02. DIFFERENTIATION 2
2.1.
and
notation
2.2. Greatest and least values
2.3. Application of Greatest and least values
Exercise 2A
2.4. Maxima and minima
2.5. Maxima and minima problems
Exercise 2B
2.6. Curve sketching
Exercise 2C
2.7. Second derivative
2.8. Stationary point and second derivative test
Exercise 2D
2.9. Small changes
Exercise 2E
Examination Questions
Chapter Answers
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02. Differentiation 2
02. DIFFERENTIATION 2
𝟐. 𝟏.
𝒅
′
𝒅𝒙 𝒂𝒏𝒅 𝒇 𝒙
𝒏𝒐𝒕𝒂𝒕𝒊𝒐𝒏
In addition to the
notation
𝑑𝑦
= 2𝑥 .
𝑑𝑥
𝑑 2
𝑑
𝑇𝑖𝑠 𝑚𝑎𝑦 𝑏𝑒 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠
𝑥 = 2𝑥 𝑤𝑒𝑟𝑒
𝑖𝑠 𝑠𝑖𝑚𝑝𝑙𝑦
𝑑𝑥
𝑑𝑥
𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑖𝑓 𝑦 = 𝑥 2 ;
𝑑𝑦
𝑑𝑥
the gradient function
can also be denoted
𝑎𝑛 𝑖𝑛𝑠𝑡𝑟𝑢𝑐𝑡𝑖𝑜𝑛 𝑡𝑜 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒.
as
The gradient function is also referred to as the derived function or the derivative.
𝑇𝑢𝑠 𝑖𝑓 𝑓 𝑥
= 𝑥2;
or
𝑡𝑒 𝑑𝑒𝑟𝑖𝑣𝑒𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑑𝑒𝑛𝑜𝑡𝑒𝑑 𝑎𝑠 𝑓 ′ 𝑥 = 2𝑥
𝐻𝑒𝑟𝑒 𝑓 ′ 𝑥 𝑖𝑠 𝑎 𝑢𝑠𝑒𝑓𝑢𝑙 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑣𝑒 𝑛𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑡𝑜
𝑑
𝑑𝑥
𝑓’ 𝑥 .
𝑑𝑦
𝑑𝑥
𝑜𝑟
𝑑
.
𝑑𝑥
Example 1:
Differentiate:
𝑎
3𝑥 2 + 4𝑥
𝑑
3𝑥 2 + 4𝑥 = 6𝑥 + 4
𝑑𝑥
𝑏
1 4
𝑥
2
𝑑
𝑑𝑥
2
In instances, where equating the
expression to 𝑦 and then finding
𝑑𝑦
𝑑𝑥 is not desirable, i.e.:
𝑎 𝑦 = 3𝑥 2 + 4𝑥
+ 6𝑥 3
1 4
𝑥
2
+ 6𝑥 3
= 2𝑥 3 + 18𝑥 2
3
𝑐 𝑥3 + 𝑥2
2
𝑑
3
𝑥2 + 𝑥3
𝑑𝑥
3 1
2
1
= 2𝑥 2 + 3𝑥 −3
𝑑𝑦
= 6𝑥 + 4 ,
𝑑𝑥
simply quoting 𝑑 𝑑𝑥 is a short-hand
method.
𝑑 𝑓 𝑥 = 3𝑥 3 + 4𝑥 2 − 2𝑥 + 5
𝑓 ′ 𝑥 = 9𝑥 2 + 8𝑥 − 2
𝑒 𝑓 𝑥 = 𝑝𝑥 4 + 𝑞𝑥 3 + 𝑟
Where the function has been given
as 𝑓 𝑥 , the derivative is simply
quoted as 𝑓 ′ 𝑥 .
𝑓 ′ 𝑥 = 4𝑝𝑥 3 + 3𝑞𝑥 2
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02. Differentiation 2
2.2. Greatest and least values
In fig 2.1(a), as the value of 𝑥 increases to the right, the gradient is positive but decreases steadily to zero at 𝐴. Past
𝐴 the gradient changes sign to negative. 𝐴𝐵 is the greatest value of 𝑦.
When the slope of a curve is going up (as
shown on the left), the gradient is positive.
And when the slope is going down, the
gradient is negative (as shown on the right).
AB is the greatest value of y.
Fig. 2.1(a)
On the left where the slope of the curve is
going down, its gradient is negative. And on
the right where the slope is going up, the
gradient is positive. CD is the least value of
y.
Fig. 2.1(b)
In fig 2.1(b), however, the gradient changes from negative to positive through 𝐶 and 𝐶𝐷 is the least value of 𝑦.
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 𝑦 = 𝑥 2 − 𝑥 − 2;
𝑇𝑒 𝑙𝑒𝑎𝑠𝑡 𝑜𝑟 𝑔𝑟𝑒𝑎𝑡𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡𝑖𝑠 𝑐𝑢𝑟𝑣𝑒 𝑜𝑐𝑐𝑢𝑟𝑠
𝑑𝑦
𝑤𝑒𝑛 𝑖𝑡𝑠 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡
= 0;
𝑑𝑥
𝑑𝑦
= 2𝑥 − 1
𝑑𝑥
∴
2𝑥 − 1 = 0
𝑥 =
1
2
First we find the derivative
𝑑𝑦
𝑑𝑥 .
For least or greatest value, we
𝑑𝑦
equate
𝑑𝑥 to zero to find the
𝑥-value.
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02. Differentiation 2
1 2
2
1
𝐴𝑡 𝑥 = 2; 𝑦 =
1
2
−
We put 𝑥 =
−2
1
2
into the equation to find
the 𝑦-coordinate.
1
𝑦 = −24
1
,
2
𝑡𝑢𝑠 𝑡𝑒 𝑝𝑜𝑖𝑛𝑡 𝑖𝑠
1
− 24 .
𝑊𝑒 𝑛𝑜𝑤 𝑖𝑛𝑣𝑒𝑠𝑡𝑖𝑔𝑎𝑡𝑒 𝑡𝑒 𝑠𝑖𝑔𝑛 𝑜𝑓 𝑡𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡
𝑑𝑦
𝑜𝑛 𝑒𝑖𝑡𝑒𝑟 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡𝑒 𝑝𝑜𝑖𝑛𝑡
𝑑𝑥
1
1
, −24
2
𝑡𝑜
Using such a table is one method (see
working in next box); we shall see
𝑑𝑖𝑠𝑐𝑜𝑣𝑒𝑟 𝑤𝑒𝑡𝑒𝑟 𝑖𝑡 𝑖𝑠 𝑡𝑒 𝑖𝑔𝑒𝑠𝑡 𝑜𝑟 𝑙𝑜𝑤𝑒𝑠𝑡
another method later in the chapter.
𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒, 𝑢𝑠𝑖𝑛𝑔 𝑡𝑒 𝑡𝑎𝑏𝑙𝑒 𝑏𝑒𝑙𝑜𝑤:
𝑑𝑦
𝑑𝑥
= 2𝑥 − 1
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝐿
1
2
𝑅
𝑑𝑦
−
0
+
𝑠𝑖𝑔𝑛 𝑜𝑓 𝑑𝑥
From fig. 2.1(b), negative on the (L)
and positive on the (R) occurs at the
𝑙𝑒𝑎𝑠𝑡
least point. So this is the least point
on the curve.
𝑇𝑢𝑠
1
1
, −24
2
𝑖𝑠 𝑡𝑒 𝑙𝑒𝑎𝑠𝑡 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 .
