)(x−4)=0
x=1 and x=4

(b) At x=2: line=3, curve=2 → line above.
A = ∫₁⁴[(5−x)−4/x]dx = [5x−x²/2−4ln x]₁⁴
= (20−8−4ln4)−(5−½−0) = (12−4ln4)−(9/2) = 7.5−4ln4 ≈ 7.5−4ln4 sq units

(c) V=π∫₁⁴[(5−x)²−(4/x)²]dx=π∫₁⁴[25−10x+x²−16/x²]dx
=π[25x−5x²+x³/3+16/x]₁⁴
=π[(100−80+64/3+4)−(25−5+1/3+16)]=π[(24+64/3)−(37/3+16)]
=π[24+64/3−16−37/3]=π[8+27/3]=π[8+9]=17π cubic units

Q5 [4 marks] — A particle moves in a straight line. Its acceleration at time t seconds is a = 2t − 6 m/s². Initially the particle is at rest at the origin. Find expressions for v and s. Find the total distance travelled in the first 6 seconds.

v = ∫(2t−6) dt = t²−6t+c. v=0 at t=0 → c=0. v = t²−6t
s = ∫(t²−6t) dt = t³/3−3t²+c. s=0 at t=0 → c=0. s = t³/3 − 3t²
v=0: t(t−6)=0 → t=0 or t=6. Particle reverses at t=6 only in [0,6].
Wait — also check sign: for 0<t<6, v=t(t−6)<0 (moving backwards). At t=6: s=72−108=−36.
Distance = |s(6)−s(0)| = |−36| = 36 m

Exam Tip: Always check whether v changes sign in the given interval. If v is negative throughout [0,6], the particle moves in one direction only — total distance = |displacement|. Only split at zeros of v within the interval.

Q6 [3 marks] — Find dy/dx for x² + 2xy − y² = 4. Hence find the equation of the normal at (2, 2).

Differentiate implicitly: 2x + 2y + 2x(dy/dx) − 2y(dy/dx) = 0
dy/dx(2x−2y) = −2x−2y → dy/dx = −(x+y)/(x−y)
At (2,2): dy/dx = −(4)/(0) → undefined (vertical tangent)
So the tangent is vertical: x=2. The normal is horizontal: y = 2

Exam Tip: When dy/dx is undefined (division by zero), the tangent is vertical (x = constant) and the normal is horizontal (y = constant). This is a valid and often tested special case.

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