Download - add maths differentiation
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.