𝑾𝒐𝒓𝒌𝒊𝒏𝒈:
1
1. 𝐴𝑡 𝑥 = 2;
𝑑𝑦
= 2
𝑑𝑥
1
2
We find the sign of the
−1 =0
2. 𝑇𝑜 𝑡𝑒 𝑙𝑒𝑓𝑡 𝐿 ; 𝑥 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡𝑎𝑛
1
;
2
gradient on either side of
𝑤𝑒 𝑝𝑢𝑡 𝑓𝑜𝑟 𝑒𝑥𝑎𝑚𝑝𝑙𝑒 𝑥 = 0
𝑑𝑦
= 2 0 − 1 = −1 −𝑣𝑒
𝑑𝑥
1
3. 𝑇𝑜 𝑡𝑒 𝑟𝑖𝑔𝑡 𝑅 ; 𝑥 𝑖𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡𝑎𝑛 2; 𝑤𝑒 𝑝𝑢𝑡 𝑓𝑜𝑟 𝑒𝑥𝑎𝑚𝑝𝑙𝑒 𝑥 = 1
the point by substituting
two suitable values of 𝑥
𝑑𝑦
into
𝑑𝑥 ; one on the left
and the other on the right
of the point.
𝑑𝑦
= 2 1 − 1 = 1 +𝑣𝑒
𝑑𝑥
Example 2:
Find the greatest or least values of the following functions:
𝑎
𝑓 𝑥 = 𝑥 4−𝑥
𝑓 𝑥 = 4𝑥 − 𝑥 2
First, we find the derivative. Then
equate the derivative to zero, to find
′
𝑓 𝑥 = 4 − 2𝑥
𝐿𝑒𝑎𝑠𝑡 𝑜𝑟 𝑔𝑟𝑒𝑎𝑡𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤𝑒𝑛 𝑓′ 𝑥 = 0
the 𝑥-value.
3
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02. Differentiation 2
∴
We put 𝑥 = 2 into the expression for
4 − 2𝑥 = 0
𝑓 𝑥 to find the 𝑦 −coordinate.
𝑥 = 2.
𝐴𝑡 𝑥 = 2; 𝑓 2 = 2 4 − 2
We find the sign of the gradient on
= 4
either side of the point to determine
𝑝𝑜𝑖𝑛𝑡 2, 4 .
𝑓′ 𝑥 =
whether it is the greatest or least point.
4 − 2𝑥
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝐿
2
𝑅
𝑠𝑖𝑔𝑛 𝑜𝑓 𝑓 ′ 𝑥
+
0
−
For example:
𝐿 𝑝𝑢𝑡 𝑥 = 1.5 𝑖𝑛𝑡𝑜 𝑓 ′ 𝑥
(1.5 is just less than 2)
𝑔𝑟𝑒𝑎𝑡𝑒𝑠𝑡
𝑓 ′ 𝑥 = 4 − 2 1.5 = +1 [𝑒𝑛𝑐𝑒 + 𝑣𝑒)
𝑇𝑢𝑠 𝑡𝑒 𝑔𝑟𝑒𝑎𝑡𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 4 𝑎𝑛𝑑 𝑜𝑐𝑐𝑢𝑟𝑠
𝑎𝑡 𝑥 = 2.
𝑅 𝑝𝑢𝑡 𝑥 = 2.5 𝑖𝑛𝑡𝑜 𝑓 ′ 𝑥
(2.5 is just greater than 2)
𝑓 ′ 𝑥 = 4 − 2 2.5 = −1 [𝑒𝑛𝑐𝑒 − 𝑣𝑒]
𝑏 𝑓 𝑥 = 15 + 2𝑥 − 𝑥 2
First, we find the derivative. Then
𝑓 𝑥 = 15 + 2𝑥 − 𝑥
𝑓′ 𝑥 =
equate the derivative to zero, to find the
2
𝑥-value.
2 − 2𝑥
𝐿𝑒𝑎𝑠𝑡 𝑜𝑟 𝑔𝑟𝑒𝑎𝑡𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤𝑒𝑛 𝑓 ′ 𝑥 = 0
∴
We put 𝑥 = 1 into the expression for
2 − 2𝑥 = 0
𝑓 𝑥 to find the 𝑦 −coordinate.
𝑥 = 1
𝐴𝑡 𝑥 = 1; 𝑓 1 = 15 + 2 1 − 12
We find the sign of the gradient on
𝑓 1 = 16 .
either side of the point.
𝑓′ 𝑥 = 2 − 2𝑥
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝐿
1
𝑅
𝑠𝑖𝑔𝑛 𝑜𝑓 𝑓 ′ 𝑥
+
0
−
For example:
𝐿 𝑝𝑢𝑡 𝑥 = 0.5 𝑖𝑛𝑡𝑜 𝑓 ′ 𝑥
𝑓 ′ 𝑥 = 2 − 2 0.5 = +1 [𝑒𝑛𝑐𝑒 + 𝑣𝑒]
𝑅 𝑝𝑢𝑡 𝑥 = 1.5 𝑖𝑛𝑡𝑜 𝑓 ′ 𝑥
𝑔𝑟𝑒𝑎𝑡𝑒𝑠𝑡
𝑓 ′ 𝑥 = 2 − 2 1.5 = −1 [𝑒𝑛𝑐𝑒 − 𝑣𝑒]
𝑇𝑢𝑠 𝑡𝑒 𝑔𝑟𝑒𝑎𝑡𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 16 𝑎𝑛𝑑 𝑜𝑐𝑐𝑢𝑟𝑠 𝑎𝑡
𝑥 =1.
Note: the closer the value used, the less
the chances of a wrong sign.
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02. Differentiation 2
𝑐
𝑦 =
2𝑥 + 3 𝑥 − 2
𝑦 = 2𝑥 2 − 𝑥 − 6
First, we find the derivative. Then
𝑑𝑦
= 4𝑥 − 1
𝑑𝑥
find the 𝑥-value.
equate the derivative to zero, to
𝐿𝑒𝑎𝑠𝑡 𝑜𝑟 𝑔𝑟𝑒𝑎𝑡𝑒𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑐𝑐𝑢𝑟𝑠 𝑎𝑡
𝑑𝑦
=0
𝑑𝑥
∴ 4𝑥 − 1 = 0
𝑥
=
We put 𝑥 =
1
4
1 2
4
1
𝐴𝑡 𝑥 = 4; 𝑦 = 2
=−
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝑠𝑖𝑔𝑛 𝑜𝑓
𝑑𝑦
𝑑𝑥
4
into the expression for
𝑓 𝑥 to find the 𝑦 −coordinate.
1
−4 −6
49
8
𝑝𝑜𝑖𝑛𝑡
𝑑𝑦
𝑑𝑥
1
1
,
4
−
49
8
.
We find the sign of the gradient on
either side of the point.
= 4𝑥 − 1
𝐿
1
4
𝑅
−
0
+
For example:
𝐿 𝑝𝑢𝑡 𝑥 = 0 𝑖𝑛𝑡𝑜
𝑑𝑦
𝑑𝑥
𝑙𝑒𝑎𝑠𝑡
𝑇𝑢𝑠 𝑡𝑒 𝑙𝑒𝑎𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑖𝑠 −
49
8
𝑑𝑦
𝑑𝑥
= 4 0 − 1 = −1 𝑒𝑛𝑐𝑒 − 𝑣𝑒
𝑅 𝑝𝑢𝑡 𝑥 = 1 𝑖𝑛𝑡𝑜 𝑓 ′ 𝑥
𝑎𝑡 𝑡𝑒 𝑝𝑜𝑖𝑛𝑡
1
49
,− 8
4
𝑑𝑦
𝑑𝑥
.
= 4 1 − 1 = +3 [𝑒𝑛𝑐𝑒 + 𝑣𝑒]
Note: the closer the value used, the
less the chances of a wrong sign.
2.3. Application of Greatest and least values
There are scenarios in real life where the practical application of the knowledge of greatest or least values is
required, as shown in the examples below:
Example 3:
A ball is thrown vertically upwards from the ground level and its height after 𝑡 𝑠 is 15.4𝑡 − 4.9𝑡 2 𝑚. Find the
greatest height it reaches and the time it takes to get there.
𝐻𝑒𝑖𝑔𝑡 = 15.4𝑡 − 4.9𝑡 2
𝑑
= 15.4 − 9.8𝑡
𝑑𝑡
First, we find the derivative
𝑑
𝑑𝑡
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02. Differentiation 2
The ball reaches its greatest height when
𝑑
𝑑𝑡 is zero. Actually the ball is
momentarily at rest at this point before
it begins to fall.
