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Differentiation
The idea of limit.
Consider y = x + 2 . If x => 0 (as x takes on smaller and smaller values until it approaches zero),
1–x
y = x + 2 approaches the value of 2. In the limit when x = 0, y = 2.
1–x
2
2
Example D.1: What is the limit of y when y = 2x + 4xh + h as h
0 ? Find y when x = 2.
Solution:
2
0 ), y approaches 2x . And if x = 2, y = 8.
As h approaches zero (h
Example D.2 Find the limit of
2
x – 9 as x
x–3
3
Solution:
2
x – 9 = (x + 3) (x – 3) = x + 3 . As x approaches 3 (x
x–3
x–3
2
3), x – 9
x–3
(x + 3)(x – 3) = x + 3 = 6
x–3
2
We write it as lim
x – 9 = lim (x + 3) = 6
x
3 x–3
x
3
Example D.3 Find the limits of the following functions when x
(i) 5x – 2
2–x
(ii)
x
x–1
0.
(iii) x + 1
x–1
Solution:
(i) lim
x
0
5x – 2 = – 2 = – 1
2–x
2
(ii) lim
x
0
x
= 0
x–1
(iii) lim
x+1 = –1
x
0 x–1
Example D.4 Find the limits of the following functions.
3x +1
(i) lim
x
∞ x
(ii) lim
x
x
∞ x+3
(iii) lim
x +1
x
∞ x
Solution:
ignore this value when x
(i) lim 3x +1 = 3 + 1 = 3
x
∞
x
x
(iv) lim
x
2
(iv) lim
x –1
x
1 x-1
2
1
x –1
x–1
=
(ii) lim
x
x
∞ x+3
(x + 1)( x – 1) = x + 1
x-1
= x =1
x
=2
(iii) lim
x +1
x ∞ x
=x
x
∞
= 1
when x approaches ∞, this value of 3 is ignored.
1|Page
Differentiation from First Principles
0 is called
The process of differentiation by calculating the limit of ∆y as ∆x
∆x
differentiation from first principles. ∆y is read as ‘delta y over delta x’ and many books write it as δy .
∆x
δx
Let’s differentiate y = 1/x from first principles.
Let x increases by δx and y then increases by δy.
We have y + δy =
1 .
x + δx
δy =
=
dividing by δx,
1
– 1
x + δx
x
x – (x + δx)
x( x + δx)
δy
δx
=
δy
Lim
δx
0 δx
–
( note y = 1/x)
– δx
.
2
x + xδx
=
1 .
2
x + xδx
= dy = –1
2
dx
x
( dy is the same as writing lim
dx
δx
δy )
0 δx
2
Example D.5 Find from first principles, y = x + 2x.
Solution:
2
y = x + 2x.
Let x increases by δx and so, y then increases by δy.
We have: y + δy = (x + δx) + 2(x + δx) = x + 2xδx + (δx)
2
2
δy = x + 2xδx + (δx)
2
dividing by δx,
δy
δx
δy
Lim
δx
0 δx
2
+ 2x + 2δx
2
2
+ 2x + 2δx – x – 2x = 2xδx + (δx) + 2δx
2
=
2
2xδx + (δx) + 2δx =
δx
2x + 2 + δx
( dy is the same as writing lim
δy )
dx
δx
0 δx
= dy = 2x + 2
dx
Differentiation by using the rule:
d(xn ) = nx n – 1 , where n is positive, negative, fraction or zero number.
dx
2x3=6
3 x 5 = 15
5 x 9 = 45
2
1
Example: (i) d(3x ) = 6x .
