as level -level differentiation pt. 3: general differentiation...differentiation pt. 5: product rule...
TRANSCRIPT
![Page 1: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/1.jpg)
Differentiation
Pt. 3: General Differentiation
1. Find ð2ðŠ
ðð¥2 for y = 4x3 + ex (2)
2. Find ððŠ
ðð¥ for y =
3
ð¥+ 3 ln ð¥ (2)
3. Find any values of x for which ððŠ
ðð¥ = 0 when y = 5ln x â 8x (2)
4. Find the co-ordinates and the nature of any stationary points on the curve y = 7 + 2x â 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 â 3) (1)
8. Find an equation for the tangent for the curve y = cosec x â 2sin x at the point x = ð
6 in the interval 0 †x †2ð (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, 2ð¥3
A-Level
Pt. 3: General Differentiation
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
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
![Page 2: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/2.jpg)
Mark Scheme
1.
ððŠ
ðð¥= 12x2 + ex M1
ð2ðŠ
ðð¥2 = 24x + ex M1
2.
y = 3
ð¥+ 3 ln ð¥ = 3x-1 + 3 ln x M1
ððŠ
ðð¥ = -3x-2 +
3
ð¥ M1
3.
ððŠ
ðð¥ =
5
ð¥ - 8 M1
5
ð¥ â 8 = 0
5
ð¥ = 8
x = 5
8
M1
3.
ððŠ
ðð¥ = 2 -
4
ð¥ M1
Stationary point: 2 - 4
ð¥ = 0
x = 2 M1
ð2ðŠ
ðð¥2 = 4x-2 M1
x = 2, ð2ðŠ
ðð¥2 = 1
ð2ðŠ
ðð¥2 > 0, therefore minimum point M1
x = 2, y = 11 â 4ln 2 M1
Therefore (2, 11 â 4ln 2) is a minimum point
5.
x = 3, y = 10 + ln 3 M1
ððŠ
ðð¥=
1
ð¥ M1
When x = 3, ððŠ
ðð¥ =
1
3
Gradient of normal = -3 M1
Therefore equation of line is:
y â (10 + ln 3) = -3(x â 3)
y = 19 + ln 3 â 3x
M1
![Page 3: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/3.jpg)
6.
ððŠ
ðð¥= 4cos x â 2 sin x M1
7.
ððŠ
ðð¥= -2coscec2(2x â 3) M1
8.
ðâðð ð¥ =ð
6, ðŠ = 1 M1
ððŠ
ðð¥ = - cosec x cot x â 2 cos x M1
ððŠ
ðð¥ (
ð
6) = â cosec (
ð
6) cot (
ð
6) â 2 cos (
ð
6)
ððŠ
ðð¥ = 3â3
M1
y â 1 = 3â3 (x - ð
6) M1
6â3ð¥ + 2y â 2 - â3ð = 0 M1
9.
ððŠ
ðð¥ = 5 sec 5x tan 5x M1
10.
ð
ðð¥(3x) = 3x ln 3 M1
11.
ð
ðð¥(51 â x) = 51 â x ln 5 à -1 M1
ð
ðð¥(51 â x) = -(51 â x)ln 5
12.
ð
ðð¥(2ð¥3
) = 2ð¥3ln 2 à 3x2 M1
ð
ðð¥(2ð¥3
) = 3x2(2ð¥3) ln 2
![Page 4: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/4.jpg)
![Page 5: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/5.jpg)
Differentiation
Pt. 4: Chain Rule
1. Differentiate with respect to x, (x + 3)5 (1)
2. Find ð2ðŠ
ðð¥2 when y = 4 ln (1 + 2x) (2)
3. Find the value of fâ(x) when f(x) = 4â3ð¥ + 15 (2)
4. Find the coordinates of any stationary points for the curve y = 8x â 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
2âln ð¥ when x = 1. (5)
7. Find the coordinates of any stationary points on the curve y = 2 ln(x â x2). (5)
8. A curve has the equation y = â8 â ð2ð¥
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 â 2x2)3
a. Find fâ(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 â 2x, x > 0.
a. Find f â²(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) â¡ x2 â 7x + 4 ln (ð¥
2), x > 0.
a. Solve the equation f â²(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
Pt. 3: General Differentiation
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
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
![Page 6: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/6.jpg)
Mark Scheme
1.
5(x + 3)4 M1
2. ððŠ
ðð¥=
4
1+2ð¥ à 2 =
8
1+2ð¥ = 8(1 + 2x)-1 M1
ð2ðŠ
ðð¥2 = -8(1 + 2x)-2 x 2
ð2ðŠ
ðð¥2 = â16
(1+2ð¥)2 M1
3.
f(x) = 4(3x + 15)0.5 M1
fâ(x) = 2(3x + 15)-0.5 x 3
fâ(x) = 6(3x + 15)-0.5 M1
4. ððŠ
ðð¥ = 8 â 2e2x M1
8 â 2e2x = 0
2e2x = 8
e2x = 4
M1
2x = ln 4
x = 1
2ln 4 = ln 2
M1
When x = ln 2
y = 8ln 2 â 4 M1
5.
