AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find 𝑑 𝑑 when y = x2 ln (x

Differentiation

Pt. 3: General Differentiation

1. Find 𝑑2𝑦

𝑑𝑥2 for y = 4x3 + ex (2)

2. Find 𝑑𝑦

𝑑𝑥 for y =

3

𝑥+ 3 ln 𝑥 (2)

3. Find any values of x for which 𝑑𝑦

𝑑𝑥 = 0 when y = 5ln x – 8x (2)

4. Find the co-ordinates and the nature of any stationary points on the curve y = 7 + 2x – 4 lnx (5)

5. Find an equation for the normal to the cube y = 10 + ln x at the x-cordinate 3. (4)

6. Find the differential of 4sin x + 2 cos x (2)

7. Differentiate cot(2x – 3) (1)

8. Find an equation for the tangent for the curve y = cosec x – 2sin x at the point x = 𝜋

6 in the interval 0 ≤ x ≤ 2𝜋 (5)

9. Find the differential of y = sec 5x (1)

10. Differentiate with respect to x, 3x

11. Differentiate with respect to x, 51-x

12. Differentiate with respect to x, 2𝑥3

A-Level


Pt. 4: The Chain Rule

Pt. 5: The Product Rule

Pt. 6: The Quotient Rule

Pt. 7: Parametric Differentiation

Pt. 8: Implicit Differentiation

Pt. 9: Rates of Change

AS Level

Pt. 1: First Principles

Pt. 2: Differentiation




Mark Scheme

1.

𝑑𝑦

𝑑𝑥= 12x2 + ex M1

𝑑2𝑦

𝑑𝑥2 = 24x + ex M1

2.

y = 3

𝑥+ 3 ln 𝑥 = 3x-1 + 3 ln x M1

𝑑𝑦

𝑑𝑥 = -3x-2 +

3

𝑥 M1

3.

𝑑𝑦

𝑑𝑥 =

5

𝑥 - 8 M1

5

𝑥 – 8 = 0

5

𝑥 = 8

x = 5

8

M1

3.

𝑑𝑦

𝑑𝑥 = 2 -

4

𝑥 M1

Stationary point: 2 - 4

𝑥 = 0

x = 2 M1

𝑑2𝑦

𝑑𝑥2 = 4x-2 M1

x = 2, 𝑑2𝑦

𝑑𝑥2 = 1

𝑑2𝑦

𝑑𝑥2 > 0, therefore minimum point M1

x = 2, y = 11 – 4ln 2 M1

Therefore (2, 11 – 4ln 2) is a minimum point

5.

x = 3, y = 10 + ln 3 M1

𝑑𝑦

𝑑𝑥=

1

𝑥 M1

When x = 3, 𝑑𝑦

𝑑𝑥 =

1

3

Gradient of normal = -3 M1

Therefore equation of line is:

y – (10 + ln 3) = -3(x – 3)

y = 19 + ln 3 – 3x

M1

6.

𝑑𝑦

𝑑𝑥= 4cos x – 2 sin x M1

7.

𝑑𝑦

𝑑𝑥= -2coscec2(2x – 3) M1

8.

𝑊ℎ𝑒𝑛 𝑥 =𝜋

6, 𝑦 = 1 M1

𝑑𝑦

𝑑𝑥 = - cosec x cot x – 2 cos x M1

𝑑𝑦

𝑑𝑥 (

𝜋

6) = − cosec (

𝜋

6) cot (

𝜋

6) – 2 cos (

𝜋

6)

𝑑𝑦

𝑑𝑥 = 3√3

M1

y – 1 = 3√3 (x - 𝜋

6) M1

6√3𝑥 + 2y – 2 - √3𝜋 = 0 M1

9.

𝑑𝑦

𝑑𝑥 = 5 sec 5x tan 5x M1

10.

𝑑

𝑑𝑥(3x) = 3x ln 3 M1

11.

𝑑

𝑑𝑥(51 – x) = 51 – x ln 5 × -1 M1

𝑑

𝑑𝑥(51 – x) = -(51 – x)ln 5

12.

𝑑

𝑑𝑥(2𝑥3

) = 2𝑥3ln 2 × 3x2 M1

𝑑

𝑑𝑥(2𝑥3

) = 3x2(2𝑥3) ln 2

Differentiation

Pt. 4: Chain Rule

1. Differentiate with respect to x, (x + 3)5 (1)

2. Find 𝑑2𝑦

𝑑𝑥2 when y = 4 ln (1 + 2x) (2)

3. Find the value of f’(x) when f(x) = 4√3𝑥 + 15 (2)

4. Find the coordinates of any stationary points for the curve y = 8x – e2x (4)

5. Find an equation for the tangent to the curve y = 2 + ln (1 + 4x) given the x-coorinate is 0. (4)

6. Find an equation for the normal to the curve y = 1

2−ln 𝑥 when x = 1. (5)

7. Find the coordinates of any stationary points on the curve y = 2 ln(x – x2). (5)

8. A curve has the equation y = √8 − 𝑒2𝑥

The point P on the curve has y-coordinate 2.

a. Find the x-coordinate of P. (2)

b. Show that the tangent to the curve at P has equation 2x + y = 2 + ln 4. (2)

9. A quantity N is increasing such that at time t seconds, N = aekt.

Given that at time t = 0, N = 20 and that at time t = 8, N = 60, find

a. The values of the constants a and k (3)

b. The value of N when t = 12 (2)

c. The rate at which N is increasing when t = 12. (3)

10. f(x) = (5 – 2x2)3

a. Find f’(x) (2)

b. Find the coordinates of the stationary points of the curve y = f(x). (4)

c. Find the equation for the tangent to the curve y = f(x) at the point with x-coordinate 3

2 giving your answer in the

form ax + by + c = 0, where a, b and c are integers. (3)

11. The diagram shows the curve y = f(x) where f(x) = 3 ln 5x − 2x, x > 0.

a. Find f ′(x). (1)

b. Find the x-coordinate of the point on the curve at which the gradient of the normal to the curve is -1

4 . (2)

c. Find the coordinates of the maximum turning point of the curve. (3)

d. Write down the set of values of x for which f(x) is a decreasing function. (1)

12. f(x) ≡ x2 − 7x + 4 ln (𝑥

2), x > 0.

a. Solve the equation f ′(x) = 0, giving your answers correct to 2 decimal places. (3)

b. Find an equation for the tangent to the curve y = f(x) at the point on the curve where x = 2. (2)

A-Level








AS Level






Mark Scheme

1.