dx
3
(ii) d(5x ) = 15x
dx
2
5
(iii) d(9x ) = 45x
dx
4
2|Page
2
3
2
Example D.6 : Differentiate with respect to x: (i) 3 , (ii) 3x , (iii) 3x , (iv) 3x , (v) 3x + 4x
3
2
2
2
(vii) 1/ x (viii) x√x (xi) (x + 2)
(vi) 3x + 4x + 5
Solution:
(i) d(3) = 0
dx
(iii) d(3x = 6x
dx
3
(iv) d(3x ) = 9x
dx
2
2
(v) d(3x + 4x) = 6x + 4
dx
0
(x = 1)
(differentiating a
constant always
gives a zero)
3
2
(ii) d(3x) = 3
dx
2
(vi) y = 3x + 4x + 5
(vii) y = 1 = x
2
x
2
dy = 9x + 8x
dx
dy = –2x
dx
–2
–3
(viii) y = x√x = x
=–2
3
x
dy = 3x
dx
2
1/2
3/2
2
(xi) y = (x + 2)
2
= x + 4x + 4
= 3
2√x
dy = 2x + 4
dx
Tangents and normals
2
Example D.7 Find the gradient of the curve y = 6x – x – 12 at the points where the curve cuts
the x- axis. Find the equations of the tangents at these points.
Solution: (when curve cuts x-axis, y = 0)
2
y = 6x – x – 12= 0
(2x – 3)(3x + 4) = 0
x = 3/2 or x = – 4/3
–4
3
Thererfore x- intercepts are at points ( 3, 0) and (– 4, 0)
2
3
Gradient of curve = dy = 12x –1. Gradient of curve at ( 3 , 0) = 17
dx
2
3
2
–12
Equation of tangent at (3, 0) is y = 17( x – 3) = 17x – 51
2
2
2
Equation of tangent is:
2y = 34x – 51
Gradient of curve = dy = 12x –1.
dx
Gradient of curve at (– 4 , 0) = – 17
3
Equation of tangent at (– 4, 0) is y = – 17 ( x + 4) = –17x – 68
3
3
3
Equation of tangent is:
3y + 51x + 68 = 0
3|Page
3
Example D.8 Find the equation of the tangent to the curve y = x at the point ( 3, 4 ).
Solution:
Curve: y = x
3
2
Gradient of curve = dy = 3x . Gradient of curve at (3, 4) = 27
dx
Equation of tangent at (3, 4) is y – 4 = 27( x – 3) = 27x – 81
Equation of tangent is:
y = 27x – 77
2
Example D.9 Find the equation of the tangent and normal to the curve y = 6x – x – 12 at the
point ( – 4, 0 ).
3
Solution:
2
Curve: y = 6x – x – 12
normal
Gradient of curve = dy = 12x –1.
dx
Gradient of curve at (– 4 , 0) = – 17
3
(– 4, 0)
3
Equation of tangent at (– 4, 0) is y = – 17 ( x + 4) = –17x – 68
3
3
3
Equation of tangent is:
3y + 51x + 68 = 0
Equation of normal is:
51y – 3x + c = 0 and at ( – 4, 0 ), c = – 4.
3
Equation of normal is:
51y – 3x – 4 = 0
tangent
( A normal is a line perpendicular to the tangent)
3
Example D.10 Find the equation of the tangent and normal to the curve y = x – 3x at the
point ( 2, 0)
Solution:
(tangent)
y = 9x – 18
3
Curve: y = x – 3x
2
Gradient of curve = dy = 3x – 3 . Gradient of curve at (2, 0) = 9
dx
Equation of tangent at (2, 0) is y = 9( x – 2) = 9x – 18
Equation of tangent is:
(2, 0)
y = 9x – 18
9y + x – 2 = 0
(normal)
Equation of normal is:
9y = – x + c
and at ( 2, 0 ), c = 2.
Equation of normal is:
9y + x – 2 = 0
( A normal is a line perpendicular to the tangent)
4|Page
Function of a function (chain rule)
4
dy = dy . du
dx
du dx
Differentiate y = (3x + 5) .