When x = 0
y = 2 + ln (1 + 0) = 2 M1
ððŠ
ðð¥=
1
1+4ð¥ à 4 =
4
1+4ð¥ M1
When x = 0, ððŠ
ðð¥ = 4 M1
y = 4x + 2 M1
6.
When x = 1,
y = 1
2âln 1 =
1
2
M1
ððŠ
ðð¥ = -(2 â ln x)-2 à -(
1
ð¥) =
1
ð¥(2âððð¥)2 M1
When x = 1, ððŠ
ðð¥ =
1
1(2âln 1)2 = -4 M1
y â 1
2= â4(ð¥ â 1) M1
y = 9
2â 4ð¥ M1
7. ððŠ
ðð¥ =
2
ð¥â ð¥2 à (1 â 2ð¥) = 2(1â2ð¥)
ð¥â ð¥2 M1
At stationary point ððŠ
ðð¥ = 0
2(1â2ð¥)
ð¥â ð¥2 = 0
2(1 â 2x) = 0
M1
x = 1
2 M1
![Page 7: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/7.jpg)
When x = 1
2
y = 2 ln(1
2 â
1
22).
y = -4 ln 2
M1
Stationary point: (1
2, -4 ln 2) M1
8a.
â8 â ð2ð¥ = 2
8 â e2x = 4 M1
x = 1
2ln 4 = ln 2 M1
8b. ððŠ
ðð¥=
1
2(8 â e2x)-0.5 x (-2e2x) M1
ððŠ
ðð¥ =
âð2ð¥
â8â ð2ð¥ M1
When x = ln 2 ððŠ
ðð¥ =
âð2ln 2
â8â ð2ln 2 = -2
M1
y â 2 = -2(x â 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. ðð
ðð¡= 20 Ã 0.1373e0.1373t = 2.747e0.1373t M1
t = 12, ðð
ðð¡ = 14.3 M1
N is increasing at 14.3 per second (3 s.f) M1
10a. ððŠ
ðð¥ = 3(5 â 2x2)2 à (â4ð¥) M1
ððŠ
ðð¥ = -12x(5 â 2x2)2 M1
10b.
At stationary point ððŠ
ðð¥ = 0
-12x(5 â 2x2)2 = 0 M1
x = 0
f(0) = (5 â 2(0)2)3 = (125) M1
x = ±1
2â10
f(+1
2â10) = 0
M1
![Page 8: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/8.jpg)
f(-1
2â10) = 0 M1
(-1
2â10, 0)
(0, 125)
(1
2â10, 0)
10c.
x = 3
2
f(3
2) = (5 â 2(
3
2)2)3 =
1
8
M1
Gradient = -18 x 1
4 = -
9
2 M1
8y â 1 = -36x + 54
36x + 8y â 55 = 0 M1
11a.
fâ(x) = 3
ð¥ - 2 M1
11b.
Gradient of curve = 4 M1 3
ð¥ â 2 = 4
3
ð¥ = 6
x = 1
2
M1
11c.
Stationary point, fâ(x) = 0 M1 3
ð¥ â 2 = 0
x = 3
2
M1
f(3
2) = 3 ln 5(
3
2) â 2(
3
2)
f(3
2) = 3 ln
15
2â 3
M1
(3
2, 3 ln
15
2â 3)
11d.
x ⥠3
2 M1
12a.
fâ(x) = 2x â 7 + 4
ð¥ = 0 M1
2x2 â 7x + 4 = 0
x = 7± â17
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
![Page 9: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/9.jpg)
Differentiation
Pt. 5: Product Rule
1. Differentiate xex with respect to x. (3)
2. Find ððŠ
ðð¥ when y = x2 ln (x + 6) (3)
3. Find the coordinates of any stationary points of the curve y = xâð¥ + 12 (4)
4. Find an equation for the tangent to the curve y = (4x â 1)ln2x (5)
5. Find an equation for the normal to the curve y = (x2 â 1)e3x (5)
6. The diagrams shows part of the curve with equation y = xðð¥2 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
3ð. (3)
7. A curve has the equation y = eâx sin x.
a. Find ððŠ
ðð¥ and
ð2ðŠ
ðð¥2 (4)
b. Find the exact coordinates of the stationary points of the curve in the interval -ð †ð¥ †ð and determine their
nature. (8)
8. The diagram shows the curve y = f(x) in the interval 0 †ð¥ †2ð, where f(x) = cos x sin 2x
a. Show that fâ(x) = 2 cos x(1 â 3sin2x) (2)
b. Find the x-coordinates of the stationary points of the curve in the interval 0 †ð¥ †2ð (4)
c. Show that the maximum value of f(x) in the interval 0 †ð¥ †2ð is 4
9â3 (4)
d. Explain why this is the maximum value of f(x) for all real values of x. (2)
A-Level
Pt. 3: General Differentiation
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
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
![Page 10: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/10.jpg)
Mark Scheme
1.