𝐺𝑟𝑒𝑎𝑡𝑒𝑠𝑡 𝑒𝑖𝑔𝑡 𝑜𝑐𝑐𝑢𝑟𝑠 𝑎𝑡
𝑑
= 0
𝑑𝑡
We equate 𝑑 𝑑𝑡 to zero at the
greatest point, to find the value
15.4 − 9.8𝑡 = 0
𝑡
=
15.4
9.8
𝑡
=
1.6 𝑠 .
of 𝑡.
𝑇𝑢𝑠 𝑔𝑟𝑒𝑎𝑡𝑒𝑠𝑡 𝑒𝑖𝑔𝑡 𝑖𝑠;
= 15.4 1.6 − 4.9 1.6
=
12.1 𝑚 .
We put 𝑡 = 1.6 into the
2
expression for to find the
greatest height.
Example 4:
A farmer has 100𝑚 of metal railing with which to form two adjacent sides of a rectangular enclosure, the other
two sides being two existing walls of the yard, meeting at right angles. What dimensions will give him the
maximum possible area?
This is a plot of Area against length 𝑥. The
area of the enclosure is maximum at the top
of the curve when 𝑑𝐴 𝑑𝑥 = 0
6
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02. Differentiation 2
𝐿𝑒𝑡 𝑜𝑛𝑒 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡𝑒 𝑚𝑒𝑡𝑎𝑙 𝑙𝑒𝑛𝑔𝑡 𝑏𝑒 𝑥 𝑚
𝑇𝑒 2𝑛𝑑 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡𝑒 𝑚𝑒𝑡𝑎𝑙 𝑤𝑖𝑑𝑡 𝑤𝑖𝑙𝑙
Black is the wall and blue is the metal
railing.
𝑏𝑒 100 − 𝑥
𝐴𝑟𝑒𝑎
100 − 𝑥
𝐴 = 𝑥 100 − 𝑥
𝐴 =
100𝑥 − 𝑥 2
𝑑𝐴
=
𝑑𝑥
100 − 2𝑥
𝑥
Area for a rectangle is given by
𝐴 = 𝑙𝑒𝑛𝑔𝑡 × 𝑤𝑖𝑑𝑡
𝑑𝐴
= 0;
𝑑𝑥
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑟𝑒𝑎 𝑜𝑐𝑐𝑢𝑟𝑠 𝑤𝑒𝑛
We obtain 𝑑𝐴 𝑑𝑥 and equate it to zero
to find the value of𝑥 .
∴ 100 − 2𝑥 = 0
𝑥 = 50 𝑚 .
𝑇𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 𝑡𝑎𝑡 𝑡𝑒 𝑎𝑟𝑒𝑎 𝑖𝑠 max 𝑤𝑒𝑛
𝑥 = 50 𝑚 .
∴ 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑓𝑜𝑟 𝐴𝑚𝑎𝑥 𝑤𝑖𝑙𝑙 𝑏𝑒 50𝑚 𝑏𝑦 50𝑚;
𝐴𝑚𝑎𝑥
= 2500 𝑚2 .
One may put 𝑥 = 50 into the formula
for 𝐴 above, or simply multiply 50 ×
50.
Example 5:
A rectangular sheep pen is to be made out of 1000𝑚 of fencing using an existing straight hedge for one of the
sides. Find the maximum area possible and the dimensions needed to achieve this.
𝐿𝑒𝑡 𝑡𝑒 𝑙𝑒𝑛𝑔𝑡 𝑜𝑓 𝑡𝑒 𝑓𝑒𝑛𝑐𝑒 𝑏𝑒 𝑥
𝑇𝑒 𝑤𝑖𝑑𝑡 = 1000 − 𝑥 − 𝑥
=
Black is the hedge and blue is the
fencing.
𝑥
1000 − 2𝑥 𝑚
1000 − 2𝑥
𝐴𝑟𝑒𝑎 𝐴 = 𝑥 1000 − 2𝑥
𝐴 =
1000𝑥 − 2𝑥 2
𝑑𝐴
=
𝑑𝑥
1000 − 4𝑥 .
𝐴𝑡 𝐴𝑚𝑎𝑥
∴
𝑥
Area for a rectangle is given by
𝐴 = 𝑙𝑒𝑛𝑔𝑡 × 𝑤𝑖𝑑𝑡
𝑑𝐴
= 0
𝑑𝑥
1000 − 4𝑥 = 0
𝑥 = 250 𝑚 .
We obtain 𝑑𝐴 𝑑𝑥 and equate it to
zero to find the value of 𝑥.
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02. Differentiation 2
𝐴𝑚𝑎𝑥
=
Put 𝑥 into the expression for 𝐴 to get
250 1000 − 500
= 125,000 𝑚
maximum area 𝐴𝑚𝑎𝑥
2
and the
dimensions can be substituted from
𝑤𝑖𝑡 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 250 𝑚 𝑏𝑦 500 𝑚 .
the rectangle above.
Example 6:
An aircraft flying level at 250 𝑚 above the ground suddenly swoops down to drop supplies and then regains its
former altitude. It is 𝑚 above the ground 𝑡 𝑠 after beginning its dive, where = 8𝑡 2 − 80𝑡 + 250. Find the
least altitude during this operation and the interval of time during which it is losing height.
The plane drops to its least altitude when 𝑑 𝑑𝑡
is zero. Actually the plane is momentarily at rest
at this point before it begins to rise again.
= 8𝑡 2 − 80𝑡 + 250
We obtain 𝑑 𝑑𝑡 and equate it to
zero to find the value of 𝑡.
𝑑
= 16𝑡 − 80
𝑑𝑡
𝑑
𝑓𝑜𝑟 𝑙𝑒𝑎𝑠𝑡 𝑎𝑙𝑡𝑖𝑡𝑢𝑑𝑒
= 0;
𝑑𝑡
∴
16𝑡 − 80 = 0
𝑡 = 5𝑠 .
𝑙𝑒𝑎𝑠𝑡 𝑒𝑖𝑔𝑡, = 8 52 − 80 5
We put 𝑡 = 5 into the expression
+ 250
for to find the least height.
= 50 𝑚 .
Exercise 2A:
1. Find 𝑑 𝑑𝑥 or 𝑓 ′ 𝑥
(a) 2𝑥 2 − 3𝑥 + 5
of the following functions:
(b) 3𝑥 3 + 4𝑥 − 12
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02. Differentiation 2
(c) 𝑝𝑥 3 − 𝑞𝑥 2 + 2𝑟
(h) 𝑓 𝑥 = −5𝑥 3 − 3𝑥 2 + 2𝑥 − 10
(d) 4𝑥 4 − 2𝑥 3 + 4𝑥 2
(i) 𝑓 𝑥 = 2𝑥 − 3 𝑥 + 4
(e)
𝑥2 − 2 𝑥 + 3
(j) 𝑓 𝑥 = 3𝑥 2 + 4 𝑥 − 5
(f)
2𝑥 − 5 𝑥 2 + 1
(k) 𝑓 𝑥 = 𝑥 + 2 𝑥 − 1 𝑥 + 3
(l) 𝑓 𝑥 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
(g) 𝑓 𝑥 = 4𝑥 2 + 2𝑥 − 3
2.
Find the coordinates of the points on the following curves where the gradient is zero:
(a) 𝑦 = 𝑥 2 − 2𝑥
(e) 𝑓 𝑥 = 3𝑥 2 − 2𝑥
(b) 𝑦 = 5𝑥 2 − 3𝑥 + 10
(f) 𝑓 𝑥 = 6 − 4𝑥 − 𝑥 2
(c) 𝑦 =
2 𝑥3
3
(g) 𝑓 𝑥 = 𝑥 3 + 3 𝑥 2 + 3 𝑥
+ 5 𝑥 2 + 12 𝑥
(h) 𝑓 𝑥 = 2 𝑥 3 + 5 𝑥 2 − 4 𝑥
(d) 𝑦 = 𝑥 𝑥 2 − 2 𝑥 − 4
3.