5(x + 3)4 M1

2. 𝑑𝑦

𝑑𝑥=

4

1+2𝑥 × 2 =

8

1+2𝑥 = 8(1 + 2x)-1 M1

𝑑2𝑦

𝑑𝑥2 = -8(1 + 2x)-2 x 2

𝑑2𝑦

𝑑𝑥2 = −16

(1+2𝑥)2 M1

3.

f(x) = 4(3x + 15)0.5 M1

f’(x) = 2(3x + 15)-0.5 x 3

f’(x) = 6(3x + 15)-0.5 M1

4. 𝑑𝑦

𝑑𝑥 = 8 – 2e2x M1

8 – 2e2x = 0

2e2x = 8

e2x = 4

M1

2x = ln 4

x = 1

2ln 4 = ln 2

M1

When x = ln 2

y = 8ln 2 – 4 M1

5.

When x = 0

y = 2 + ln (1 + 0) = 2 M1

𝑑𝑦

𝑑𝑥=

1

1+4𝑥 × 4 =

4

1+4𝑥 M1


𝑑𝑥 = 4 M1

y = 4x + 2 M1

6.

When x = 1,

y = 1

2−ln 1 =

1

2

M1

𝑑𝑦

𝑑𝑥 = -(2 – ln x)-2 × -(

1

𝑥) =

1

𝑥(2−𝑙𝑛𝑥)2 M1


𝑑𝑥 =

1

1(2−ln 1)2 = -4 M1

y – 1

2= −4(𝑥 − 1) M1

y = 9

2− 4𝑥 M1

7. 𝑑𝑦

𝑑𝑥 =

2

𝑥− 𝑥2 × (1 − 2𝑥) = 2(1−2𝑥)

𝑥− 𝑥2 M1

At stationary point 𝑑𝑦

𝑑𝑥 = 0

2(1−2𝑥)

𝑥− 𝑥2 = 0

2(1 – 2x) = 0

M1

x = 1

2 M1

When x = 1

2

y = 2 ln(1

2 –

1

22).

y = -4 ln 2

M1

Stationary point: (1

2, -4 ln 2) M1

8a.

√8 − 𝑒2𝑥 = 2

8 – e2x = 4 M1

x = 1

2ln 4 = ln 2 M1

8b. 𝑑𝑦

𝑑𝑥=

1

2(8 − e2x)-0.5 x (-2e2x) M1

𝑑𝑦

𝑑𝑥 =

−𝑒2𝑥

√8− 𝑒2𝑥 M1

When x = ln 2 𝑑𝑦

𝑑𝑥 =

−𝑒2ln 2

√8− 𝑒2ln 2 = -2

M1

y – 2 = -2(x – ln 2)

2x + y = 2 + 2 ln 2 M1

2x + y = 2 + ln 22

2x + y = 2 + ln 4 M1

9a.

t = 0, N = 20

Therefore a = 20 M1

t = 8, N = 60

60 = 20e8k M1

k = 1

8ln 3 = 0.137 M1

9b.

N = 20e0.1373t M1

t = 12, N = 104 M1

9c. 𝑑𝑁

𝑑𝑡= 20 × 0.1373e0.1373t = 2.747e0.1373t M1

t = 12, 𝑑𝑁

𝑑𝑡 = 14.3 M1

N is increasing at 14.3 per second (3 s.f) M1

10a. 𝑑𝑦

𝑑𝑥 = 3(5 – 2x2)2 × (−4𝑥) M1

𝑑𝑦

𝑑𝑥 = -12x(5 – 2x2)2 M1

10b.


𝑑𝑥 = 0

-12x(5 – 2x2)2 = 0 M1

x = 0

f(0) = (5 – 2(0)2)3 = (125) M1

x = ±1

2√10

f(+1

2√10) = 0

M1

f(-1

2√10) = 0 M1

(-1

2√10, 0)

(0, 125)

(1

2√10, 0)

10c.

x = 3

2

f(3

2) = (5 – 2(

3

2)2)3 =

1

8

M1

Gradient = -18 x 1

4 = -

9

2 M1

8y – 1 = -36x + 54

36x + 8y – 55 = 0 M1

11a.

f’(x) = 3

𝑥 - 2 M1

11b.

Gradient of curve = 4 M1 3

𝑥 – 2 = 4

3

𝑥 = 6

x = 1

2

M1

11c.

Stationary point, f’(x) = 0 M1 3

𝑥 – 2 = 0

x = 3

2

M1

f(3

2) = 3 ln 5(

3

2) – 2(

3

2)

f(3

2) = 3 ln

15

2− 3

M1

(3

2, 3 ln

15

2− 3)

11d.

x ≥ 3

2 M1

12a.

f’(x) = 2x – 7 + 4

𝑥 = 0 M1

2x2 – 7x + 4 = 0

x = 7± √17

4

M1

x = 0.72

x = 2.78 M1

12b.

x = 2

Therefore, y = -10

gradient = -1

M1

y + 10 = -(x-2)

y = -x - 8 M1

Differentiation

Pt. 5: Product Rule

1. Differentiate xex with respect to x. (3)

2. Find 𝑑𝑦

𝑑𝑥 when y = x2 ln (x + 6) (3)

3. Find the coordinates of any stationary points of the curve y = x√𝑥 + 12 (4)

4. Find an equation for the tangent to the curve y = (4x – 1)ln2x (5)

5. Find an equation for the normal to the curve y = (x2 – 1)e3x (5)

6. The diagrams shows part of the curve with equation y = x𝑒𝑥2 and the tangent to the curve at the point P with x-

coordinate 1.