Here, y is a function of x
Let u = 3x +5
Here, u is a function of x
du = 3
dx
4
Hence y = u
Here, y is a function of u
3
dy = 4u
du
y is a function of u, and u is a function of x
what we have here is called a ‘function of a function’
therefore
dy = dy x du
dx
du
dx
dy =
dx
3
3
4u x 3 = 12u = 12(3x + 5)
3
9
Now it will be laborious if we differentiate y = (3x + 5) .
d(3x + 5)
dx
Hence we shorten the steps as follows:
9
Decrease the index by 1
8
= 9(3x + 5) (3) = 27(3x + 5)
index
the term is 3x + 5
8
differentiating 3x + 5 gives 3
1. Multiply the term (3x + 5) by the index and reduce the index by 1.
2. Then differentiate the term 3x + 5.
3. Multiply these two items.
3
Example D.11 Differentiate (i) (3x – 4x)
4
(ii) (3x + 5)
–3
(ii) d (3x + 5)
dx
–3
4
3
2
–4
(3) =
(iii) (2x – 6x + 5x + 2)
3
Solution:
3
4
3
3
2
(i) d (3x – 4x) = 4(3x – 4x) .(9x – 4)
dx
4
3
2
(iii) d (2x – 6x + 5x + 2)
dx
3
4
3
2
2
3
= – 3(3x + 5)
–9 .
4
(3x + 5)
2
= 3(2x – 6x + 5x + 2) ( 8x – 18x + 10x)
Rates of Change
dy compares the rate of change of y with respect to x. If dy = 3, y is increasing 3 times as fast as x is increasing.
dx
dx
If x and y are distances, and if x is increasing at 2 m/sec, then y is increasing at 6 m/sec.
If dy = – 5, then y is decreasing 5 times as fast as x is increasing.
dx
5|Page
3
Example D.12 If 2 cm of air is blown into a spherical balloon every second, how quickly will the
(We are told to find dr , the rate of change of radius with time).
radius grow when r = 3 cm ?
dt
Solution:
3
Volume of sphere , V = 4 r
3
2
dV = 4 r
dr
And here we are given: dV = 2.
dt
r
Hence dV = dV x dr
dt
dr
dt
2
2 = 4 r x dr
dt
dr =
2
dt
4 (3)
= 0.053 cm/sec
Example D.13 The radius of a spherical ink blob is increasing at the rate of 0.2 cm/sec.
Find the rate of increase of the area of the blog when the radius is 5 cm.
Solution:
Here we are given dr = 0.2 and we are to find dA when r = 5.
dt
dt
Area of blob, A =
dA =
dr
r = 10
r
2
(the area increases 10
Therefore dA = dA x dr =
dt
dr
dt
times as faster as the radius increases)
(5) x 0.2 = 2
2
cm /sec
3
Example D.14: The volume of a cube is increasing at the rate of 2 cm /sec. Find the rate of
change of side of the base when its length is 12 cm
Solution: Let the volume be V cm 3 and the side of the cube be x cm.
Volume of cube, V = x
3
2
dV = 3x = 3 x 144 = 432
dx
rate of change of side , dx = dx x dV
dt
dV
dt
=
1
432
x 2 = 0.00463 cm/sec
6|Page
Example D.15: A container in the shape of a right circular cone of height 10cm and base radius
3
1 cm is receiving milk shake at the rate of 4 cm /sec. Find the rate at which the depth of the milk
shake is increasing when the milk shake is already filled half-way up the cone.
3
Solution: Let the volume be V cm , height h, and the radius of the surface area be r cm when half-filled.
2
2
By similar triangles, r = h , then r = h
1 10
100
2
Volume of cone, V = 1 r h = 1
3
3
2
)h =
100
1
1
3
h
300
r
10 cm
2
dV =
h
dh
100
h
given: dV = 4,
dt
Now dh = dh x dV
dt
dV
dt
= 100 x 4
2
h
When cone is half-filled, h = 5, therefore
dh =
dt
100 x 4 = 400
2
h
25
= 16 cm/sec
Example D.16: A kite is 30 m above the ground, and has 50 m of string out. If the kite is moving
horizontally at 2 m/sec directly away from the person who is flying the kite, at what rate is the
string being rolled out?
Solution: Let the horizontal distance of kite from the person at time t be x m, and string out be s m.
Horizontal velocity = dx = 2 m/sec.
dt
2
2
2
2
2 1/2
By Pythagoras’ theorem,
s = x + 30 , and s = (x + 30 )
2
ds =
dx
2
(x + 30 )
2
– 1/2
K
s
(2x ) =
x
.
2
2 1/2
(x + 30 )
dx = 2
dt
30
P
O
x
ds = ds . dx
dt
dx dt
=
x
. ( 2) =
2x
.