Find the greatest or least values of the following functions:
(a) 𝑦 = 𝑥 2 + 2
(d) 𝑓 𝑥 = 𝑥 2 + 4𝑥 + 4
(b) 𝑦 = 1 − 2𝑥 − 𝑥 2
(e) 𝑓 𝑥 = 3 − 2𝑥 − 4𝑥 2
(c) 𝑦 = 3𝑥 2 − 6𝑥 + 5
(f) 𝑓(𝑥) = 2𝑥 − 1 2 − 𝑥
4.
The curve 𝑦 = 15 + 𝑏𝑥 – 𝑎𝑥 2 has the greatest point as 1, 16 . Find the values of 𝑎 and 𝑏.
5.
The curve 𝑓 𝑥 = 𝑎𝑥 2 − 𝑏𝑥 + 3 has the least value at the point −2, 5 . Find the values of 𝑎 and 𝑏.
6.
The curve 𝑦 = 𝑎𝑥 2 − 4𝑥 + 𝑐 has the greatest point
7.
A stone is thrown vertically upwards from the ground level and its height after 𝑡 𝑠 is
1
3
,
25
3
. Find the values of 𝑎 and 𝑐.
39.2𝑡 − 4.9𝑡 2 𝑚. Find the greatest height it reaches and the time it takes to get there.
8. A point moves along a straight line so that at the end of 𝑡 𝑠 its distance from a fixed point is 𝑠 = 4 + 8𝑡 −
8𝑡 2 𝑚. Find the least distance of the particle from the fixed point.
9.
A rectangular enclosure is formed by using 1200 𝑚 of fencing. Find the greatest possible area that can be
enclosed in this way and the corresponding dimensions of the rectangle.
10. A train is to make a voyage of 200 𝑘𝑚 at a constant speed. When the speed of the ship is 𝑣 𝑘𝑚 the cost is
$ 𝑣2 +
1024
𝑣
per hour. Find the least cost of the voyage.
2.4. Maxima and Minima
From the previous section;
1.
The greatest value of a function occurs at a maximum point. This maximum value is called the maxima.
2.
The least value of a function occurs at a minimum point. This minimum value is called the minima.
3.
A maximum point such as 𝐴 (fig. 2.2) is obviously not the greatest point on the curve, but is greatest in
relation to the points close to it. Such a point may be called a local maximum. Similarly a point such
as 𝐵 may be called a local minimum.
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02. Differentiation 2
4.
5.
A minimum or maximum point can also be called a turning point.
𝑑𝑦
At any point on a curve where the gradient
𝑑𝑥 is zero, 𝑦 is said to have a stationary value. This
point may also be called a stationary point.
6.
A point where the gradient of the curve is zero but does not change sign is called a point of inflexion 𝑃
(fig. 2.3).
Point A is relatively higher than the points
close to it, so we call it a Local maximum.
Similarly, point B is lower in relation to the
points close it and we call it a Local minimum.
Fig. 2.2
The gradient of the curve simply keeps rising
but at point P, it is momentarily zero, after
which it continues to rise again. Such a point
as P is called an inflexion.
Fig. 2.3
Example 7:
Find the turning points of the curves:
𝑎
𝑦 = 𝑥 𝑥 2 − 12
𝑦 = 𝑥 3 − 12𝑥
𝑑𝑦
= 3𝑥 2 − 12
𝑑𝑥
𝐴𝑡 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡
We obtain
𝑑𝑦
𝑑𝑥
𝑑𝑦
= 0
𝑑𝑥
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02. Differentiation 2
Then equate
∴ 3𝑥 2 − 12 = 0
𝑑𝑥 to zero to find the 𝑥-
value.
𝑥2 = 4
𝑥 = 2 𝑜𝑟 𝑥 = −2 .
Put 𝑥 into the equation of the curve to
𝑦 = 2 22 − 12 = −16
𝐴𝑡 𝑥 = 2;
𝑑𝑦
get the 𝑦 −coordinate.
𝑝𝑜𝑖𝑛𝑡 2, −16
𝐴𝑡 𝑥 = −2; 𝑦 = − 2 −2
2
− 12
= 16
We use the table to investigate the
nature of the turning points above.
𝑝𝑜𝑖𝑛𝑡 −2, 16
𝑑𝑦
𝑑𝑥
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝑠𝑖𝑔𝑛 𝑜𝑓
𝐿
𝑑𝑦
−
𝑑𝑥
= 3𝑥 2 − 12
2
𝑅
0
For 𝑥 = 2;
𝐿
+
−2
+
0
𝑅
𝑑𝑦
𝑑𝑥
−
2
= 3 1.5
𝑚𝑎𝑥𝑖𝑚𝑢𝑚
𝑑𝑦
𝑑𝑥
= 3 2.5
2
𝑑𝑥
− 12 = −5.25 [𝑒𝑛𝑐𝑒 − 𝑣𝑒]
𝑅 𝑝𝑢𝑡 𝑥 = 2.5 𝑖𝑛𝑡𝑜
𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝑑𝑦
𝐿 𝑝𝑢𝑡 𝑥 = 1.5 𝑖𝑛𝑡𝑜
𝑑𝑦
𝑑𝑥
− 12 = 6.75 [𝑒𝑛𝑐𝑒 + 𝑣𝑒]
For 𝑥 = −2;
𝑡𝑒 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑟𝑒; 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 −2, 16
𝐿 𝑝𝑢𝑡 𝑥 = −2.5 𝑖𝑛𝑡𝑜
𝑑𝑦
𝑑𝑥
𝑎𝑛𝑑 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 2, 16 .
= 3 −2.5
2
𝑏
3
= 3 −1.5
2
𝑑𝑥
− 12 = 6.75
𝑅 𝑝𝑢𝑡 𝑥 = −1.5 𝑖𝑛𝑡𝑜
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑒𝑛𝑐𝑒 + 𝑣𝑒
𝑑𝑦
𝑑𝑥
− 12 = −5.25 [𝑒𝑛𝑐𝑒 − 𝑣𝑒]
2
𝑓 𝑥 = 2𝑥 − 3𝑥 − 12𝑥 − 7
𝑓′ 𝑥
= 6𝑥 2 − 6𝑥 − 12
𝐴𝑡 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 𝑓 ′ 𝑥
Obtain 𝑓 ′ 𝑥 and equate it to zero to find
= 0;
the 𝑥-values of the coordinates.
6𝑥 2 − 6𝑥 − 12 = 0
𝑥
2
We solve the quadratic equation by
factorization.
− 𝑥 − 2 = 0
𝑥−2
𝑥 = 2
𝑥+1 = 0
𝑜𝑟 𝑥 = −1
𝐴𝑡 𝑥 = 2; 𝑓 2 = 2 23 − 3 22 − 12 2 − 7
𝑓 2 = −27
𝐴𝑡 𝑥 = −1; 𝑓 −1 = 2 −1
𝑓 −1 = 0
𝑝𝑜𝑖𝑛𝑡 2, −27
3
− 3 −1
2
− 12 −1 − 7
Put 𝑥 into the equation of the curve to
get the 𝑦 −coordinate.
We use the table below to determine
the nature of the turning points.
𝑝𝑜𝑖𝑛𝑡 −1, 0 .