u = x ðð¢
ðð¥ = 1
v = ex ðð£
ðð¥ = ex
M1
ð
ðð¥(ð¥ðx) = 1 à ex + x à ex M1
ð
ðð¥(ð¥ðx) = ex(1 + x) M1
2.
u = x2 ðð¢
ðð¥ = 2x
v = ln (x + 6) ðð£
ðð¥ =
1
ð¥+6
M1
ððŠ
ðð¥ = 2x x ln(x + 6) + x2 x
1
ð¥+6 M1
ððŠ
ðð¥ = 2x ln (x + 6) +
ð¥2
ð¥+6 M1
3.
u = x ðð¢
ðð¥ = 1
v = âð¥ + 12 = (x + 12)0.5 ðð£
ðð¥ = 0.5(x + 12)-0.5
M1
ððŠ
ðð¥ = 1 à âð¥ + 12 + x à 0.5(x + 12)-0.5
ððŠ
ðð¥ = 0.5(x + 12)-0.5 [2(x + 12) + x]
M1
At stationary point ððŠ
ðð¥ = 0,
3(ð¥ + 8)
2âð¥ + 12= 0
x = -8
M1
When x = -8,
y(-8) = (-8)â(â8) + 12 = 16 M1
(-8, -16)
4.
x = 1
2
y(1
2) = 0
M1
u = 4x - 1 ðð¢
ðð¥ = 4
v = ln 2ð¥ ðð£
ðð¥ =
1
ð¥
ððŠ
ðð¥= 4 à ðð2ð¥ + (4ð¥ â 1) Ã
1
ð¥
M1
ððŠ
ðð¥ = 4 ln 2x + 4 -
1
ð¥ M1
![Page 11: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/11.jpg)
ððŠ
ðð¥ (
1
2) = 4 ln 2(
1
2) + 4 -
11
2
= 2 M1
y â 0 = 2(x - 1
2)
y = 2x - 1 M1
5.
When x = 0, y = -1 M1
u = (x2 â 1) ðð¢
ðð¥ = 2x
v = e3x ðð£
ðð¥ = 3e3x
ððŠ
ðð¥= 2ð¥ à ð3ð¥ + (ð¥2 â 1) à 3ð3ð¥
M1
ððŠ
ðð¥ = ð3ð¥[2x + 3(ð¥2 â 1)]
= ð3ð¥(3x2 + 2x â 3) M1
ððŠ
ðð¥ (0) = -3
Therefore gradient of normal = 1
3
M1
y + 1 = 1
3 (x â 0)
x â 3y â 3 = 0 M1
6a.
When x = 1
y (1) = xðð¥2= (1)ð(1)2
= e M1
u = x ðð¢
ðð¥ = 1
v = ðð¥2
ðð£
ðð¥ = 2xðð¥2
ððŠ
ðð¥= 1 à ðð¥2
+ ð¥ à ðð¥2 à 2ð¥
M1
ððŠ
ðð¥ = ðð¥2
(1 + 2ð¥2) M1
ððŠ
ðð¥ (1) = ð(1)2
(1 + 2(1)2) = 3e M1
y â e = 3e(x â 1)
y = e(3x â 2) M1
6b.
When x = 0
y = -2e M1
When y = 0,
x = 2
3 M1
Therefore area = 1
2 Ã 2ð Ã
2
3=
2
3ð M1
7a.
![Page 12: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/12.jpg)
u = e-x
ðð¢
ðð¥ = -e-x
v = sin x ðð£
ðð¥ = cos x
ððŠ
ðð¥= -e-x à sin ð¥ + ð-x à cos x
M1
ððŠ
ðð¥ = e-x(cos x â sin x) M1
ð
ðð¥ e-x(cos x â sin x)
u = e-x
ðð¢
ðð¥ = -e-x
v = cos x â sin x ðð£
ðð¥ = - sin x â cos x
ð2ðŠ
ðð¥2 = -e-x à (cos x â sin x) + e-x à (- sin x â cos x)
M1
ð2ðŠ
ðð¥2 = â2e-x cos x M1
7b.
At stationary point, ððŠ
ðð¥ = 0
e-x(cos x â sin x) = 0 M1
cos x â sin x = 0
tan x = 1 M1
x = ð
4
ð2ðŠ
ðð¥2 = ââ2 ðâð
4
ð2ðŠ
ðð¥2 < 0 , therefore maximum point
M1
M1
x = â3ð
4
ð2ðŠ
ðð¥2 = â2 ð3ð
4
ð2ðŠ
ðð¥2 > 0 , therefore minimum point
M1
M1
When x = ð
4, y =
1
â2ðâ
ð
4 , maximum point M1
When x = â3ð
4, y = â
1
â2ð
3ð
4 , minimum point M1
8a.
fâ(x) = - sin x à sin 2x + cos x à 2cos 2x M1
fâ(x) = 2cos x(1 â 2sin2x) â 2sin2x cos x
= 2 cos x(1 â 3 sin2x) M1
8b.
