Topic 9

differentiation

523 xy

let u = 3x + 2

dx

du

du

dy

dx

dy

5uy

and 45u

du

dy

35 4 u

3dx

du

415u4)23(15 x

4)76( xy

5)43(

3

xy

231 xy

23 23 xxy

3

23xdx

du )3()23(2 1 x

dx

dv

dx

dy

12[2x )1()23( x

3xu 2)23( xvlet and

)]23(6[3 xx )]3()23[( 22 xx

)23(6 x

x )1( )23( x ]

)23(3 2 xx ]232[ xx

)25)(23(3 2 xxx

2)15(4 xxy

64 )21( xxy

3)43)(16( xxy

35

4

x

xy

)5)(4( x

4dx

du 5dx

dv

xu 4 35 xvlet and

dx

dy 4)35( x

2)35(

201220

x

xx

2)35(

12

x

2)35( x

1 nt mm

dx

dym gent tan

)( 11 xxmyy t ),( 11 yx

),( 11 yx

)( 11 xxmyy n

5

38 x

xxy 34 2 at point (1,1).

dx

dy

When x = 1, 3)1(8 dx

dy

Gradient of tangent = 5

dx

dy

)2(87 xy

1687 xy

x4

12 2 xy at point (2,7).

When x = 2, 8)2(4 dx

dy

98 xy

dx

dy

12 nm

)( 11 xxmyy n

x26

26 xxy at point (2,8).

When x = 2,

2)2(26 dx

dymt

2162 xy1 nt mm

2

1nm

)2(2

18 xy

)2(1)8(2 xy

182 xy

82 3 xy

xxy 62

34

x

y

93 xy

xxy 62

)1)(12( xxy

xy

4

23 xy

xxy

12

)1(7)2(5)3(4 2 xxdx

dy

)1(10)2(122

2

xdx

yd

71012 2 xx

1024 x

331 x

23 2 x

23 1 x

A 60m wire is bent into the shape of a rectangle. The length of one of its sides is x cm and its area is A cm².

(i) Show that A = 30x - x².

(ii) Hence, find the value of x such that the area of the rectangle is a maximum. Determine the maximum area.

2.

A 60m wire is bent into the shape of a rectangle. The length of one of its sides is x cm and its area is A cm².

(i) Show that A = 30x - x².

x

2

260 xx30

Area of rectangle = length X width

A = x ( 30 - x )

= 30x -x² (shown)

A 60m wire is bent into the shape of a rectangle. The length of one of its sides is x cm and its area is A cm².

(ii) Hence, find the value of x such that the area of the rectangle is a maximum. Determine the maximum area.

A = 30x -x²

For maximum area

0dx

dA0230 x

dx

dA

x = 15

Maximum area occurred when x = 15

A = 30(15) - 15²

= 225 cm²

3

Diagram shows a semicircle with centre O. The diameter PQ can be adjusted and R moves on the circumference that PR + QR = 10 cm. Given that QR = x cm and A is the area of triangle PQR. Find the expression of in terms of x. Hence, find the maximum value of the area of triangle.

dx

dA

3Diagram shows a semicircle with centre O. The diameter PQ can be adjusted and R moves on the circumference that PR + QR = 10 cm. Given that QR = x cm and A is the area of triangle PQR. Find the expression of in terms of x. Hence, find the maximum value of the area of triangle.

dx

dA

)10(2

1 2xxA

)210(2

1x

dx

dA

2

heightbaseA

2

)10( xxA

x

Area of triangle

PR + QR = 10

PR + x = 10

PR = 10 - x

10 - x

x5

For maximum area

0dx

dA

05 x5x

Maximum value of the area of triangle

]5)5(10[2

1 2A 5.12

4Diagram shows a cuboid ABCDEFGH with a

square base EFGH. The volume of the cuboid is 27cm².

(i) If A is the total area of the cuboid, show that

(ii) Find the value of x so that the total surface area is minimum.

xxA

1082 2

5Diagram shows a composite solid which consists of

a cone and a cylinder of radius r cm. Given the slant height of the cone is 3r and the volume of cylinder is 32π cm³.

(i) Prove that the total surface area A of the solid is given by

(i) calculate the minimum value of the surface area.

rrA

644 2

2)8(3

192

= 192 x 0.2

= 38.4 cm³sˉ¹

=0.2

= ?

x = 8

r = 7 = 0.5

dr

dA

r

A

7

)7(2 5.014 14

If y = f(x),

the new value of y,

yyy ab

yyy ab

Find the turning point for each of the curve below.

Determine whether the turning point is a maximum

or a minimum point.

2245 xxy

262 xxy

xxy 82

632 xxy

1

2

3

4

Find the turning point for each of the curve below.

Determine whether the turning point is a maximum or a minimum point.