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02. Differentiation 2
For 𝑥 = 2;
′
𝐿 𝑝𝑢𝑡 𝑥 = 1.5 𝑖𝑛𝑡𝑜 𝑓 ′ 𝑥 :
2
𝑓 𝑥
= 6𝑥 − 6𝑥 − 12
2
6 1.5
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝐿
2
𝑅
𝐿
−1
𝑅
𝑠𝑖𝑔𝑛 𝑜𝑓 𝑓 ′ 𝑥
−
0
+
+
0
−
𝑚𝑖𝑛𝑖𝑚𝑢𝑚
− 6 1.5 − 12 − 7.5 [𝑒𝑛𝑐𝑒 − 𝑣𝑒]
𝑅 𝑝𝑢𝑡 𝑥 = 2.5 𝑖𝑛𝑡𝑜 𝑓 ′ 𝑥 :
6 2.5
2
− 6 2.5 − 12 = 10.5 [𝑒𝑛𝑐𝑒 + 𝑣𝑒]
For 𝑥 = −1;
𝐿 𝑝𝑢𝑡 𝑥 = −1.5 𝑖𝑛𝑡𝑜 𝑓 ′ 𝑥 :
𝑚𝑎𝑥𝑖𝑚𝑢𝑚
6 −1.5
𝑡𝑒 𝑡𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑟𝑒; 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 −1, 0 𝑎𝑛𝑑
𝑚𝑖𝑛𝑖𝑚𝑢𝑚 2, −27 .
2
− 6 −1.5 − 12 = 10.5 [𝑒𝑛𝑐𝑒 + 𝑣𝑒]
𝑅 𝑝𝑢𝑡 𝑥 = −0.5 𝑖𝑛𝑡𝑜 𝑓 ′ 𝑥 :
6 −0.5
2
Obtain
𝑑𝑦
− 6 0.5 − 12 = −13.5 [𝑒𝑛𝑐𝑒 − 𝑣𝑒]
Example 8:
𝐼𝑛𝑣𝑒𝑠𝑡𝑖𝑔𝑎𝑡𝑒 𝑡𝑒 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑡𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑐𝑢𝑟𝑣𝑒𝑠:
𝑎
𝑦 = 𝑥3 2 − 𝑥
𝑦 = 2𝑥 3 − 𝑥 4
𝑑𝑦
= 6𝑥 2 − 4𝑥 3
𝑑𝑥
𝑑𝑦
𝐴𝑡 𝑡𝑒 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒;
= 0
𝑑𝑥
Put 𝑥 into the equation of the curve to get
the 𝑦 −coordinate. We use the table below
= 0
𝑜𝑟
3 3
2
3
𝐴𝑡 𝑥 = 2; 𝑦 =
𝑠𝑖𝑔𝑛 𝑜𝑓
𝑑𝑦
𝑑𝑥
𝐿
+
= 0
For 𝑥 = 0;
0, 0
2 −
𝑝𝑜𝑖𝑛𝑡
points.
3
2
𝑥 =
𝑦 = 0 2 − 0
𝑑𝑦
𝑑𝑥
to determine the nature of the turning
𝑜𝑟 6 − 4𝑥 = 0
𝑝𝑜𝑖𝑛𝑡
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
We’ve solved the equation by factorization.
𝑥 2 6 − 4𝑥
𝑥=0
𝐴𝑡 𝑥 = 0;
𝑥-values of the coordinates.
6𝑥 2 − 4𝑥 3 = 0
𝑒𝑖𝑡𝑒𝑟 𝑥 2 = 0
∴
and equate it to zero to find the
𝑑𝑥
3
2
=
27
16
.
0
𝑅
+
𝐿
+
− 4 −0.5
6 0.5
2
− 4 0.5
3
𝑑𝑥
= 2 [𝑒𝑛𝑐𝑒 + 𝑣𝑒]
𝑑𝑥
= 1 [𝑒𝑛𝑐𝑒 + 𝑣𝑒]
For 𝑥 = 3 2 ;
3
2
0
𝑅
−
𝐿 𝑝𝑢𝑡 𝑥 = 1 𝑖𝑛𝑡𝑜
𝑚𝑎𝑥𝑖𝑚𝑢𝑚
𝑑𝑦
𝑑𝑥
6 1 − 4 1 = 2 [𝑒𝑛𝑐𝑒 + 𝑣𝑒]
𝑑𝑦
𝑅 𝑝𝑢𝑡 𝑥 = 2 𝑖𝑛𝑡𝑜
𝑖𝑛𝑓𝑙𝑒𝑥𝑖𝑜𝑛
3
𝑑𝑦
𝑅 𝑝𝑢𝑡 𝑥 = 0.5 𝑖𝑛𝑡𝑜
= 6𝑥 2 − 4𝑥 3
0
2
6 −0.5
3 27
,
2 16
𝑑𝑦
𝐿 𝑝𝑢𝑡 𝑥 = −0.5 𝑖𝑛𝑡𝑜
6 2
2
−4 2
3
𝑑𝑥
= −8 [𝑒𝑛𝑐𝑒 − 𝑣𝑒]
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02. Differentiation 2
𝑡𝑒 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒𝑠 𝑎𝑟𝑒 0 𝑎𝑛𝑑
3
𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑓𝑙𝑒𝑥𝑖𝑜𝑛 𝑎𝑛𝑑 𝑝𝑜𝑖𝑛𝑡
2
,
27
16
;
27
0, 0 𝑖𝑠 𝑎
𝑖𝑠 𝑡𝑒
16
As noted above, a point of inflexion is
one where the sign of the gradient
remains the same at either side.
𝑚𝑎𝑥𝑖𝑚𝑢𝑚.
𝑏
𝑦 = 3𝑥 4 + 16𝑥 3 + 24𝑥 2 + 3
𝑑𝑦
= 12𝑥 3 + 48𝑥 2 + 48𝑥
𝑑𝑥
𝑑𝑦
𝐴𝑡 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑝𝑜𝑖𝑛𝑡
= 0
𝑑𝑥
Obtain
12𝑥 𝑥 + 4𝑥 + 4
𝑒𝑖𝑡𝑒𝑟 12𝑥 = 0
=
We’ve solved the equation by
factorization.
0
2
𝑜𝑟
𝑥= 0
and equate it to zero to find
𝑑𝑥
the 𝑥-values of the coordinates.
12𝑥 3 + 48𝑥 2 + 48𝑥 = 0
2
𝑑𝑦
𝑥 + 4𝑥 + 4 = 0
𝑜𝑟
𝑥+2
2
= 0
𝑥 = −2
Put 𝑥 into the equation of the curve to
𝐴𝑡 𝑥 = 0; 𝑦 = 0 + 0 + 0 + 3
𝑦 = 3
𝐴𝑡 𝑥 = −2; 𝑦 = 3 −2
𝑦 =
𝑑𝑦
𝑑𝑥
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝑠𝑖𝑔𝑛 𝑜𝑓
𝑑𝑦
𝑑𝑥
get the 𝑦-coordinate. We use the table
𝑝𝑜𝑖𝑛𝑡 0, 3
4
+ 16 −2
3
+ 24 −2
2
+3
below to determine the nature of the
turning points.
19 𝑝𝑜𝑖𝑛𝑡 −2, 19 .
For 𝑥 = 0;
= 12𝑥 3 + 48𝑥 2 + 48𝑥
𝑑𝑦
𝐿 𝑝𝑢𝑡 𝑥 = −0.5 𝑖𝑛𝑡𝑜
𝐿
0
𝑅
𝐿
−2
𝑅
−
0
+
−
0
−
3
12 −0.5
𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝑖𝑛𝑓𝑙𝑒𝑥𝑖𝑜𝑛
𝑑𝑦
𝑅 𝑝𝑢𝑡 𝑥 = 0.5 𝑖𝑛𝑡𝑜
12 0.5
3
+ 48 0.5
𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑎𝑛𝑑
0, 3 𝑖𝑠 𝑎
−2, 19 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑓𝑙𝑒𝑥𝑖𝑜𝑛 .
𝑑𝑥
+ 48 −0.5 = −13.5
𝑑𝑥
2
+ 48 0.5 = + 37.5
For 𝑥 = −2
𝑑𝑦
𝐿 𝑝𝑢𝑡 𝑥 = −2.5 𝑖𝑛𝑡𝑜
𝑡𝑒 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑣𝑎𝑙𝑢𝑒𝑠 𝑎𝑟𝑒 19 𝑎𝑛𝑑 3;
2
+ 48 −0.5
12 −2.5
3
+ 48 −2.5
𝑅 𝑝𝑢𝑡 𝑥 = −1.5 𝑖𝑛𝑡𝑜
12 1.5
3
+ 48 1.5
2
2
𝑑𝑦
𝑑𝑥
+ 48 −2.5 = −7.5
𝑑𝑥
+ 48 1.5 = + 220.5
2.5. Maximum and minimum problems
Problems involving maximum and minimum points arise often in real life, in which case the first task is to derive an
equation relating the two variables. The stationary points can then be determined and their practical significance
interpreted. The following examples illustrate these case scenarios.