a. Find an equation for the tangent to the curve at P. (5)

b. Show that the area of the triangle bounded by this tangent and the coordinate axes is 2

3𝑒. (3)

7. A curve has the equation y = e−x sin x.

a. Find 𝑑𝑦

𝑑𝑥 and

𝑑2𝑦

𝑑𝑥2 (4)

b. Find the exact coordinates of the stationary points of the curve in the interval -𝜋 ≤ 𝑥 ≤ 𝜋 and determine their

nature. (8)

8. The diagram shows the curve y = f(x) in the interval 0 ≤ 𝑥 ≤ 2𝜋, where f(x) = cos x sin 2x

a. Show that f’(x) = 2 cos x(1 – 3sin2x) (2)

b. Find the x-coordinates of the stationary points of the curve in the interval 0 ≤ 𝑥 ≤ 2𝜋 (4)

c. Show that the maximum value of f(x) in the interval 0 ≤ 𝑥 ≤ 2𝜋 is 4

9√3 (4)

d. Explain why this is the maximum value of f(x) for all real values of x. (2)

A-Level








AS Level






Mark Scheme

1.

u = x 𝑑𝑢

𝑑𝑥 = 1

v = ex 𝑑𝑣

𝑑𝑥 = ex

M1

𝑑

𝑑𝑥(𝑥𝑒x) = 1 × ex + x × ex M1

𝑑

𝑑𝑥(𝑥𝑒x) = ex(1 + x) M1

2.

u = x2 𝑑𝑢

𝑑𝑥 = 2x

v = ln (x + 6) 𝑑𝑣

𝑑𝑥 =

1

𝑥+6

M1

𝑑𝑦

𝑑𝑥 = 2x x ln(x + 6) + x2 x

1

𝑥+6 M1

𝑑𝑦

𝑑𝑥 = 2x ln (x + 6) +

𝑥2

𝑥+6 M1

3.

u = x 𝑑𝑢

𝑑𝑥 = 1

v = √𝑥 + 12 = (x + 12)0.5 𝑑𝑣

𝑑𝑥 = 0.5(x + 12)-0.5

M1

𝑑𝑦

𝑑𝑥 = 1 × √𝑥 + 12 + x × 0.5(x + 12)-0.5

𝑑𝑦

𝑑𝑥 = 0.5(x + 12)-0.5 [2(x + 12) + x]

M1


𝑑𝑥 = 0,

3(𝑥 + 8)

2√𝑥 + 12= 0

x = -8

M1

When x = -8,

y(-8) = (-8)√(−8) + 12 = 16 M1

(-8, -16)

4.

x = 1

2

y(1

2) = 0

M1

u = 4x - 1 𝑑𝑢

𝑑𝑥 = 4

v = ln 2𝑥 𝑑𝑣

𝑑𝑥 =

1

𝑥

𝑑𝑦

𝑑𝑥= 4 × 𝑙𝑛2𝑥 + (4𝑥 − 1) ×

1

𝑥

M1

𝑑𝑦

𝑑𝑥 = 4 ln 2x + 4 -

1

𝑥 M1

𝑑𝑦

𝑑𝑥 (

1

2) = 4 ln 2(

1

2) + 4 -

11

2

= 2 M1

y – 0 = 2(x - 1

2)

y = 2x - 1 M1

5.

When x = 0, y = -1 M1

u = (x2 – 1) 𝑑𝑢

𝑑𝑥 = 2x

v = e3x 𝑑𝑣

𝑑𝑥 = 3e3x

𝑑𝑦

𝑑𝑥= 2𝑥 × 𝑒3𝑥 + (𝑥2 − 1) × 3𝑒3𝑥

M1

𝑑𝑦

𝑑𝑥 = 𝑒3𝑥[2x + 3(𝑥2 − 1)]

= 𝑒3𝑥(3x2 + 2x – 3) M1

𝑑𝑦

𝑑𝑥 (0) = -3

Therefore gradient of normal = 1

3

M1

y + 1 = 1

3 (x – 0)

x – 3y – 3 = 0 M1

6a.

When x = 1

y (1) = x𝑒𝑥2= (1)𝑒(1)2

= e M1

u = x 𝑑𝑢

𝑑𝑥 = 1

v = 𝑒𝑥2

𝑑𝑣

𝑑𝑥 = 2x𝑒𝑥2

𝑑𝑦

𝑑𝑥= 1 × 𝑒𝑥2

+ 𝑥 × 𝑒𝑥2 × 2𝑥

M1

𝑑𝑦

𝑑𝑥 = 𝑒𝑥2

(1 + 2𝑥2) M1

𝑑𝑦

𝑑𝑥 (1) = 𝑒(1)2

(1 + 2(1)2) = 3e M1

y – e = 3e(x – 1)

y = e(3x – 2) M1

6b.

When x = 0

y = -2e M1

When y = 0,

x = 2

3 M1

Therefore area = 1

2 × 2𝑒 ×

2

3=

2

3𝑒 M1

7a.

u = e-x

𝑑𝑢

𝑑𝑥 = -e-x

v = sin x 𝑑𝑣

𝑑𝑥 = cos x

𝑑𝑦

𝑑𝑥= -e-x × sin 𝑥 + 𝑒-x × cos x

M1

𝑑𝑦

𝑑𝑥 = e-x(cos x – sin x) M1

𝑑

𝑑𝑥 e-x(cos x – sin x)

u = e-x

𝑑𝑢

𝑑𝑥 = -e-x

v = cos x – sin x 𝑑𝑣

𝑑𝑥 = - sin x – cos x

𝑑2𝑦

𝑑𝑥2 = -e-x × (cos x – sin x) + e-x × (- sin x – cos x)

M1

𝑑2𝑦

𝑑𝑥2 = −2e-x cos x M1

7b.