At stationary point fâ(x) = 0
2 cos x(1 â 3 sin2x) = 0 M1
Cos x = 0
sin x = = ±1
â3
M1
![Page 13: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/13.jpg)
x = 0.615, ð
2, 2.53, 3.76,
3ð
2, 5.76 M1 M1
8c.
f(x) at max when sin x = 1
â3 M1
cos2x = 1 â sin2x = 1 â 1
3 =
2
3 M1
f(x) = cos x à 2 sin ð¥ cos ð¥ = 2 sin x cos2x
M1
Therefore maximum = 2 Ã1
â3 Ã
2
3 =
4
9â3 M1
8d.
Period of cos x = 2ð
Period of sin x = ð M1
Therefore period of f(x) = 2ð
Values of f(x) in this interval are repeated. M1
![Page 14: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/14.jpg)
Differentiation
Pt. 6: Quotient Rule
1. Differentiate ðð¥
ð¥â4 with respect to xand simplify your answer (2)
2. Find ððŠ
ðð¥ of y =
ln(3ð¥â1)
ð¥+2 (2)
3. Find the co-ordinates of any stationary point on the curve y = ð¥+5
â2ð¥ +1 (4)
4. Find the equation of the tangent to the curve y = 3ð¥+4
ð¥2+1 at the point when x = -1 (5)
5. The curve C has the equation y = 3+sin 2ð¥
2+cos 2ð¥
a. Show that ððŠ
ðð¥=
6 sin 2ð¥+4 cos 2ð¥+2
(2+cos 2ð¥)2 (4)
b. Find an equaton of the tangent to C at the point where x = ð
2. Write your answer in the form y = ax+ b, where a
and b are exact constants. (4)
6. f(x) = ð¥â5
2ð¥+3+
2ð¥+4
2ð¥2+7ð¥+6
a. Express f(x) as a fraction in its simplest form (2)
b. Hence find fâ(x) in its simplest form (3)
7. Differentiate cos 2ð¥
âð¥ with respect to x (3)
8. Given that y = 3 sin ð¥
2 sin ð¥+2 cos ð¥, show that
ððŠ
ðð¥=
ðŽ
1+sin 2ð¥ where A is a rational constant to be found. The value of xlies
in the domain âð
4 < ð¥ <
3ð
4 (5)
9. Find equation of the normal to the curve y = ââð¥
5 cos 5ð¥ at the x-cordiante 16. (7)
A-Level
Pt. 3: General Differentiation
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
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
![Page 15: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/15.jpg)
Mark Scheme
1. ð
ðð¥(
ðð¥
ð¥â4 )
= ðð¥ Ã(ð¥â4)â ðð¥ Ã1
(ð¥â4)2 M1
= ðð¥ (ð¥â5)
(ð¥â4)2 M1
2. ð
ðð¥(
ln(3ð¥â1)
ð¥+2) )
=
3
3ð¥â1Ã(ð¥+2)âln (3ð¥â1)Ã1
(ð¥+2)2 M1
= 3
(3ð¥â1)(ð¥+2)â
ln (3ð¥â1)
(ð¥+2)2 M1
3.
ððŠ
ðð¥=
1Ãâ2ð¥+1â(ð¥+5)Ã1
2(2ð¥+1)
12Ã2
2ð¥+1 M1
ððŠ
ðð¥=
(2ð¥+1)â(ð¥+5)
(2ð¥+1)32
=ð¥â4
(2ð¥+1)32
M1
At stationary points, ððŠ
ðð¥ = 0
ð¥â4
(2ð¥+1)32
= 0
xâ 4 = 0
x= 4
M1
When x= 4
y(4) = 4+5
â2(4) +1 = 3 M1
Stationary point = (4, 3)
4. ððŠ
ðð¥=
3 Ã(ð¥2+1)â(3ð¥+4)Ã2ð¥
(ð¥2+1)2 M1
ððŠ
ðð¥ =
3â8ð¥â3ð¥2
(ð¥2+1)2 M1
ððŠ
ðð¥(â1) = 2 M1
When x= -1
y(-1) = 1
2
M1
y â 1
2= 2(ð¥ + 1) =
y = 2x+ 5
2
M1
5a. ððŠ
ðð¥=
(2+cos 2ð¥)(2 cos 2ð¥)â(3+sin 2ð¥)(â2ð ðð2ð¥)
(2+ððð 2ð¥)2 M1
ððŠ
ðð¥ =
4 cos 2ð¥ + 2ððð 22ð¥ + 6ð ðð2ð¥ + 2ð ðð22ð¥
(2+ððð 2ð¥)2 M1
ððŠ
ðð¥=
6 sin 2ð¥ + 4 cos 2ð¥ + 2(ððð 22ð¥ + ð ðð22ð¥)
(2+ððð 2ð¥)2 M1
ððŠ
ðð¥ =
6 sin 2ð¥ + 4 cos 2ð¥ + 2
(2+ððð 2ð¥)2 M1
![Page 16: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/16.jpg)
5b.