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02. Differentiation 2
Example 9:
A chemical factory wishes to make a cylindrical container of thin metal to hold 10 𝑐𝑚3 using the least possible
area of metal. If the outside surface area is 𝑠 𝑐𝑚2 and the radius is 𝑟 𝑐𝑚, show that 𝑠 = 2𝜋𝑟 2 +
20
𝑟
and hence
find the required radius and height of the container (leave 𝜋 in your answer).
𝒔 = 𝟐𝝅𝒓𝟐 +
𝑺𝒉𝒐𝒘𝒊𝒏𝒈 𝒕𝒉𝒂𝒕
𝟐𝟎
𝒓
We put the value 𝑣 = 10 into the
formula for volume.
𝑣 = 𝜋𝑟 2 ; 𝑏𝑢𝑡 𝑔𝑖𝑣𝑒𝑛 𝑣 = 10 𝑐𝑚3
∴ 𝜋𝑟 2 = 10
10
𝜋𝑟 2
=
We make the subject
𝑇𝑜𝑡𝑎𝑙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎, 𝑠 = 2𝜋𝑟 2 + 2𝜋𝑟
𝑠 = 2𝜋𝑟 2 +
20
𝑟
We substitute for in the formula
10
𝜋𝑟 2
∴ 𝑠 = 2𝜋𝑟 2 + 2𝜋𝑟 ×
for 𝑠 and hence eliminate .
𝑒𝑛𝑐𝑒 𝑠𝑜𝑤𝑛.
𝑭𝒊𝒏𝒅𝒊𝒏𝒈 𝒕𝒉𝒆 𝒓𝒂𝒅𝒊𝒖𝒔 𝒓 𝒂𝒏𝒅 𝒉𝒆𝒊𝒈𝒉𝒕 𝒉
𝑑𝑠
= 4𝜋𝑟 + 20 −𝑟 −2
𝑑𝑟
𝑑𝑠
= 4𝜋𝑟 −
𝑑𝑟
𝑑𝑠
= 0;
𝑑𝑟
𝐹𝑜𝑟 𝑠𝑚𝑖𝑛
∴ 4𝜋𝑟 −
20
𝑟2
10
=
𝑟
=
𝜋×
5 2 3
𝜋
=
occurs.
20
𝑟2
= 0
= 4𝜋𝑟
𝑟3
𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑟 𝑖𝑛 =
=
For least surface area 𝑠𝑚𝑖𝑛 , we
find 𝑑𝑠 𝑑𝑟 and equate it to zero to
find the value of 𝑟 at which it
20
𝑟2
5
𝜋
3 5
𝜋
𝑐𝑚
We substitute the value of
𝑟 obtained into the expression for
10
𝜋𝑟 2
obtained above.
2 × 51 × 5−2
𝜋 1 × 𝜋 −2 3
= 2
𝑡𝑢𝑠 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑎𝑑𝑖𝑢𝑠 𝑖𝑠
3 5
𝜋
3 5
𝜋
3
=
2 × 51
𝜋1 3
3
Here we use the Algebra of indices
to simplify the value of .
𝑐𝑚 .
𝑐𝑚 𝑎𝑛𝑑 𝑒𝑖𝑔𝑡 2
[One could as well simply use a calc
3 5
𝜋
unless stated otherwise]
𝑐𝑚 .
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02. Differentiation 2
Example 10:
The figure below represents a rectangular sheet of metal 8𝑐𝑚 𝑏𝑦 5𝑐𝑚. Equal squares of side 𝑥 𝑐𝑚 are removed
from each corner and the edges are then turned to make an open box of volume 𝑣 𝑐𝑚3 . Show that 𝑣 =
40𝑥 − 26𝑥 2 + 4𝑥 3 .
Hence find the maximum possible volume and the corresponding value of 𝑥.
The box produced is
shown. Both length and
width are less by 2𝑥
5𝑐𝑚
𝒙
5 − 2𝑥
𝑥
𝒙
the two squares of
𝑥 cm removed from
8 𝑐𝑚
each corner.
8 − 2𝑥
𝑣 = 𝑙×𝑤×
𝑣 =
that corresponds to
We quote the formula for finding the
8 − 2𝑥 × 5 − 2𝑥 × 𝑥
volume of a box, substitute and expand.
𝑣 = 40𝑥 − 16𝑥 2 − 10𝑥 2 + 4𝑥 3
𝑣 = 40𝑥 − 26𝑥 2 + 4𝑥 3 ; 𝑒𝑛𝑐𝑒 𝑠𝑜𝑤𝑛.
𝑑𝑣
= 40 − 52𝑥 + 12𝑥 2
𝑑𝑥
Maximum volume occurs when 𝑑𝑣 𝑑𝑥 = 0.
So we find the derivative and equate it
𝑑𝑣
= 0;
𝑑𝑥
𝐴𝑡 𝑣𝑚𝑎𝑥
to zero to obtain the values of 𝑥.
∴ 12𝑥 2 − 52𝑥 + 40 = 0
We can solve the equation by
3𝑥 2 − 13𝑥 + 10 = 0
𝑥−1
3𝑥 − 10
𝑥 = 1
𝑑𝑣
𝑑𝑥
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝑠𝑖𝑔𝑛 𝑜𝑓
𝑑𝑣
𝑑𝑥
factorization
= 0
10
3
𝑜𝑟 𝑥 =
We use the table to determine which of
= 40 − 52𝑥 + 12𝑥
the two values of 𝑥 obtained produces
2
the maximum volume.
𝐿
10
3
𝑅
𝐿
1
𝑅
−
0
+
+
0
−
To get the sign of
𝑑𝑣
𝑑𝑥
, we substitute
close values of 𝑥 on either side of the
𝑚𝑖𝑛𝑖𝑚𝑢𝑚
𝑚𝑎𝑥𝑖𝑚𝑢𝑚
𝑇𝑢𝑠 𝑣𝑚𝑎𝑥 𝑜𝑐𝑐𝑢𝑟𝑠 𝑎𝑡 𝑥 = 1;
𝑣𝑚𝑎𝑥 = 40 − 26 + 4
𝑣𝑚𝑎𝑥 = 18 𝑐𝑚
3
point.
We put 𝑥 = 1 into the expression for 𝑣
above to find the maximum volume.
.
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02. Differentiation 2
Example 11:
A piece of wire of length of length 𝑙 is cut into two parts 𝑥 and 𝑙 − 𝑥. The former is bent into the shape of a
square and the latter into a rectangle of which the base is double the height. Find an expression for the sum of the
areas of the two figures. Prove that the only value of 𝑥 for which this sum is a maximum or minimum is 𝑥 =
8𝑙
17
and find which it is.
RECTANGLE
SQUARE
1
𝑥
4
1
𝑥
4
𝑦
2𝑦
If the height (or width)
is 𝑦, the length is double,
𝑙𝑒𝑡 𝑡𝑒 𝑤𝑖𝑑𝑡 𝑏𝑒 𝑦
2 2𝑦 + 𝑦
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒
2
1
𝑥
4
𝐴𝑠 =
𝐴𝑠 =
1 2
𝑥
16
= 𝑙−𝑥
6𝑦
= 𝑙−𝑥
𝑦
=
1
6
hence 2𝑦. But the
perimeter is equal to the
part 𝑙 − 𝑥 . Perimeter is
𝑙−𝑥
given by 2(𝑙𝑒𝑛𝑔𝑡 +
𝑤𝑖𝑑𝑡).