At stationary point, 𝑑𝑦

𝑑𝑥 = 0

e-x(cos x – sin x) = 0 M1

cos x – sin x = 0

tan x = 1 M1

x = 𝜋

4

𝑑2𝑦

𝑑𝑥2 = −√2 𝑒−𝜋

4

𝑑2𝑦

𝑑𝑥2 < 0 , therefore maximum point

M1

M1

x = −3𝜋

4

𝑑2𝑦

𝑑𝑥2 = √2 𝑒3𝜋

4

𝑑2𝑦

𝑑𝑥2 > 0 , therefore minimum point

M1

M1

When x = 𝜋

4, y =

1

√2𝑒−

𝜋

4 , maximum point M1

When x = −3𝜋

4, y = −

1

√2𝑒

3𝜋

4 , minimum point M1

8a.

f’(x) = - sin x × sin 2x + cos x × 2cos 2x M1

f’(x) = 2cos x(1 – 2sin2x) – 2sin2x cos x

= 2 cos x(1 – 3 sin2x) M1

8b.

At stationary point f’(x) = 0

2 cos x(1 – 3 sin2x) = 0 M1

Cos x = 0

sin x = = ±1

√3

M1

x = 0.615, 𝜋

2, 2.53, 3.76,

3𝜋

2, 5.76 M1 M1

8c.

f(x) at max when sin x = 1

√3 M1

cos2x = 1 – sin2x = 1 – 1

3 =

2

3 M1

f(x) = cos x × 2 sin 𝑥 cos 𝑥 = 2 sin x cos2x

M1

Therefore maximum = 2 ×1

√3 ×

2

3 =

4

9√3 M1

8d.

Period of cos x = 2𝜋

Period of sin x = 𝜋 M1

Therefore period of f(x) = 2𝜋

Values of f(x) in this interval are repeated. M1

Differentiation

Pt. 6: Quotient Rule

1. Differentiate 𝑒𝑥

𝑥−4 with respect to xand simplify your answer (2)

2. Find 𝑑𝑦

𝑑𝑥 of y =

ln(3𝑥−1)

𝑥+2 (2)

3. Find the co-ordinates of any stationary point on the curve y = 𝑥+5

√2𝑥 +1 (4)

4. Find the equation of the tangent to the curve y = 3𝑥+4

𝑥2+1 at the point when x = -1 (5)

5. The curve C has the equation y = 3+sin 2𝑥

2+cos 2𝑥

a. Show that 𝑑𝑦

𝑑𝑥=

6 sin 2𝑥+4 cos 2𝑥+2

(2+cos 2𝑥)2 (4)

b. Find an equaton of the tangent to C at the point where x = 𝜋

2. Write your answer in the form y = ax+ b, where a

and b are exact constants. (4)

6. f(x) = 𝑥−5

2𝑥+3+

2𝑥+4

2𝑥2+7𝑥+6

a. Express f(x) as a fraction in its simplest form (2)

b. Hence find f’(x) in its simplest form (3)

7. Differentiate cos 2𝑥

√𝑥 with respect to x (3)

8. Given that y = 3 sin 𝑥

2 sin 𝑥+2 cos 𝑥, show that

𝑑𝑦

𝑑𝑥=

𝐴

1+sin 2𝑥 where A is a rational constant to be found. The value of xlies

in the domain −𝜋

4 < 𝑥 <

3𝜋

4 (5)

9. Find equation of the normal to the curve y = −√𝑥

5 cos 5𝑥 at the x-cordiante 16. (7)

A-Level








AS Level






Mark Scheme

1. 𝑑

𝑑𝑥(

𝑒𝑥

𝑥−4 )

= 𝑒𝑥 ×(𝑥−4)− 𝑒𝑥 ×1

(𝑥−4)2 M1

= 𝑒𝑥 (𝑥−5)

(𝑥−4)2 M1

2. 𝑑

𝑑𝑥(

ln(3𝑥−1)

𝑥+2) )

=

3

3𝑥−1×(𝑥+2)−ln (3𝑥−1)×1

(𝑥+2)2 M1

= 3

(3𝑥−1)(𝑥+2)−

ln (3𝑥−1)

(𝑥+2)2 M1

3.

𝑑𝑦

𝑑𝑥=

1×√2𝑥+1−(𝑥+5)×1

2(2𝑥+1)

12×2

2𝑥+1 M1

𝑑𝑦

𝑑𝑥=

(2𝑥+1)−(𝑥+5)

(2𝑥+1)32

=𝑥−4

(2𝑥+1)32

M1

At stationary points, 𝑑𝑦

𝑑𝑥 = 0

𝑥−4

(2𝑥+1)32

= 0

x– 4 = 0

x= 4

M1

When x= 4

y(4) = 4+5

√2(4) +1 = 3 M1

Stationary point = (4, 3)

4. 𝑑𝑦

𝑑𝑥=

3 ×(𝑥2+1)−(3𝑥+4)×2𝑥

(𝑥2+1)2 M1

𝑑𝑦

𝑑𝑥 =

3−8𝑥−3𝑥2

(𝑥2+1)2 M1

𝑑𝑦

𝑑𝑥(−1) = 2 M1

When x= -1

y(-1) = 1

2

M1

y – 1

2= 2(𝑥 + 1) =

y = 2x+ 5

2

M1

5a. 𝑑𝑦

𝑑𝑥=

(2+cos 2𝑥)(2 cos 2𝑥)−(3+sin 2𝑥)(−2𝑠𝑖𝑛2𝑥)

(2+𝑐𝑜𝑠2𝑥)2 M1

𝑑𝑦

𝑑𝑥 =

4 cos 2𝑥 + 2𝑐𝑜𝑠22𝑥 + 6𝑠𝑖𝑛2𝑥 + 2𝑠𝑖𝑛22𝑥

(2+𝑐𝑜𝑠2𝑥)2 M1

𝑑𝑦

𝑑𝑥=

6 sin 2𝑥 + 4 cos 2𝑥 + 2(𝑐𝑜𝑠22𝑥 + 𝑠𝑖𝑛22𝑥)

(2+𝑐𝑜𝑠2𝑥)2 M1

𝑑𝑦

𝑑𝑥 =

6 sin 2𝑥 + 4 cos 2𝑥 + 2

(2+𝑐𝑜𝑠2𝑥)2 M1

5b.