2
2 1/2
2
2 1/2
(x + 30 )
(x + 30 )
When s = 50, x = 40, and
ds = 2 (40)
= 1.6 m/sec
2
2 1/2
dt
(40 + 30 )
For advanced students only ( you could shorten your working by differentiating with respect to t )
2
2
2
s = x + 30
2sds = 2xdx , therefore ds = x dx
dt
dt
dt
s dt
when s = 50, x = 40 (by Pythagoras’ theorem)
and ds = 40 x 2 = 1.6 m/sec
dt
50
7|Page
Product Rule
Let y be the product of two functions u and v of a variable x. Then y = u.v , and...
dy = v du + u dv
dx
dx
dx
2
2
3
2
Example D.17: Differentiate with respect to x: (i) (x – 2)(x + 1) (ii) x (x + 3) (iii) (x + 3) (x + 1)
Solution:
u
3
v
2
(i) y = (x – 2)(x + 1)
2
(ii)
2
2
y = x (x + 3)
3
dy = (x + 1) d (x – 2) + (x – 2 )d (x + 1)
dx
dx
dx
2
2
2
2
2
dy = (x + 3) d(x ) + x d (x + 3)
dx
dx
dx
3
= (x + 1)(1) + (x – 2).2x
= x + 1 + 2x – 4x
3
2
= 2x (x + 3) + 3x (x+3)
2
3
2
2
= 3x – 4x + 1
= x(x + 3) [ 2(x + 3) + 3x ]
2
= x(x + 3) (5x + 6)
2
(iii) y = (x + 3) (x + 1)
3
3
2
2
dy = (x + 1) d (x + 3) + (x + 3) d (x + 1)
dx
dx
dx
3
3
2
= 2(x + 1) (x + 3 ) + 3(x + 3) (x + 1)
2
2
= (x + 1) (x + 3) [ 2(x + 1) + 3(x + 3) ]
2
= (x + 1) (x + 3) (2x + 2 + 3x + 9)
= (x + 3)( 5x + 11)(x + 1)
2
= (5x + 26x +33) (x + 1)
2
2
Quotient Rule
Let y be the quotientt of two functions u and v of a variable x. Then y = u , and ...
v
dy
dx
=
v du – u dv 1
dx
dx v 2
8|Page
Example D.18: Differentiate with respect to x: (i) (x – 2)
2
(x + 1)
Solution:
u
v
(i) y = (x – 2)
2
(x + 1)
(ii)
(ii)
2
2
dy = (x + 1) d (x – 2) – (x – 2 )d (x + 1) 1
2
2
dx
dx
dx
(x + 1)
2
2
2
y=
2
= 1 + 4x – x
2
2
(x + 1)
2
2
x
3
(x + 3)
3
2
2
3
2
= 2x (x + 3) – 3x (x+3)
3
(x + 3)
2
2
2
= x(x + 3) [ 2(x + 3) – 3x ]
6
(x + 3)
3
3
2
(x + 3)
3
(x + 1)
dy = (x + 3) d(x ) – x d (x + 3)
1 .
6
dx
dx
dx
(x + 3)
=
(iii) y = (x + 3) (x + 1)
(iii)
3
= (x + 1)(1) – (x – 2).2x
2
2
(x + 1)
= x + 1 – 2x + 4x
2
2
(x + 1)
2
x
3
(x + 3)
2
x (6 – x )
4
(x + 3)
3
dy = (x + 1) d (x + 3) – (x + 3) d (x + 1)
1 .
6
dx
dx
dx
(x + 1)
3
2
= 2(x + 1) (x + 3 ) – 3(x + 3) (x + 1)
6
(x + 1)
2
2
= (x + 1) (x + 3) [ 2(x + 1) – 3(x + 3) ] =
6
(x + 1)
=
(x + 3)(– x – 7)
4
(x + 1)
(x + 3) (2x + 2 – 3x – 9)
4
(x + 1)
2
= – (x + 10x +21)
4
(x + 1)
Maxima and Minima
1. When an object is thrown from a point O, it will reach a maximum height AB and then falls back to the ground.
2. During the ascend between O and A, the gradient is positive, but steadily decreasing to zero at A. After passing
the point A, the particle is descending, and the gradient is negative. The shape of the curve is a parabola and the
2
equation of the curve is y = x
3. Notice that before the maximum point A, the gradient is positive, and after the point A, the gradient is negative.
Therefore, by comparing the gradients immediately before and after a turning point, we can determine whether the
turning point is a maximum turning point or a minimum turning point.