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑐𝑡𝑎𝑛𝑔𝑙𝑒
𝐴𝑟 = 𝑙𝑒𝑛𝑔𝑡 × 𝑤𝑖𝑑𝑡
Perimeter of square equals to
part 𝑥; hence each side is
1
4
𝐴𝑟 =
𝑥.
𝐴𝑟 =
2
6
𝑙−𝑥 ×
1
18
𝑙−𝑥
1
6
𝑙−𝑥
2
𝑭𝒊𝒏𝒅𝒊𝒏𝒈 𝒕𝒉𝒆 𝒔𝒖𝒎 𝑺 𝒐𝒇 𝒕𝒉𝒆 𝒕𝒘𝒐 𝒂𝒓𝒆𝒂𝒔
𝑆𝑢𝑚 𝑜𝑓 𝐴𝑟𝑒𝑎𝑠 𝑆
=
𝑆 =
𝑆 =
We find the sum of the area of the
𝐴𝑠 + 𝐴𝑟
1 2
𝑥
16
1 2
𝑥
16
+
1
18
square and area of the rectangle. We
2
𝑙−𝑥
+
1
18
1 2
𝑙
18
then expand the expression and
𝑙 2 − 2𝑥𝑙 + 𝑥 2
𝑆 =
1 2
𝑥
16
+
𝑆 =
17 2
𝑥
144
− 9𝑙𝑥 +
1
− 9𝑥𝑙 +
1
1 2
𝑥
18
1 2
𝑙
18
𝑭𝒊𝒏𝒅𝒊𝒏𝒈 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒙 𝒇𝒐𝒓 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒓 𝒎𝒊𝒏𝒊𝒎𝒖𝒎 𝑺:
𝑑𝑆
=
𝑑𝑥
17
𝑥
72
−
Note that 𝑙 is treated as a constant
since we are differentiating with
1
𝑙
9
𝐹𝑜𝑟 𝑆 𝑡𝑜 𝑏𝑒 𝑎 𝑚𝑎𝑥𝑖𝑚𝑎 𝑜𝑟 𝑚𝑖𝑛𝑖𝑚𝑎
simplify.
respect to 𝑥.
𝑑𝑆
= 0;
𝑑𝑥
16
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02. Differentiation 2
∴
17
𝑥
72
1
9
−
𝑙 = 0
1
𝑙
9
𝑥 =
We obtain 𝑥 in terms of 𝑙 .
72
17
×
8
𝑙
17
𝑥 =
𝑒𝑛𝑐𝑒 𝑝𝑟𝑜𝑣𝑒𝑛.
𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒊𝒏𝒈 𝒘𝒉𝒆𝒕𝒉𝒆𝒓 𝑺 𝒊𝒔 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒓 𝒎𝒊𝒏𝒊𝒎𝒖𝒎
𝑑𝑠
𝑑𝑥
17
𝑥
72
=
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝑠𝑖𝑔𝑛 𝑜𝑓
𝑑𝑠
𝑑𝑥
𝑅
0
+
𝑑𝑥
=−
=−
𝑑𝑥
=
=
𝐹𝑟𝑜𝑚 𝑆 =
𝑆𝑚𝑖𝑛 =
1
18
+
1 8𝑙 2
16 17
𝑙2
34
𝑆𝑚𝑖𝑛 =
𝑙−𝑥
+
1
18
𝑙
in the table
17
72
25
72
1
𝑙−
9
𝑙
𝑙 (−𝑣𝑒)
𝑅 𝑝𝑢𝑡 𝑥 = 𝑙 𝑖𝑛𝑡𝑜 𝑑𝑠 𝑑𝑥
𝑑𝑆
𝑚𝑖𝑛𝑖𝑚𝑢𝑚
1 2
𝑥
16
8
17
as previously:
𝑑𝑆
8
𝑙
17
−
𝑥=
𝐿 𝑝𝑢𝑡 𝑥 = −𝑙 𝑖𝑛𝑡𝑜 𝑑𝑠 𝑑𝑥
1
− 9𝑙
𝐿
We still deal with
17
72
1
8
𝑙 −
1
9
𝑙
𝑙 (+𝑣𝑒)
2
8𝑙 2
𝑙 − 17
We substitute for 𝑥 into the expression
.
for 𝑆 to find the minimum value.
Exercise 2B:
1.
Find the turning points of the following curves:
(a) 𝑦 = 𝑥 2 − 8𝑥
(e) 𝑦 = 2 𝑥 3 + 2 𝑥 2 − 10 𝑥 + 15
(b) 𝑦 = 2𝑥 2 + 4𝑥 + 7
3
2.
(f) 𝑓 𝑥 = 2𝑥 𝑥 − 5
2
(c) 𝑦 = 𝑥 + 4𝑥 + 4𝑥 + 2
(g) 𝑓 𝑥 = 9 − 6𝑥 − 4𝑥 2
(d) 𝑦 = 𝑥 3 + 𝑥 2 − 8 𝑥 + 15
(h) 𝑓 𝑥 = 12 + 5 𝑥 − 𝑥 2 − 𝑥 3
Investigate the stationary values of the following curves:
(a) 𝑦 = 3𝑥 2 + 5𝑥 − 7
(d) 𝑓 𝑥 = 𝑥 3 2𝑥 − 4
(b) 𝑦 = 𝑥 2 2𝑥 + 4
(e) 𝑦 = 𝑥 4 + 12 𝑥 3 + 40 𝑥 2 + 5
3
4
2
3
(c) 𝑓 𝑥 = 5 + 10𝑥 − 𝑥 2 − 𝑥 3
3.
The turning point of the curve 𝑦 = 𝑎𝑥 2 − 𝑏𝑥 + 9 𝑖𝑠 2, −11 . Find the values of 𝑎 and 𝑏.
4.
The curve 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 crosses the 𝑦 −axis at the point (0, 3) and has a stationary point (1, 2). Find the
values of 𝑎, 𝑏 and 𝑐.
5.
The gradient of the tangent to the curve 𝑦 = 𝑝𝑥 2 − 𝑞𝑥 − 𝑟 at the point (1, −2) is 1. One point of intersection
of the curve with the 𝑥 −axis is (−1, 0).
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02. Differentiation 2
Find the turning point of the curve and the other point of intersection with the 𝑥-axis and sketch the curve.
6.
𝑦 is a quadratic function of 𝑥. The line 𝑦 = 2𝑥 + 2 is a tangent to the curve at the point
2
(1, 4). The turning point on the curve occurs where 𝑥 = . Find the equation of the curve.
3
7.
The following facts are known about the function 𝑦 = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑;
(a) passes through the point (0, −3),
(b) has a stationary point at (−1, 1),
𝑑2 𝑦
𝑑𝑥 2 = 2 when 𝑥 = 1.
Find the values of 𝑎, 𝑏, 𝑐 and 𝑑.
(c)
8.
A door is in the shape of a rectangle surmounted by a semicircle whose diameter is equal to the width of the
rectangle. If the perimeter of the door is 7𝑚, and the radius of the semicircle is 𝑟 𝑚𝑒𝑡𝑟𝑒𝑠, express the height
of the rectangle in terms of 𝑟. Show that the area of the door has a maximum value when 𝑟 is
9.
7
(4+𝜋)
.
An open tank is constructed, with a square base and vertical sides, to hold 32 𝑚3 of water. Find the
dimensions of the tank if the area of the metal sheet used is to have a minimum value.
10. A farmer has a 80 𝑚 length of fencing. He wants to use it to form three sides of a rectangular enclosure
against an existing fence which provides the fourth side. Find the maximum area that he can enclose and give
its dimensions.
11. An open box is made from a square sheet of cardboard, with sides half a 𝑚𝑒𝑡𝑟𝑒 long, by cutting out a square
from each corner, folding up the sides and joining the cut edges. Find the maximum capacity of the box.
12. A cylinder has a radius 𝑟 metres and a height metres. The sum of the radius and height is 2 𝑐𝑚. Find an
expression for the volume 𝑉 in 𝑐𝑚3 of the cylinder in terms of 𝑟 only. Hence find the maximum volume.