When x= 𝜋

2

𝑑𝑦

𝑑𝑥 =

6 sin 2(𝜋

2) + 4 cos 2(

𝜋

2) + 2

(2+𝑐𝑜𝑠2(𝜋

2))2

= -2 M1

When x= 𝜋

2

y(𝜋

2) =

3+sin 2(𝜋

2)

2+cos 2(𝜋

2) = 3

M1

y – 3 = -2(x- 𝜋

2) M1

y = -2x+ 𝜋 + 3 M1

6a.

f(x) = 𝑥−5

2𝑥+3+

2(𝑥+2)

(2𝑥+3)(𝑥+2) M1

= 𝑥−5

2𝑥+3+

2

2𝑥+3

= 𝑥−3

2𝑥+3

M1

6b.

f’(x) = 2𝑥+3−2(𝑥−3)

(2𝑥+3)2 M1

M1

f’(x) = 2𝑥+3−2𝑥+6

(2𝑥+3)2

f’(x) = 9

(2𝑥+3)2 M1

7.

u = cos 2x 𝑑𝑢

𝑑𝑥= −2 sin 2𝑥

v = 𝑥1

2 𝑑𝑣

𝑑𝑥 =

1

2𝑥−

1

2

M1

𝑑

𝑑𝑥(

𝑢

𝑣) =

𝑥12(−2 sin 2𝑥)−𝑐𝑜𝑠2𝑥(

1

2𝑥

−12)

(𝑥12)2

M1

𝑑

𝑑𝑥(

𝑢

𝑣) =

−2𝑥12 sin 2𝑥 −

1

2𝑥

−12 cos 2𝑥

𝑥 M1

8.

u = 3 sin x 𝑑𝑢

𝑑𝑥= 3 cos 𝑥

v = 2 sin x+ 2 cos x 𝑑𝑣

𝑑𝑥 = 2 cos x– 2 sin x

M1

𝑑

𝑑𝑥(

𝑢

𝑣) =

3 cos 𝑥(2 sin 𝑥+2 cos 𝑥)−(2 cos 𝑥−2 sin 𝑥)(3 sin 𝑥)

(2 sin 𝑥+2 cos 𝑥)2 M1

= 6

4 ×

cos 𝑥 sin 𝑥 + 𝑐𝑜𝑠2𝑥−cos 𝑥 sin 𝑥+ 𝑠𝑖𝑛2𝑥

𝑠𝑖𝑛2𝑥+2 sin 𝑥 cos 𝑥+ 𝑐𝑜𝑠2𝑥

USING:

(𝑠𝑖𝑛2𝑥 + 𝑐𝑜𝑠2𝑥 = 1)

(sin 2x= 2 sin xcos x)

M1

M1

𝑑

𝑑𝑥(

𝑢

𝑣) =

3

2

1+sin 2𝑥 M1

9.

u = -√𝑥 = -x0.5

𝑑𝑢

𝑑𝑥= −0.5 𝑥-0.5

v = 5cos 5x 𝑑𝑣

𝑑𝑥 = -5 sin 5x× 5 = -25 sin 5x

M1

𝑑

𝑑𝑥(

𝑢

𝑣) =

(5 cos 5𝑥) (−0.5𝑥−0.5)−(−𝑥0.5)(−25 sin 5𝑥)

(5 cos 5𝑥)2 M1

At the point x= 1 , 𝑑𝑦

𝑑𝑥=

(5 cos 5) (−0.5−0.5)−(−10.5)(−25 sin 5)

(5 cos 5)2

𝑑𝑦

𝑑𝑥 = 10.9…

M1

M1

Gradient of normal = -0.1 M1

When x= 1, y = −√1

5 cos 5 = -0.71 M1

y = mx+ c

-0.71 = (-0.1)(1) + c

c = -0.61

M1

y = -0.1x– 0.61

Differentiation


1. A curve is given by the parametric equations x = t2 + 1 and y = 4

𝑡

a. Write down the coorindates of the point on the curve when t = 2 (2)

b. Find the value of t at the point on the curve with coordinates (5

4, −8) (1)

2. Find the cartesian equation for the curve, given its parametric equations are x = 3t and y = t2 (2)

3. Find the cartesian equations for the curve, given its parametric equations are x = 2t – 1 and y = 2

𝑡2 (2)

4. Write down the parametic equation for a circle with centre (6, -1) and radius 2. (2)

5. A curve is given by the parametric equations x = 2 + t and y = t2 – 1

a. Write down expression for 𝑑𝑥

𝑑𝑡 and

𝑑𝑦

𝑑𝑡 (1)

b. Hence, show that 𝑑𝑦

𝑑𝑥= 2𝑡 (1)

6. Find and simplify an expression for 𝑑𝑦

𝑑𝑥 in terms of the parameter t when x = 3t – 1 and y = 2 -

1

𝑡 (3)

7. Find and simplify an expression for 𝑑𝑦

𝑑𝑥 in terms of the parameter t when x = sin2t and y = cos3t (3)

8. A curve has the parametric equations x = t3 and y = 3t2. Find, in the form y = mx + c, an equation for the tangent

to the curve at the point with the given value of the parameter t. (4)

9. Show that the normal to the curve with parametric equations, x = sec 𝜃, y = tan 𝜃, 0 ≤ 𝜃 < 𝜋

2 at the point where

𝜃 = 𝜋

3, has the equation √3𝑥 + 4𝑦 = 10√3 (6)

10. A curve has parametric equations x = sin 2t and y = sin2t, 0 < t < 𝜋


𝑑𝑥 =

1

2tan 2𝑡 (2)

b. Find an equation for the tangent to the curve at the point where t = 𝜋

6 (3)

11. A curve is given by the parametric equations

x = t 2, y = t(t − 2), t ≥ 0.