When x= ð
2
ððŠ
ðð¥ =
6 sin 2(ð
2) + 4 cos 2(
ð
2) + 2
(2+ððð 2(ð
2))2
= -2 M1
When x= ð
2
y(ð
2) =
3+sin 2(ð
2)
2+cos 2(ð
2) = 3
M1
y â 3 = -2(x- ð
2) M1
y = -2x+ ð + 3 M1
6a.
f(x) = ð¥â5
2ð¥+3+
2(ð¥+2)
(2ð¥+3)(ð¥+2) M1
= ð¥â5
2ð¥+3+
2
2ð¥+3
= ð¥â3
2ð¥+3
M1
6b.
fâ(x) = 2ð¥+3â2(ð¥â3)
(2ð¥+3)2 M1
M1
fâ(x) = 2ð¥+3â2ð¥+6
(2ð¥+3)2
fâ(x) = 9
(2ð¥+3)2 M1
7.
u = cos 2x ðð¢
ðð¥= â2 sin 2ð¥
v = ð¥1
2 ðð£
ðð¥ =
1
2ð¥â
1
2
M1
ð
ðð¥(
ð¢
ð£) =
ð¥12(â2 sin 2ð¥)âððð 2ð¥(
1
2ð¥
â12)
(ð¥12)2
M1
ð
ðð¥(
ð¢
ð£) =
â2ð¥12 sin 2ð¥ â
1
2ð¥
â12 cos 2ð¥
ð¥ M1
8.
u = 3 sin x ðð¢
ðð¥= 3 cos ð¥
v = 2 sin x+ 2 cos x ðð£
ðð¥ = 2 cos xâ 2 sin x
M1
ð
ðð¥(
ð¢
ð£) =
3 cos ð¥(2 sin ð¥+2 cos ð¥)â(2 cos ð¥â2 sin ð¥)(3 sin ð¥)
(2 sin ð¥+2 cos ð¥)2 M1
= 6
4 Ã
cos ð¥ sin ð¥ + ððð 2ð¥âcos ð¥ sin ð¥+ ð ðð2ð¥
ð ðð2ð¥+2 sin ð¥ cos ð¥+ ððð 2ð¥
USING:
(ð ðð2ð¥ + ððð 2ð¥ = 1)
(sin 2x= 2 sin xcos x)
M1
M1
ð
ðð¥(
ð¢
ð£) =
3
2
1+sin 2ð¥ M1
![Page 17: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/17.jpg)
9.
u = -âð¥ = -x0.5
ðð¢
ðð¥= â0.5 ð¥-0.5
v = 5cos 5x ðð£
ðð¥ = -5 sin 5xà 5 = -25 sin 5x
M1
ð
ðð¥(
ð¢
ð£) =
(5 cos 5ð¥) (â0.5ð¥â0.5)â(âð¥0.5)(â25 sin 5ð¥)
(5 cos 5ð¥)2 M1
At the point x= 1 , ððŠ
ðð¥=
(5 cos 5) (â0.5â0.5)â(â10.5)(â25 sin 5)
(5 cos 5)2
ððŠ
ðð¥ = 10.9âŠ
M1
M1
Gradient of normal = -0.1 M1
When x= 1, y = ââ1
5 cos 5 = -0.71 M1
y = mx+ c
-0.71 = (-0.1)(1) + c
c = -0.61
M1
y = -0.1xâ 0.61
![Page 18: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/18.jpg)
Differentiation
Pt. 7: Parametric Differentiation
1. A curve is given by the parametric equations x = t2 + 1 and y = 4
ð¡
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, â8) (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 â 1 and y = 2
ð¡2 (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 â 1
a. Write down expression for ðð¥
ðð¡ and
ððŠ
ðð¡ (1)
b. Hence, show that ððŠ
ðð¥= 2ð¡ (1)
6. Find and simplify an expression for ððŠ
ðð¥ in terms of the parameter t when x = 3t â 1 and y = 2 -
1
ð¡ (3)
7. Find and simplify an expression for ððŠ
ðð¥ 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 ð, y = tan ð, 0 †ð < ð
2 at the point where
ð = ð
3, has the equation â3ð¥ + 4ðŠ = 10â3 (6)
10. A curve has parametric equations x = sin 2t and y = sin2t, 0 < t < ð
a. Show that ððŠ
ðð¥ =
1
2tan 2ð¡ (2)
b. Find an equation for the tangent to the curve at the point where t = ð
6 (3)
11. A curve is given by the parametric equations
x = t 2, y = t(t â 2), t ⥠0.