4. For a maximum turning point, the gradient changes from positive to negative and for a minimum turning point,
the gradient changes from negative to positive.
O
X
P
A
+
–
–
+
Gradient is zero
+
–
O
B
C
Maximum turning point
–
+
Y
Minimum turning point
9|Page
Example D.19: Find the coordinates of the points on the following curves where the gradient is
2
2
2
(ii) y = 3x + 2x – 4 (iii) y = 4x .
zero. (i) y = 3x – 2x
Solution:
2
2
(i) y = 3x – 2x
(iii) y = 4x
(ii) y = 3x + 2x – 4
dy = 3 – 4x = 0 (for turning points)
dx
dy = 6x + 2 = 0 (for turning points)
dx
x = 3 , y = 27
4
16
dy = 8x = 0
dx
x = –1 , y = –13
3
3
Coordinates ≡ ( 3 , 27 )
4 16
2
x = 0, y = 0
Coordinates ≡ ( –1 , – 13)
3
3
Coordinates ≡ (0,0)
Example D.20: Find the turning points of the following curves and indicate whether they are
2
2
x .
maximum or minimum turning points. (i) y = x – 5x +6 (ii) y = 4 – 18 x – 3x (iii) y =
2
x +4
Solution:
2
(i) y = x – 5x +6
(ii) y = 4 – 18 x – 3x
dy = 2x – 5 = 0 (for turning points)
dx
dy = – 6x – 18
dx
x =5,y= –1
2
4
= 0 (for turning points)
x = – 3, y = – 77
Coordinates ≡ ( –3 , – 77)
Coordinates ≡ ( 5 , – 1 )
2
4
2
2
d y = 2 which is positive.
2
dx
d y = – 6 which is negative
2
dx
therefore ( – 3, – 77 ) is a maximum point
therefore ( 5 , – 1 ) is a minimum point.
2
2
4
Differentiating the gradient a second time to check the gradient immediate after the turning point.
1. If it is positive, then the turning point is a minimum point.
2. If it is negative, then the turning point is a maximum point.
(iii) y =
dy =
dx
x
2
x +4
it is laborious to differentiate a second time, so we take a value
slightly greater, and also a value slightly less than the calculated
value of x.
2
2
(x + 4) d(x) – x d((x + 4)
dx
2
= (x + 4) – x(2x) =
2
(x + 4)
=
x=2
2
1
2
(x + 4)
dx
2
2
x + 4 – 2x
2
2
(x + 4)
2
4–x
= 0 (for turning point)
2
2
(x + 4)
2
For x slightly < – 2, (say x = – 2.5), dy = (–) = (–)
dx
(+)
For x slightly > – 2, (say x = – 1.5), dy = (+) = (+)
dx
(+)
Hence ( – 2, – ¼
) is a minimum point.
For x slightly < 2, (say x = 1.5), dy = (+) = (+)
dx
(+)
or x = – 2
For x slightly > 2, (say x = 2.5), dy = (–) = (–)
x = 2 , and y = ¼ , and x = – 2 , y = – ¼
dx (+)
Hence ( 2, ¼ ) is a maximum point
10 | P a g e
3
Example D.21: The volume of a cylinder is 500 cm . Show that the total surface area of the cylinder
2
2
is equal to 2 r + 1000/r cm , where r is the radius of the base. Hence find the value of r which
makes the surface area a minimum.
Solution:
2
Volume of cylinder = r h = 500
Hence h = 500
2
r
r
Total surface area of cylinder, S = 2 rh + 2 r
2
2
2
= 2 r (500/ r ) + 2 r
h
2
= 2 r + 1000/r
2
dS = 4 r – 1000/r = 0 ( for minimum value).
dr
r = 1000/4 , therefore r = √(250/ ) = 4.3 cm
3
3
3
Example D.22: An open cardboard box with a square base has a capacity of 4 m . Find the
dimensions if the area of the cardboard used is as small as possible.