a. Find the coordinates of any points where the curve meets the coordinate axes. (3)

b. Find 𝑑𝑦

𝑑𝑥 in terms of x by first finding

𝑑𝑦

𝑑𝑥 in terms of t. (3)

c. Find 𝑑𝑦

𝑑𝑥 in terms of x by first finding a cartesian equation for the curve. (2)

12. A curve has parametric equations x = sin2t and y = tan t, −𝜋

2 < t <

𝜋

2

a. Show that the tangent to the curve at the point where t = 𝜋

4 passes through the origin. (6)

b. Find a cartesian equation for the curve in the form y2 = f(x). (2)

A-Level








AS Level






Mark Scheme

1a.

x(2) = 22 + 1 = 5 M1

y(2) = 4

2 = 2 M1

1b. 4

𝑡= −8

t = -1

2

M1

2.

x = 3t → t = 𝑥

3 M1

y = (𝑥

3)2 =

𝑥2

9 M1

3.

t = 1

2(x + 1) M1

y = 2

1

4(𝑥+1)2

=8

(𝑥+1)2 M1

4.

x = 6 + 2 cos x M1

y = -1 + 2 sin x M1

0 ≤ x < 2𝜋

5a. 𝑑𝑥

𝑑𝑡= 1

𝑑𝑦

𝑑𝑡= 2𝑡

M1

5b. 𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡 ÷

𝑑𝑥

𝑑𝑡

𝑑𝑦

𝑑𝑥 = 2t ÷ 1 = 2t

M1

6.

𝑑𝑥

𝑑𝑡= 3

𝑑𝑦

𝑑𝑡= 𝑡−2

M1

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡 ÷

𝑑𝑥

𝑑𝑡

𝑑𝑦

𝑑𝑥 = t-2 ÷ 3 =

𝑡−2

3

M1

𝑑𝑦

𝑑𝑥 =

1

3𝑡2 M1

7.

𝑑𝑥

𝑑𝑡= 2 sin 𝑡 × cos 𝑡

𝑑𝑦

𝑑𝑡= 3cos2t × (- sin t)

M1

𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑡 ÷

𝑑𝑥

𝑑𝑡

= −3𝑐𝑜𝑠2𝑡 sin 𝑡

2 sin 𝑡 cos 𝑡

M1

𝑑𝑦

𝑑𝑥 = -

3

2cos 𝑡 M1

8.

when t = 1

x(1) = 1

y(1) = 3

M1

𝑑𝑥

𝑑𝑡= 3t2

𝑑𝑦

𝑑𝑡= 6t

M1

𝑑𝑦

𝑑𝑥 =

6𝑡

3𝑡2 𝑑𝑦

𝑑𝑥 (1) = 2

M1

y – 3 = 2(x – 1)

y = 2x + 1 M1

9.

𝜃 = 𝜋

3

x(𝜋

3) = 2

y(𝜋

3) = 2√3

M1

𝑑𝑥

𝑑𝜃 = sec 𝜃 tan 𝜃

𝑑𝑦

𝑑𝜃= 2sec2 𝜃

M1

𝑑𝑦

𝑑𝑥 =

2𝑠𝑒𝑐2𝜃

sec 𝜃 tan 𝜃

𝑑𝑦

𝑑𝑥 =

2 sec 𝜃

tan 𝜃= 2 𝑐𝑜𝑠𝑒𝑐 𝜃

M1

Gradient = 4

√3

Therefore gradient of normal = -√3

4

M1

y - 2√3 = -√3

4(x – 2)

4y - 8√3 = -√3x + 2√3 M1

√3x + 4y = 10√3 M1

10a. 𝑑𝑥

𝑑𝑡= 2 cos 2𝑡

𝑑𝑦

𝑑𝑡= 2 sin 𝑡 × cos 𝑡 = sin 2𝑡

M1

𝑑𝑦

𝑑𝑥=

sin 2𝑡

2 cos 2𝑡=

1

2tan 2𝑡 M1

10b.

t = 𝜋

6

x(𝜋

6) =

√3

2

y(𝜋

6) =

1

4

M1

Gradient = √3

2

y - 1

4 =

√3

2(x -

√3

2)

M1

√3 x – 2y – 1 = 0 M1

11a.

x = 0

t2 = 0

t = 0

M1

y = 0

t(t – 2) = 0

t = 0 or t = 2

M1

Co-ordinates: (0, 0) and (4, 0) M1

11b.

𝑑𝑥

𝑑𝑡 = 2t

𝑑𝑦

𝑑𝑡= 2𝑡 − 2

M1

𝑑𝑦

𝑑𝑥=

2𝑡−2

2𝑡= 1 − t -1 M1

t = x0.5 𝑑𝑦

𝑑𝑥 = 1 – x-0.5 M1

11c.

y = 𝑥1

2(𝑥1

2 – 2) = x - 2𝑥1

2 M1

𝑑𝑦

𝑑𝑥= 1 − 𝑥−

1

2 M1

12a.

t = 𝜋

4

x(𝜋

4) =

1

2

y(𝜋

4) = 1

M1

𝑑𝑥

𝑑𝑡 = 2 sin t cos t

𝑑𝑦

𝑑𝑡= sec2 t

M1

𝑑𝑦

𝑑𝑥=

𝑠𝑒𝑐2𝑡

2 sin 𝑡 cos 𝑡=

1

2 𝑠𝑒𝑐3𝑡 𝑐𝑜𝑠𝑒𝑐 𝑡 M1

Gradient = 1

2× (√2)3 × √2 = 2 M1

y – 1 = 2(x – 1

2)

y = 2x M1

When x = 0, y = 0 therefore passes through origin M1

12b.

y2 = tan2t = 𝑠𝑖𝑛2𝑡

𝑐𝑜𝑠2𝑡 =

𝑠𝑖𝑛2𝑡

1− 𝑠𝑖𝑛2𝑡 M1

y2 = 𝑥

1−𝑥 M1

Differentiation


1. Differentiate with respect to x, y3 (1)

2. Find 𝑑𝑦

𝑑𝑥 in terms of x and y for the curve x2 + y2 + 3x – 4y = 9 (2)