a. Find the coordinates of any points where the curve meets the coordinate axes. (3)
b. Find ððŠ
ðð¥ in terms of x by first finding
ððŠ
ðð¥ in terms of t. (3)
c. Find ððŠ
ðð¥ 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, âð
2 < t <
ð
2
a. Show that the tangent to the curve at the point where t = ð
4 passes through the origin. (6)
b. Find a cartesian equation for the curve in the form y2 = f(x). (2)
A-Level
Pt. 3: General Differentiation
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
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
![Page 19: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/19.jpg)
Mark Scheme
1a.
x(2) = 22 + 1 = 5 M1
y(2) = 4
2 = 2 M1
1b. 4
ð¡= â8
t = -1
2
M1
2.
x = 3t â t = ð¥
3 M1
y = (ð¥
3)2 =
ð¥2
9 M1
3.
t = 1
2(x + 1) M1
y = 2
1
4(ð¥+1)2
=8
(ð¥+1)2 M1
4.
x = 6 + 2 cos x M1
y = -1 + 2 sin x M1
0 †x < 2ð
5a. ðð¥
ðð¡= 1
ððŠ
ðð¡= 2ð¡
M1
5b. ððŠ
ðð¥=
ððŠ
ðð¡ ÷
ðð¥
ðð¡
ððŠ
ðð¥ = 2t ÷ 1 = 2t
M1
6.
ðð¥
ðð¡= 3
ððŠ
ðð¡= ð¡â2
M1
ððŠ
ðð¥=
ððŠ
ðð¡ ÷
ðð¥
ðð¡
ððŠ
ðð¥ = t-2 ÷ 3 =
ð¡â2
3
M1
ððŠ
ðð¥ =
1
3ð¡2 M1
![Page 20: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/20.jpg)
7.
ðð¥
ðð¡= 2 sin ð¡ Ã cos ð¡
ððŠ
ðð¡= 3cos2t à (- sin t)
M1
ððŠ
ðð¥=
ððŠ
ðð¡ ÷
ðð¥
ðð¡
= â3ððð 2ð¡ sin ð¡
2 sin ð¡ cos ð¡
M1
ððŠ
ðð¥ = -
3
2cos ð¡ M1
8.
when t = 1
x(1) = 1
y(1) = 3
M1
ðð¥
ðð¡= 3t2
ððŠ
ðð¡= 6t
M1
ððŠ
ðð¥ =
6ð¡
3ð¡2 ððŠ
ðð¥ (1) = 2
M1
y â 3 = 2(x â 1)
y = 2x + 1 M1
9.
ð = ð
3
x(ð
3) = 2
y(ð
3) = 2â3
M1
ðð¥
ðð = sec ð tan ð
ððŠ
ðð= 2sec2 ð
M1
ððŠ
ðð¥ =
2ð ðð2ð
sec ð tan ð
ððŠ
ðð¥ =
2 sec ð
tan ð= 2 ððð ðð ð
M1
Gradient = 4
â3
Therefore gradient of normal = -â3
4
M1
y - 2â3 = -â3
4(x â 2)
4y - 8â3 = -â3x + 2â3 M1
â3x + 4y = 10â3 M1
![Page 21: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/21.jpg)
10a. ðð¥
ðð¡= 2 cos 2ð¡
ððŠ
ðð¡= 2 sin ð¡ Ã cos ð¡ = sin 2ð¡
M1
ððŠ
ðð¥=
sin 2ð¡
2 cos 2ð¡=
1
2tan 2ð¡ M1
10b.
t = ð
6
x(ð
6) =
â3
2
y(ð
6) =
1
4
M1
Gradient = â3
2
y - 1
4 =
â3
2(x -
â3
2)
M1
â3 x â 2y â 1 = 0 M1
11a.
x = 0
t2 = 0
t = 0
M1
y = 0
t(t â 2) = 0
t = 0 or t = 2
M1
Co-ordinates: (0, 0) and (4, 0) M1
11b.
ðð¥
ðð¡ = 2t
ððŠ
ðð¡= 2ð¡ â 2
M1
ððŠ
ðð¥=
2ð¡â2
2ð¡= 1 â t -1 M1
t = x0.5 ððŠ
ðð¥ = 1 â x-0.5 M1
11c.
y = ð¥1
2(ð¥1
2 â 2) = x - 2ð¥1
2 M1
ððŠ
ðð¥= 1 â ð¥â
1
2 M1
![Page 22: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/22.jpg)
12a.
t = ð
4
x(ð
4) =
1
2
y(ð
4) = 1
M1
ðð¥
ðð¡ = 2 sin t cos t
ððŠ
ðð¡= sec2 t
M1
ððŠ
ðð¥=
ð ðð2ð¡
2 sin ð¡ cos ð¡=
1
2 ð ðð3ð¡ ððð ðð ð¡ M1
Gradient = 1
2Ã (â2)3 Ã â2 = 2 M1
y â 1 = 2(x â 1
2)
y = 2x M1
When x = 0, y = 0 therefore passes through origin M1
12b.
