1. Find the value of f’(x) when, a. f(x) = 3x + ex (1) b. f(x) = 𝑥𝑥

12 + 2 ln 𝑥𝑥 (1)

c. f(x) = 4√𝑥𝑥 + 14

ln𝑥𝑥 (1) __________________________________________________________________________________________

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2. Find the value of x for which f’(x) = -1 when f(x) = 𝑥𝑥

2

8− 2𝑥𝑥 + ln𝑥𝑥 (3)

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3. Find the equation for the normal to the curve y = 4 + 3ex when x = 0 (4) __________________________________________________________________________________________

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4. Find he equation for the tangent to the curve y = 𝑥𝑥13 − 3𝑒𝑒𝑥𝑥 at the point when x = 1. (4)

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A-Level Starter Activity

Topic: Differentiating Logarithms and Exponentials Chapter Reference: Pure 2, Chapter 9

8 minutes

Solutions

1a. f’(x) = 3 + ex M1

1b. f’(x) = 1

2𝑥𝑥−

12 + 2

𝑥𝑥 M1

1c. f’(x) = 2𝑥𝑥−

12 + 1

4𝑥𝑥 M1

2.

f’(x) = 14𝑥𝑥 – 2 + 1

𝑥𝑥

14𝑥𝑥 – 2 + 1

𝑥𝑥= −1

M1

x2 – 4x + 4 = 0

(x – 2)2 = 0

M1

x = 2 M1 3.

When x = 0, y = 7 M1 𝑑𝑑𝑑𝑑𝑑𝑑𝑥𝑥

= 3𝑒𝑒x

Gradient = 3 M1

Gradient of normal = -13 M1

y = 7 - 13𝑥𝑥 M1

4.

x = 1 y = 1 – 3e M1 𝑑𝑑𝑑𝑑𝑑𝑑𝑥𝑥

= 13𝑥𝑥−

23 - 3𝑒𝑒𝑥𝑥

Gradient = 13− 3𝑒𝑒

M1

y – (1 – 3e) = (13− 3𝑒𝑒)(𝑥𝑥 − 1) M1

y = (13− 3𝑒𝑒)𝑥𝑥 + 2

3 M1

1. Differentiate: a. y = cos (3x – 2) (1) b. y = 4 sin (𝜋𝜋

3− 𝑥𝑥) (1)

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2. Find the coordinate of any stationary points on the curve y = x + 2 sin x in the interval 0 ≤ x ≤ 2𝜋𝜋 (4) __________________________________________________________________________________________

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3. Find the equation for the tangent to the curve y = cos x at the point x = 𝜋𝜋

3 (5)

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A-Level Starter Activity

Topic: Differentiating sin and cos Chapter Reference: Pure 2, Chapter 9

8 minutes

Solutions

1a. 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= −3 sin(3𝑥𝑥 − 2) M1 1b.

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= -4 cos (𝜋𝜋3− x) M1

2.

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 1 + 2 cos 𝑥𝑥 M1 S.P: 1 + 2 cos x = 0 cos x = -1

2 M1

x = 𝜋𝜋 − 𝜋𝜋3 = 2𝜋𝜋

3

y = 2𝜋𝜋3

+ √3 M1

x = 𝜋𝜋 + 𝜋𝜋3 = 4𝜋𝜋

3

y = 4𝜋𝜋3− √3

M1

3.

𝑥𝑥 = 𝜋𝜋3

y = 12

M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= − sin𝑥𝑥 M1

gradient = -sin (𝜋𝜋3) = -√3

2 M1

y - 12

= −√32

(𝑥𝑥 − 𝜋𝜋3

) M1

3√3𝑥𝑥 + 6y – 3 - √3𝜋𝜋 = 0 M1

1. Differentiate 3 cosec (x + 𝜋𝜋

6) (1)

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2. Differentiate sec22x (1) __________________________________________________________________________________________

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3. Differentiate ln (tan 4x) (2) __________________________________________________________________________________________

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4. Differentiate 2 cot x2 (1) __________________________________________________________________________________________

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5. Find the equation for the tangent to the curve y = cosec x – 2 sin x at the point x = 𝜋𝜋

6 (4)

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A-Level Starter Activity

Topic: Differentiating tan, sec, cosec, and cot Chapter Reference: Pure 2, Chapter 9

7 minutes

Solutions

1. Chain Rule 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= −3 cosec 3x cot 2x M1

2.