3. Find 𝑑𝑦

𝑑𝑥 in terms of x and y for the curve 2e3x + e-2y + 7 = 0 (3)

4. Find 𝑑𝑦

𝑑𝑥 in terms of x and y for the curve 2xy2 – x3y = 0 (4)

5. Find 𝑑𝑦

𝑑𝑥 in terms of x and y for the curve cos 2x sec 3y + 1 = 0 (3)

6. A curve has the equation x2 + 4xy − 3y2 = 36.

a. Find an equation for the tangent to the curve at the point P (4, 2). (5)

Given that the tangent to the curve at the point Q on the curve is parallel to the tangent at P,

b. Find the coordinates of Q. (4)

7. A curve has the equation x2 – 4xy + y2 = 24.


𝑑𝑥=

𝑥−2𝑦

2𝑥−𝑦 (3)

b. Find an equation for the tangent to the curve at the point P(2, 10) (2)

The The tangent to the curve at Q is prallallel to the tangent at P

c. Find the coordinates of Q. (5)

A-Level








AS Level






Mark Scheme

1. 𝑑

𝑑𝑥(𝑦3) = 3y2 𝑑𝑦

𝑑𝑥 M1

2. 𝑑

𝑑𝑥(x2 + y2 + 3x – 4y = 9) → 2x + 2y

𝑑𝑦

𝑑𝑥 + 3 - 4

𝑑𝑦

𝑑𝑥 = 0 M1

𝑑𝑦

𝑑𝑥 (4 – 2y) = 2x + 3

𝑑𝑦

𝑑𝑥 =

2𝑥+3

4−2𝑦

M1

3. 𝑑

𝑑𝑥(2e3x + e-2y + 7 = 0) → 6e3x – 2e-2y

𝑑𝑦

𝑑𝑥 = 0 M1

6e3x = 2e-2y 𝑑𝑦

𝑑𝑥 M1

𝑑𝑦

𝑑𝑥 =

3𝑒3𝑥

𝑒−2𝑦 = 3e3x + 2y M1

4. 𝑑

𝑑𝑥(2xy2) → product rule

u = 2x

u’ = 2

v = y2

v’ = 2y 𝑑𝑦

𝑑𝑥

𝑑

𝑑𝑥(2xy2) → (2 × y2) + (2x × 2y

𝑑𝑦

𝑑𝑥)

M1

𝑑

𝑑𝑥(𝑥3y) → product rule

u = x3

u’ = 3x2

v = y

v’ = 𝑑𝑦

𝑑𝑥

𝑑

𝑑𝑥(𝑥3y) → (3x2 × y) + (x3 ×

𝑑𝑦

𝑑𝑥)

M1

𝑑

𝑑𝑥(2xy2 – x3y) = (2 × y2) + (2x × 2y

𝑑𝑦

𝑑𝑥) – ((3x2 × y) + (x3 ×

𝑑𝑦

𝑑𝑥))

2y2 – 3x2y = 𝑑𝑦

𝑑𝑥(x3 – 4xy)

M1

𝑑𝑦

𝑑𝑥=

2𝑦2−3𝑥2𝑦

𝑥3−4𝑥𝑦 M1

5. 𝑑

𝑑𝑥(cos 2x sec 3y + 1) →

𝑑

𝑑𝑥(cos 2x sec 3y) + 0 → Product Rule

u = cos 2x

u’ = -2sin 2x

v = sec 3y

v’ = 3𝑑𝑦

𝑑𝑥 sec 3y tan 3y

𝑑

𝑑𝑥(cos 2x sec 3y) → (-2 sin 2x × sec 3y + cos 2x × 3

𝑑𝑦

𝑑𝑥 sec 3y tan 3y)

M1

3𝑑𝑦

𝑑𝑥 cos 2x sec 3y tan 3y = 2sin 2x sec 3y M1

𝑑𝑦

𝑑𝑥 =

2 sin 2𝑥

3 cos 2𝑥 tan 3𝑦=

2

3tan 2𝑥 cot 3𝑦 M1

6a. 𝑑

𝑑𝑥 (x2 + 4xy − 3y2 = 36)

2x + 4 × y + 4x × 𝑑𝑦

𝑑𝑥 – 6y

𝑑𝑦

𝑑𝑥 = 0

M1

2x + 4y = 𝑑𝑦

𝑑𝑥(6𝑦 − 4𝑥) M1

𝑑𝑦

𝑑𝑥=

𝑥+2𝑦

3𝑦−2𝑥 M1

𝑑𝑦

𝑑𝑥(4, 2) = −4 M1

y – 2 = -4(x – 4)

y = 18 – 4x M1

6b.

At Q, 𝑥+2𝑦

3𝑦−2𝑥 = -4

x + 2y = -4(3y – 2x)

x = 2y

M1

Sub into equation of the curve:

(2y)2 + 4y(2y) – 3y2 = 36 M1

y2 = 4

y = 2 or -2 M1

Therefore Q: (-4, -2) M1

7a. 𝑑

𝑑𝑥 (x2 – 4xy + y2 = 24) → 2x – 4 × y – 4x ×

𝑑𝑦

𝑑𝑥 + 2y

𝑑𝑦

𝑑𝑥 = 0 M1

2x – 4y = 𝑑𝑦

𝑑𝑥(4x – 2y) M1

𝑑𝑦

𝑑𝑥 =

2𝑥−4𝑦

4𝑥−2𝑦=

𝑥 − 2𝑦

2𝑥 − 𝑦 M1

7b.