y2 = tan2t = ð ðð2ð¡
ððð 2ð¡ =
ð ðð2ð¡
1â ð ðð2ð¡ M1
y2 = ð¥
1âð¥ M1
![Page 23: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/23.jpg)
Differentiation
Pt. 8: Implicit Differentiation
1. Differentiate with respect to x, y3 (1)
2. Find ððŠ
ðð¥ in terms of x and y for the curve x2 + y2 + 3x â 4y = 9 (2)
3. Find ððŠ
ðð¥ in terms of x and y for the curve 2e3x + e-2y + 7 = 0 (3)
4. Find ððŠ
ðð¥ in terms of x and y for the curve 2xy2 â x3y = 0 (4)
5. Find ððŠ
ðð¥ in terms of x and y for the curve cos 2x sec 3y + 1 = 0 (3)
6. A curve has the equation x2 + 4xy â 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 â 4xy + y2 = 24.
a. Show that ððŠ
ðð¥=
ð¥â2ðŠ
2ð¥âðŠ (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
Pt. 3: General Differentiation
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
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
![Page 24: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/24.jpg)
Mark Scheme
1. ð
ðð¥(ðŠ3) = 3y2 ððŠ
ðð¥ M1
2. ð
ðð¥(x2 + y2 + 3x â 4y = 9) â 2x + 2y
ððŠ
ðð¥ + 3 - 4
ððŠ
ðð¥ = 0 M1
ððŠ
ðð¥ (4 â 2y) = 2x + 3
ððŠ
ðð¥ =
2ð¥+3
4â2ðŠ
M1
3. ð
ðð¥(2e3x + e-2y + 7 = 0) â 6e3x â 2e-2y
ððŠ
ðð¥ = 0 M1
6e3x = 2e-2y ððŠ
ðð¥ M1
ððŠ
ðð¥ =
3ð3ð¥
ðâ2ðŠ = 3e3x + 2y M1
4. ð
ðð¥(2xy2) â product rule
u = 2x
uâ = 2
v = y2
vâ = 2y ððŠ
ðð¥
ð
ðð¥(2xy2) â (2 à y2) + (2x à 2y
ððŠ
ðð¥)
M1
ð
ðð¥(ð¥3y) â product rule
u = x3
uâ = 3x2
v = y
vâ = ððŠ
ðð¥
ð
ðð¥(ð¥3y) â (3x2 à y) + (x3 Ã
ððŠ
ðð¥)
M1
ð
ðð¥(2xy2 â x3y) = (2 à y2) + (2x à 2y
ððŠ
ðð¥) â ((3x2 à y) + (x3 Ã
ððŠ
ðð¥))
2y2 â 3x2y = ððŠ
ðð¥(x3 â 4xy)
M1
ððŠ
ðð¥=
2ðŠ2â3ð¥2ðŠ
ð¥3â4ð¥ðŠ M1
![Page 25: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/25.jpg)
5. ð
ðð¥(cos 2x sec 3y + 1) â
ð
ðð¥(cos 2x sec 3y) + 0 â Product Rule
u = cos 2x
uâ = -2sin 2x
v = sec 3y
vâ = 3ððŠ
ðð¥ sec 3y tan 3y
ð
ðð¥(cos 2x sec 3y) â (-2 sin 2x à sec 3y + cos 2x à 3
ððŠ
ðð¥ sec 3y tan 3y)
M1
3ððŠ
ðð¥ cos 2x sec 3y tan 3y = 2sin 2x sec 3y M1
ððŠ
ðð¥ =
2 sin 2ð¥
3 cos 2ð¥ tan 3ðŠ=
2
3tan 2ð¥ cot 3ðŠ M1
6a. ð
ðð¥ (x2 + 4xy â 3y2 = 36)
2x + 4 à y + 4x à ððŠ
ðð¥ â 6y
ððŠ
ðð¥ = 0
M1
2x + 4y = ððŠ
ðð¥(6ðŠ â 4ð¥) M1
ððŠ
ðð¥=
ð¥+2ðŠ
3ðŠâ2ð¥ M1
ððŠ
ðð¥(4, 2) = â4 M1
y â 2 = -4(x â 4)
y = 18 â 4x M1
6b.
At Q, ð¥+2ðŠ
3ðŠâ2ð¥ = -4
x + 2y = -4(3y â 2x)
x = 2y
M1
Sub into equation of the curve:
(2y)2 + 4y(2y) â 3y2 = 36 M1
y2 = 4
y = 2 or -2 M1
Therefore Q: (-4, -2) M1
7a. ð
ðð¥ (x2 â 4xy + y2 = 24) â 2x â 4 à y â 4x Ã
ððŠ
ðð¥ + 2y
ððŠ
ðð¥ = 0 M1
2x â 4y = ððŠ
ðð¥(4x â 2y) M1
ððŠ
ðð¥ =
2ð¥â4ðŠ
4ð¥â2ðŠ=
ð¥ â 2ðŠ
2ð¥ â ðŠ M1
7b.