Chain Rule 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2 sec 2𝑥𝑥 × 2 sec 2𝑥𝑥 tan 2𝑥𝑥 = 4 sec22x tan 2x

M1

3.

Chain Rule 1

tan4𝑑𝑑 × 4𝑠𝑠𝑠𝑠𝑠𝑠24𝑥𝑥 M1

= cos4𝑑𝑑sin4𝑑𝑑

× 4𝑐𝑐𝑐𝑐𝑐𝑐24𝑑𝑑

= 4 sec 4x cosec 4x

M1

4.

Chain Rule -2 cosec2x2 × 2𝑥𝑥 = -4x cosec2 x2

M1

5.

x = 𝜋𝜋6

y = 1 M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= −𝑠𝑠𝑐𝑐𝑠𝑠𝑠𝑠𝑠𝑠 𝑥𝑥 cot 𝑥𝑥 − 2 cos 𝑥𝑥 M1 Gradient = -3√3 M1 y – 1 = -3√3 (x - 𝜋𝜋

6)

6√3x + 2y – 2 - √3𝜋𝜋 = 0 M1

1. Differentiate 4

(2𝑥𝑥+3)3 (2)

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2. Differentiate with respect x, 3ln(4 - √𝑥𝑥) (2) __________________________________________________________________________________________

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3. Find the coordinates of any station points on the curve y = 8x – e2x (4) __________________________________________________________________________________________

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4. Find an equation for the normal to each curve at the point on the curve when y = 𝑒𝑒4−𝑥𝑥2- 10 when x = -2 (4) __________________________________________________________________________________________

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A-Level Starter Activity

Topic: The Chain Rule Chapter Reference: Pure 2, Chapter 9

8 minutes

Solutions 1. -12 (2x + 3)-4 x 2 M1 = -24(2x + 3)-4 M1

2.

34−√𝑥𝑥

× (−12𝑥𝑥−

12) M1

= 32𝑥𝑥−8√𝑥𝑥

M1 3.

𝑑𝑑𝑑𝑑𝑑𝑑𝑥𝑥

= 8 − 2𝑒𝑒2𝑥𝑥 S.P: 8 − 2𝑒𝑒2𝑥𝑥 = 0

M1

e2x = 4 M1 x = 1

2ln 4 = ln 2 M1

y = 8ln2 – 4 M1 (ln 2, 8 ln2 – 4)

4.

x = -2 y = -9 M1 𝑑𝑑𝑑𝑑𝑑𝑑𝑥𝑥

= 𝑒𝑒4−𝑥𝑥2 × (−2𝑥𝑥) = −2𝑥𝑥𝑒𝑒4−𝑥𝑥2 M1 Gradient = 4 Gradient of normal = -1

4 M1

y + 9 = -14

(𝑥𝑥 + 2)

y = -14𝑥𝑥 − 19

2

M1

1. Find 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 when y = (x + 1) ln (x2 – 1) (3)

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2. Find the values of f’(x) at the point x = 1

4 when f(x) = 𝑥𝑥

12(1 − 2𝑥𝑥)3 (4)

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3. Find the stationary points on the curve y = 2 + x2e-4x (4) __________________________________________________________________________________________

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5. Find the value of f’(x) at the point x = 𝜋𝜋4 when f(x) = sin 3x cos 5x (3)