Gradient = 3 M1

y – 10 = 3(x – 2)

y = 3x + 4 M1

7c. 𝑥 − 2𝑦

2𝑥 − 𝑦= 3 M1

x – 2y = 3(2x – y)

y = 5x M1

x2 – (4x)(5x) + (5x)2 = 24 M1

x2 = 4

x = or x = -2 M1

Therefore, (-2, -10) M1

Differentiation


1. A biological culture is growing exponentally such that the number of bacteria present, N, at the time t minutes is

given by N = 8000(1.04)t. Find the rate at which the number of bacteria is increasing when there are 4000 bacteria

present. (3)

2. The radius of a circle is increasing at a constant rate of 0.2 cm s−1.

a. Show that the perimeter of the circle is increasing at the rate of 0.4π cm s−1. (3)

b. Find the rate at which the area of the circle is increasing when the radius is 10 cm. (3)

c. Find the radius of the circle when its area is increasing at the rate of 20 cm2 s−1. (2)

3. 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 the rate of 2 mms−1 (4)

4. The diagram shows the cross-section of a right-circular paper cone being used as a filter funnel. The volume of

liquid in the funnel is V cm3 when the depth of the liquid is h cm. Given that the angle between the sides of the

funnel in the cross-section is 60° as shown.

a. Show that V = 1

9𝜋h3 (2)

Given also that at time t seconds after liquid is put in the funnel, V = 600e−0.0005t

b. Show that after two minutes, the depth of liquid in the funnel is approximately 11.7 cm (2)

c. Find the rate at which the depth of liquid is decreasing after two minutes. (6)

5. The volume, V m3, of liquid in a container is given by: V = (3h2 + 4)3

2 – 8 where h m is the depth of the liquid.

a. Find the value of 𝑑𝑉

𝑑ℎ when h = 0.6, giving your answer correct to 2 decimal places. (4)

b. Liquid is leaking from the container. It is oberseved that, when the depth of the liquid is 0.6m, the depth is

decreasing at a rate of 0.015m per hour. Find the rate at which the volume in the containiner is decreasing at the

instant when the depth is 0.6m (3)

A-Level








AS Level






Mark Scheme

1. 𝑑𝑁

𝑑𝑡= 800(1.04)t x ln 1.04 M1

N = 4000

4000 = 800(1.04)t

(1.04)t = 5

M1

𝑑𝑁

𝑑𝑡= 800 × 5 × ln 1.04 = 157

Therefore, growing at a rate of 157 per minute. M1

2a. 𝑑𝑃

𝑑𝑡=

𝑑𝑃

𝑑𝑟×

𝑑𝑟

𝑑𝑡

𝑑𝑟

𝑑𝑡= 0.2

M1

P = 2𝜋𝑟 𝑑𝑃

𝑑𝑟= 2𝜋

M1

𝑑𝑃

𝑑𝑡= 2𝜋 × 0.2 = 0.4𝜋

Therefore perimeter is increasing at 0.4𝜋 cm s-1 M1

2b. 𝑑𝐴

𝑑𝑡=

𝑑𝐴

𝑑𝑟×

𝑑𝑟

𝑑𝑡

A = 𝜋r2 𝑑𝐴

𝑑𝑟= 2𝜋𝑟

M1

When r = 10, 𝑑𝐴

𝑑𝑟= 20𝜋 M1

𝑑𝐴

𝑑𝑡= 20𝜋 × 0.2 = 4𝜋 cm2 s-1 M1

2b.

20 = 2𝜋𝑟 × 0.2 M1

r = 50

𝜋 = 15.9 cm M1

3a. 𝑑𝑉

𝑑𝑡=

𝑑𝑉

𝑑𝑙 ×

𝑑𝑙

𝑑𝑡

𝑑𝑉

𝑑𝑡= 3.5, V = l3

𝑑𝑉

𝑑𝑙= 3l2

M1

200 = l3

l = 5.848 M1

𝑑𝑉

𝑑𝑙= 3 × 5.8482 = 102.6 M1

3.5 = 102.6 x 𝑑𝑙

𝑑𝑡 M1

𝑑𝑙

𝑑𝑡= 3.5 ÷ 102.6 = 0.0341 𝑐𝑚s-1 M1

3b.

2 mm s-1 = 0.2 cm s-1 M1

3.5 = 𝑑𝑉

𝑑𝑙 × 0.2

𝑑𝑉

𝑑𝑙 = 17.5

M1

17.5 = 3l2

l = 2.415 M1

V = l3 = 14.1 cm3 M1

4a.

tan 30 = 1

√3 =

𝑟

ℎ

r = ℎ

√3

M1

V = 1

3𝜋r2h =

1

3𝜋 ×

1

3ℎ2 × ℎ =

1

9𝜋ℎ3 M1

4b.

t = 120, V = 565.06

h = √9 ×565.06

𝜋

3

M1

h = 11.7cm M1

4c. h = 11.74 𝑑𝑉

𝑑ℎ= 144.4

M1

𝑑𝑉

𝑑𝑡=

𝑑𝑉

𝑑ℎ×

𝑑ℎ

𝑑𝑡

𝑑𝑉

𝑑ℎ=

1

3𝜋h2

M1

𝑑𝑉

𝑑𝑡= 600 × (−0.0005)e-0.0005t

= -0.3e-0.0005t M1

t = 120 𝑑𝑉

𝑑𝑡= −0.2825

M1

0.2825 = 144.4 x 𝑑ℎ

𝑑𝑡

𝑑ℎ

𝑑𝑡= −0.00196

M1

The depth is decreasing at 0.00196 cm s-1 M1

5a. 𝑑𝑉

𝑑ℎ=

3

2(3ℎ2 + 4)

1

2(6ℎ) = 3(3ℎ2 + 4)1

2(3ℎ) M1

= 9h(3h2 + 4)1

2 M1

When h = 0.6, 𝑑𝑉

𝑑ℎ= 9(0.6)(3(0.62) + 4)

12 M1

𝑑𝑉

𝑑ℎ= 12.17 M1

5b. 𝑑𝑉

𝑑𝑡=

𝑑𝑉

𝑑ℎ×

𝑑ℎ

𝑑𝑡 M1

When h = 0.6, 𝑑𝑉

𝑑𝑡 = (12.170) x (-0.015) M1

𝑑𝑉

𝑑𝑡 = -0.1825

The volume is decreasing at a rate of 0.18m3 per hour. M1

AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find 𝑑 𝑑 when y = x2 ln (x

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