Solution: Let the height of the box be h m and its length and breadth be x m.
x
2
Volume, V = x h = 4, therefore h = 4
2
x
2
2
x + 4xh = x + 4x(4)
2
x
2
= x + 16
x
Total area of cardboard used, A =
h
dA = 2x – 16 = 0 (for minimum area)
2
dx
x
x = √8 = 2 m, and hence h = 4/(2) = 1 m.
3
2
therefore the dimensions are length = 2 m, breadth = 2 m , height = 1 m
11 | P a g e
Small changes (Approximations)
We have learnt that, as δx
dy . Therefore, if δx is small, then
dx
0, δy
δx
δy ≈ dy
δx
dx
δy ≈ dy.δx
dx
and
Example D.22: The side of a square is 15 cm. Find the increase in the area of the square when the
side expands by 0.02 cm.
2
Solution: Let the area of the square be A cm and the side be x cm.
2
A = x , therefore, dA = 2x.
dx
δA ≈ dA. δx ≈ 2x(δx).
dx
When x = 15, and δx = 0.02
δA ≈ dA. δx ≈ 2x(δx). ≈ 2(15)(0.02) ≈ 0.6 cm
dx
2
Example D.23: An error of 1% is made by Ah Bean when he measures the radius of a sphere. Find
the percentage error that occurs when calculating the surface area.
Solution: Let the area of the sphere be A and the radius be r .
A=4
2
, therefore, dA =
dr
δA ≈ dA. δx ≈ 8 (δr).
dr
At 1% error, δr = 0.01r
δA ≈ dA. δr ≈ 8
dr
δr). ≈ 8(0.01)
2
≈ 0.08
2
Percentage error in surface area ≈ δA (100) = 0.08
2
A
4
2
(100) = 2%.
Example D.24: Find an approxximate value for √16.02.
Solution: Let y = √x, and dy = 1/(2√x)
dx
δy ≈ dy. δx ≈ (δx).
dx
2√x
√16.02 = √(16 + 0.02)
when x = 16 and δx = 0.02
δy ≈ (0.02) ≈
2√16
0.0025
therefore √16.02 ≈ 4.0025
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Example D.25:
Find an approxximate value for √627.
Solution: Let y = √x, and dy = 1
dx (2√x)
δy ≈ dy. δx ≈ (δx).
x
δx
dx
2√x
let us write √627 = √(625 + 2)
(2) ≈
2√625
Solution: Let y = 1
2
x
= x
–2
1 .
2
(2.97)
, and dy = – 2
3
dx
x
δy ≈ dy. δx ≈ – 2(δx).
3
dx
x
δx
x
let us write 2.97 = 3 – 0.03
when x = 625 and δx = 2
δy ≈
Example D.26: Find an approximate value for
when x = 3 and δx = – 0.03
0.04
δy
y
therefore √627 ≈ √625 + 0.04
δy ≈ – 2( – 0.03) ≈ 0.002222
3
3
2
y is inversely proportional to x . When x decreases, y increases.
δy
y
≈ 25.04
Therefore
(when x = 625, δx = 2 and δy = 0.04)
1
=
2
(2.97)
1 + 0.00222 = 0.1133
2
3
(when x = 3, δx = 0 – 0.03 and δy = 0.00222)
Example D.27: The volume of a sphere increases by 5%. Find the corresponding percentage
increase in surface area.
Solution: Let the surface area of the sphere be A , volume be V and the radius be r
3
2
V=4
, therefore, dV = 4
,
3
dr
2
δr, but δV = 0.05V
δV ≈ dV.δr = 4
dr
therefore, 4
δr = 0.05r
3
2
δr = 0.05 (4
3
)
surface area, A = 4
2
,
δA ≈ dA. δx ≈ 8 (δr).
dr
% surface area ≈ δA (100) = 8
A
4
(100) .
2
3
=
2(0.05)(100) = 3⅓ %
3
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