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A-Level Starter Activity

Topic: The Product Rule Chapter Reference: Pure 2, Chapter 9

8 minutes

Solutions

1. 1 x ln(x2 – 1) + (x + 1) x 1

𝑑𝑑2−1 x 2x M1

= ln(x2 – 1) + 2𝑑𝑑(𝑑𝑑+1)(𝑑𝑑+1)(𝑑𝑑−1) M1

= ln(x2 – 1) + 2𝑑𝑑𝑑𝑑−1

M1 2.

f’(x) = 12𝑥𝑥−

12 × (1 − 2𝑥𝑥)3 + 𝑥𝑥

12 × 3(1 − 2𝑥𝑥)2 × −2 M1

= 12𝑥𝑥−

12(1 − 2𝑥𝑥)2[(1 – 2x) – 12x]

= 12𝑥𝑥−

12(1 – 14x)(1 − 2𝑥𝑥)2

M1

f’(14) = 1

2× 2 × (−5

2) × 1

4 M1

f’(14) = -5

8 M1

3.

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 1 × (𝑥𝑥 − 4)3 + 𝑥𝑥 × 3(𝑥𝑥 − 4)2 = (x – 4)2[(x – 4) + 3x]

M1

4(x - 1)(x – 4)2 = 0 M1 x = 1 y = -27 M1

x = 0 y = 4 M1

4.

f’(x) = 3 cos 3x × cos 5𝑥𝑥 + sin 3𝑥𝑥 × (−5 sin 5𝑥𝑥) M1 f’(x) = 3 cos 3x cos 5x – 5 sin 3x sin 5x M1 f’(𝜋𝜋

4) = 3(- 1

√2)( 1√2

) – 5 × 1√2

(- 1√2

) = 4

M1

1. Find 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 when y = 1−𝑑𝑑

𝑑𝑑3+2 (2)

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2. Find the coordinates of any stationary points on the curve y = 𝑒𝑒

4𝑥𝑥

2𝑑𝑑−1 (5)

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3. Find the value of f’(x) when f(x) = ln (2cos𝑑𝑑)

sin𝑑𝑑 at the point x = 𝜋𝜋

3 (3)

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A-Level Starter Activity

Topic: The Quotient Rule Chapter Reference: Pure 2, Chapter 9

7 minutes

Solutions

1. 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= = −1×�𝑑𝑑3+2�−(1−𝑑𝑑)×3𝑑𝑑2

(𝑑𝑑3+2)2 M1

= 2𝑑𝑑3−3𝑑𝑑2−2

(𝑑𝑑3+2)2 M1

2.

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 4𝑒𝑒4𝑥𝑥×(2𝑑𝑑−1)−𝑒𝑒4𝑥𝑥×2(2𝑑𝑑−1)2

M1 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2𝑒𝑒4𝑥𝑥(4𝑑𝑑−3)(2𝑑𝑑−1)2

M1 2𝑒𝑒4𝑥𝑥(4𝑑𝑑−3)

(2𝑑𝑑−1)2 = 0 M1

x = 34

y = 2e3 M1

(34, 2e3) M1

3.

f’(x) = 1

2 cos𝑥𝑥×(−2sin𝑑𝑑) ×(sin𝑑𝑑)−ln (2cos𝑑𝑑) ×cos𝑑𝑑

𝑠𝑠𝑠𝑠𝑠𝑠2𝑑𝑑 M1

= -sec x - cos𝑑𝑑 ln(2cos𝑑𝑑)𝑠𝑠𝑠𝑠𝑠𝑠2𝑑𝑑

M1 f’(𝜋𝜋

3) = -2 – 0 = -2 M1

1. The diagram below shows a closed box used by a show for packing pieces of cake. The box is a right prism of height h cm. The cross section is a sector of a circle. The sector has radius r cm and angle 1 radian. The volume of the box is 300 cm3 a. Show that the surface area of the box, S cm2, is given by, S = r2 + 1800

𝑟𝑟 (5)

b. Use calculus to find the value of r for which S is stationary. (4) c. Prove that this value of r gives a minimum value of S. (2) d. Find, to the nearest cm2, this minimum value of S. (2) __________________________________________________________________________________________