Gradient = 3 M1
y â 10 = 3(x â 2)
y = 3x + 4 M1
![Page 26: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/26.jpg)
7c. ð¥ â 2ðŠ
2ð¥ â ðŠ= 3 M1
x â 2y = 3(2x â y)
y = 5x M1
x2 â (4x)(5x) + (5x)2 = 24 M1
x2 = 4
x = or x = -2 M1
Therefore, (-2, -10) M1
![Page 27: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/27.jpg)
Differentiation
Pt. 9: Rates of Change
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 sâ1.
a. Show that the perimeter of the circle is increasing at the rate of 0.4Ï cm sâ1. (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 sâ1. (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 mmsâ1 (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
9ðh3 (2)
Given also that at time t seconds after liquid is put in the funnel, V = 600eâ0.0005t
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 â 8 where h m is the depth of the liquid.
a. Find the value of ðð
ðâ 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
Pt. 3: General Differentiation
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
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
![Page 28: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/28.jpg)
Mark Scheme
1. ðð
ðð¡= 800(1.04)t x ln 1.04 M1
N = 4000
4000 = 800(1.04)t
(1.04)t = 5
M1
ðð
ðð¡= 800 Ã 5 Ã ln 1.04 = 157
Therefore, growing at a rate of 157 per minute. M1
2a. ðð
ðð¡=
ðð
ððÃ
ðð
ðð¡
ðð
ðð¡= 0.2
M1
P = 2ðð ðð
ðð= 2ð
M1
ðð
ðð¡= 2ð Ã 0.2 = 0.4ð
Therefore perimeter is increasing at 0.4ð cm s-1 M1
2b. ððŽ
ðð¡=
ððŽ
ððÃ
ðð
ðð¡
A = ðr2 ððŽ
ðð= 2ðð
M1
When r = 10, ððŽ
ðð= 20ð M1
ððŽ
ðð¡= 20ð Ã 0.2 = 4ð cm2 s-1 M1
2b.
20 = 2ðð Ã 0.2 M1
r = 50
ð = 15.9 cm M1
3a. ðð
ðð¡=
ðð
ðð Ã
ðð
ðð¡
ðð
ðð¡= 3.5, V = l3
ðð
ðð= 3l2
M1
200 = l3
l = 5.848 M1
ðð
ðð= 3 Ã 5.8482 = 102.6 M1
3.5 = 102.6 x ðð
ðð¡ M1
ðð
ðð¡= 3.5 ÷ 102.6 = 0.0341 ððs-1 M1
3b.
2 mm s-1 = 0.2 cm s-1 M1
3.5 = ðð
ðð Ã 0.2
ðð
ðð = 17.5
M1
17.5 = 3l2
l = 2.415 M1
V = l3 = 14.1 cm3 M1
![Page 29: AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule 1. Differentiate xex with respect to x. (3) 2. Find ð ð when y = x2 ln (x](https://reader035.vdocuments.mx/reader035/viewer/2022062417/61260bfcbe62d860ec62f44f/html5/thumbnails/29.jpg)
4a.
tan 30 = 1
â3 =
ð
â
r = â
â3
M1
V = 1
3ðr2h =
1
3ð Ã
1
3â2 Ã â =
1
9ðâ3 M1
4b.
t = 120, V = 565.06
h = â9 Ã565.06
ð
3
M1
h = 11.7cm M1
4c. h = 11.74 ðð
ðâ= 144.4
M1
ðð
ðð¡=
ðð
ðâÃ
ðâ
ðð¡
ðð
ðâ=
1
3ðh2
M1
ðð
ðð¡= 600 Ã (â0.0005)e-0.0005t
= -0.3e-0.0005t M1
t = 120 ðð
ðð¡= â0.2825
M1
0.2825 = 144.4 x ðâ
ðð¡
ðâ
ðð¡= â0.00196
M1
The depth is decreasing at 0.00196 cm s-1 M1
5a. ðð
ðâ=
3
2(3â2 + 4)
1
2(6â) = 3(3â2 + 4)1
2(3â) M1
= 9h(3h2 + 4)1
2 M1
When h = 0.6, ðð
ðâ= 9(0.6)(3(0.62) + 4)
12 M1
ðð
ðâ= 12.17 M1
5b. ðð
ðð¡=
ðð
ðâÃ
ðâ
ðð¡ M1
When h = 0.6, ðð
ðð¡ = (12.170) x (-0.015) M1
ðð
ðð¡ = -0.1825
The volume is decreasing at a rate of 0.18m3 per hour. M1