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A-Level Starter Activity

Topic: Radians Chapter Reference: Pure 2, Chapter 9

8 minutes

Solutions

1a. Arc length = r𝜃𝜃 = r x 1 = r M1 Sector area = 1

2r2𝜃𝜃 = 1

2(1)2𝜃𝜃 = 𝑟𝑟

2

2 M1

Surface area = 2 sectors + 2 rectangles + curved face. = 𝑟𝑟2 + 3𝑟𝑟ℎ M1

Volume = 300 = 12𝑟𝑟2ℎ M1

𝑆𝑆 = 𝑟𝑟2 + 3 × 3 × 600𝑟𝑟

𝑆𝑆 = 𝑟𝑟2 + 1800𝑟𝑟

M1

1b.

𝑑𝑑𝑑𝑑𝑑𝑑𝑟𝑟

= 2𝑟𝑟 − 1800𝑟𝑟2

M1

Maximum when 𝑑𝑑𝑑𝑑𝑑𝑑𝑟𝑟

= 0 M1

2𝑟𝑟 − 1800𝑟𝑟2

= 𝑟𝑟3 = 900

M1

𝑟𝑟 = 9.7 M1 1c.

𝑑𝑑2𝑑𝑑𝑑𝑑𝑟𝑟2

= 2 + 3600𝑟𝑟3

> 0 M1 Therefore, a minimum point. M1

1d.

Smin = (9.65)2 + 18009.65

M1 Smin = 279.65 M1

1. Find an expression for 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 when x = t2 – 1, y = 2t3 + t2 (2)

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2. Find an expression for 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 when x = et + 1 and y = e2t – 1 (3)

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3. Find in the form y = mx + c, find an equation for the tangent to the curve at the point t = 𝜋𝜋

3 when x = 2 sin t and

y = 1 – 4 cos t. (4) __________________________________________________________________________________________

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A-Level Starter Activity

Topic: Parametric Differentiation Chapter Reference: Pure 2, Chapter 9

8 minutes

Solutions

1. 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 6𝑡𝑡2 + 2t M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 6𝑑𝑑2+2𝑑𝑑2𝑑𝑑

= 3𝑡𝑡 + 1 M1 2.

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= et + 1 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2𝑒𝑒2t - 1 M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2𝑒𝑒2𝑡𝑡−1

𝑒𝑒𝑡𝑡+1 M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2et - 2 M1 3.

t = 𝜋𝜋3

x = √3 y = -1

M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2 cos 𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 4 sin 𝑡𝑡 M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 4sin 𝑑𝑑2cos𝑑𝑑

= 2 tan 𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

(√3) = 2√3 M1

y + 1 = 2√3(x - √3) y = 2√3x - 7

M1

1. Find 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 in terms of x and y for the curve, 2e3x + e-2y + 7 = 0. Leave your answers in the form 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 = aebx + cy where

a, b, and c are constants. (3) __________________________________________________________________________________________

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2. Find 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 in terms of x and y given x sin y + x2 cos y = 1 (3)

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3. Find an equation for the tangent to the curve at (𝜋𝜋

3, 𝜋𝜋6

) on the curve, 4 sin y – sec x = 0 (5) __________________________________________________________________________________________

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A-Level Starter Activity

Topic: Implicit Differentiation Chapter Reference: Pure 2, Chapter 9

8 minutes

Solutions 1. 6𝑒𝑒3x – 2𝑒𝑒2y𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑= 0 M1

6𝑒𝑒3x = 2𝑒𝑒-2𝑦𝑦 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 3𝑒𝑒3𝑥𝑥

𝑒𝑒−2𝑦𝑦

M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 3e3x + 2y M1 2.

1 × sin𝑦𝑦 + 𝑥𝑥 × 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

cos 𝑦𝑦 + 2𝑥𝑥 × cos 𝑦𝑦 + 𝑥𝑥2 × (− sin𝑦𝑦) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 0 M1

sin y + 2x cos y = 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

(𝑥𝑥2 sin y – x cos y) M1 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= sin𝑑𝑑+2𝑑𝑑𝑥𝑥𝑥𝑥𝑥𝑥 𝑑𝑑𝑑𝑑2 sin𝑑𝑑−𝑑𝑑 cos𝑑𝑑

M1 3.

4𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

cos 𝑦𝑦 − sec 𝑥𝑥 tan 𝑥𝑥 = 0 M1

4𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

cos 𝑦𝑦 = sec𝑥𝑥 tan 𝑥𝑥 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= sec𝑑𝑑 tan𝑑𝑑4 cos𝑑𝑑

M1

Gradient = 2×√3

4×√32

= 1 M1

y - 𝜋𝜋6

= 𝑥𝑥 − 𝜋𝜋3 M1

y = x - 𝜋𝜋6 M1

1. The area of a circle is decreasing at a constant rate of 0.5 cm2 s-1. a. Find the rate at which the radius of the circle is decreasing when the radius is 8 cm. (4) b. Find the rate at which the perimeter of the circle is decreasing when the radius is 8 cm. (3) __________________________________________________________________________________________

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2. The volume of a cube is increasing at a constant rate of 3.5 cm3 s-1. Find, a. The rate at which the length of one side of the cube is increasing when the volume is 200 cm3. (5) b. The volume of the cube when the length of one side is increasing at a rate of 2 mm s-1. (4) __________________________________________________________________________________________

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A-Level Starter Activity

Topic: Connected Rates of Change Chapter Reference: Pure 2, Chapter 9

10 minutes

Solutions

1a. 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

× 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

M1 A = 𝜋𝜋r2 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2𝜋𝜋r M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= -0.5, r = 8 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 16𝜋𝜋 M1

-0.5 = 16𝜋𝜋 x 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= − 132𝜋𝜋

= -0.00995 cm s-1 M1

Therefore, the radius is decreasing at 0.00995cm s-1 (3 s.f) 1b.

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

× 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

M1 P = 2𝜋𝜋r 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2𝜋𝜋 M1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2𝜋𝜋 × − 132𝜋𝜋

= - 116

Therefore, perimeter decreases at a rate of 0.0625 cm s-1

M1

2a. 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

× 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 3.5, V = l3 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 3l2

M1

200 = l3 l = 5.848 M1 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 3 × 5.8482 = 102.6 M1

3.5 = 102.6 x 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

M1 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 3.5 ÷ 102.6 = 0.0341 𝑐𝑐𝑐𝑐s-1 M1

2b. 2 mm s-1 = 0.2 cm s-1 M1 3.5 = 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 × 0.2

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 17.5 M1

17.5 = 3l2 l = 2.415 M1

V = l3 = 14.1 cm3 M1

1. Given that y = x(3x – 1)3 a. Find 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 and 𝑑𝑑

2𝑑𝑑𝑑𝑑𝑑𝑑2

(3) b. Find the points of inflection of y (3) __________________________________________________________________________________________

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A-Level Starter Activity

Topic: Using Second Derivatives Chapter Reference: Pure 2, Chapter 9

6 minutes

Solutions 1. y = x(3x – 1)5 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 15𝑥𝑥(3𝑥𝑥 − 1)4 + (3x – 1)5 M1

𝑑𝑑2𝑑𝑑𝑑𝑑𝑑𝑑2

= 15(3x – 1)4 + 15(3x – 1)4 + 180x(3x – 1)3 = 30(3x – 1)4 + 180x(3x -1)3

M1

= 30(3x – 1)3(9x – 1) M1 2.

At points of inflection, 𝑑𝑑2𝑑𝑑

𝑑𝑑𝑑𝑑2 = 0

30(3x – 1)3(9x – 1) = 0 M1

x = 13

y = 13

× (1 − 1)5 = 0 M1

x = 19

y = 19

(13− 1)5 = - 32

2187

M1

Points of inflection are: (19, - 322187

) and (13, 0)