13. The height, metres of an object thrown vertically upwards at time 𝑡 seconds after it is released is given
by = 45 − 𝑡 − 2𝑡 2 .
(a) Calculate how long it takes for the object to return to its point of projection.
(b) Find the value of 𝑡 at which the object is momentarily stationary, and hence calculate the maximum
height reached by the object.
14. The displacement 𝑠 𝑚𝑒𝑡𝑟𝑒𝑠 of an object from some fixed point 𝑂 at time 𝑡 seconds is given by 𝑠 = 𝑡 3 − 3𝑡 2 +
4𝑡 + 5.
(a) Find the expression for the velocity 𝑣 and acceleration 𝑎 at time 𝑡.
(b) Show that the object is never stationary.
(c) Calculate the average speed of the object between 𝑡 = 0 and 𝑡 = 3 𝑠.
2.6. Curve sketching
We shall now apply the idea of maxima and minima as an introduction to sketching curves of functions. However,
in 𝑐𝑎𝑝𝑡𝑒𝑟 16, which is completely dedicated to curve sketching further methods are explored.
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02. Differentiation 2
Example 12:
Find where the following curves meet the axes. Find also the coordinates of their stationary points and use these
results to sketch the curves.
1. 𝑦 = 3𝑥 2 − 𝑥 3
𝑎 𝑇𝑜 𝑓𝑖𝑛𝑑 𝑤𝑒𝑟𝑒 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 𝑚𝑒𝑒𝑡𝑠 𝑡𝑒 𝑥 − 𝑎𝑥𝑖𝑠
𝑝𝑢𝑡 𝑦 = 0;
3𝑥 2 − 𝑥 3 = 0
𝑥2 3 − 𝑥
𝑥 =0
At the 𝑥-intercept, put 𝑦 = 0 to
= 0
𝑜𝑟
find the values of 𝑥.
𝑥 =3
∴ 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 𝑚𝑒𝑒𝑡𝑠 𝑡𝑒 𝑎𝑥𝑖𝑠 𝑎𝑡 0, 0 𝑎𝑛𝑑
3, 0 .
𝑏 𝑇𝑜 𝑓𝑖𝑛𝑑 𝑤𝑒𝑟𝑒 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 𝑚𝑒𝑒𝑡𝑠 𝑡𝑒 𝑦 − 𝑎𝑥𝑖𝑠
𝑝𝑢𝑡 𝑥 = 0;
𝑦 = 3 0 − 0
At the 𝑦-intercept, put 𝑥 = 0 to
find the value of 𝑦.
𝑦 = 0
∴ 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 𝑚𝑒𝑒𝑡𝑠 𝑡𝑒 𝑦 − 𝑎𝑥𝑖𝑠 𝑎𝑡 0, 0 .
𝑐 𝑇𝑜 𝑓𝑖𝑛𝑑 𝑡𝑒 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑛𝑑 𝑡𝑒𝑖𝑟 𝑛𝑎𝑡𝑢𝑟𝑒;
2
𝑦 = 3𝑥 − 𝑥
3
𝑥=0
curve to find the 𝑦-values.
= 0
We use the table to determine
the nature of the points.
𝑜𝑟 𝑥 = 2
𝐴𝑡 𝑥 = 0;
𝑦 = 0
𝐴𝑡 𝑥 = 2;
𝑦 = 4 𝑝𝑜𝑖𝑛𝑡 2, 4
𝑑𝑦
𝑑𝑥
𝑝𝑜𝑖𝑛𝑡 0, 0
= 6𝑥 − 3𝑥 2
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝐿
0
𝑅
𝐿
2
𝑅
𝑑𝑦
𝑑𝑥
−
0
+
+
0
−
𝑠𝑖𝑔𝑛 𝑜𝑓
𝑑𝑥 to zero to find the
𝑥-coordinates of the stationary
We put the 𝑥-values into the
6𝑥 − 3𝑥 2 = 0
3𝑥 2 − 𝑥
𝑑𝑦
points.
𝑑𝑦
= 6𝑥 − 3𝑥 2
𝑑𝑥
∴
Equate
∴ 𝑡𝑒 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑟𝑒
0, 0 𝑚𝑖𝑛𝑖 𝑎𝑛𝑑
2, 4 𝑚𝑎𝑥𝑖.
𝑇𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑠 𝑎𝑟𝑒 𝑡𝑒𝑛 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑠𝑘𝑒𝑡𝑐 𝑡𝑒 𝑔𝑟𝑎𝑝 𝑜𝑛 𝑡𝑒
𝑟𝑖𝑔𝑡.
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02. Differentiation 2
2. 𝑦 = 𝑥 3 − 2𝑥 2 + 𝑥
𝑎 𝑤𝑒𝑟𝑒 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 𝑚𝑒𝑒𝑡𝑠 𝑡𝑒 𝑥 − 𝑎𝑥𝑖𝑠 𝑝𝑢𝑡 𝑦 = 0;
𝑥 3 − 2𝑥 2 + 𝑥 = 0
At the 𝑥-intercept, put 𝑦 = 0 to
𝑥 𝑥 2 − 2𝑥 + 1 = 0
find the values of 𝑥.
𝑥 = 0 𝑜𝑟 𝑥 2 − 2𝑥 + 1 = 0
2
𝑥−1
𝑝𝑜𝑖𝑛𝑡𝑠
0, 0
We’ve solved the equation by
factorization.
= 0
𝑥
= 1
𝑎𝑛𝑑
1, 0 .
𝑏 𝑤𝑒𝑟𝑒 𝑐𝑢𝑟𝑣𝑒 𝑚𝑒𝑒𝑡𝑠 𝑦 − 𝑎𝑥𝑖𝑠 𝑝𝑢𝑡 𝑥 = 0;
At the 𝑦-intercept, put 𝑥 = 0 to
𝑦=0 − 0 + 0= 0
𝑝𝑜𝑖𝑛𝑡
find the value of 𝑦.
0, 0 .
𝑐 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑛𝑑 𝑡𝑒𝑖𝑟 𝑛𝑎𝑡𝑢𝑟𝑒;
𝑦 = 𝑥 3 − 2𝑥 2 + 𝑥
Equate
𝑑𝑦
= 3𝑥 2 − 4𝑥 + 1
𝑑𝑥
points.
𝑑𝑦
𝑑𝑥 to zero to find the 𝑥coordinates of the stationary
∴ 3𝑥 2 − 4𝑥 + 1 = 0
𝑥−1
3𝑥−1 = 0
𝑥 = 1 𝑜𝑟 𝑥 =
𝐴𝑡 𝑥 = 1;
1
𝐴𝑡 𝑥 = 3;
𝑦 = 0
𝑦 =
𝑑𝑦
𝑑𝑥
4
27
We put the 𝑥-values into the curve
1
3
to find the 𝑦-values.
We then use the table to
𝑝𝑜𝑖𝑛𝑡 1, 0
𝑝𝑜𝑖𝑛𝑡
1
4
,
3 27
determine the nature of the points.
.
= 3𝑥 2 − 4𝑥 + 1
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥
𝐿
1
𝑅
𝐿
1
3
𝑅
𝑑𝑦
𝑑𝑥
−
0
+
+
0
−
1
4
,
3 27
𝑚𝑎𝑥𝑖
𝑠𝑖𝑔𝑛 𝑜𝑓
𝑝𝑜𝑖𝑛𝑡𝑠 1, 0 𝑚𝑖𝑛𝑖 𝑎𝑛𝑑
𝑇𝑒𝑠𝑒 𝑟𝑒𝑠𝑢𝑙𝑡𝑠 𝑎𝑟𝑒 𝑡𝑒𝑛 𝑢𝑠𝑒𝑑 𝑡𝑜 𝑠𝑘𝑒𝑡𝑐 𝑡𝑒 𝑐𝑢𝑟𝑣𝑒 𝑜𝑛 𝑡𝑒
𝑡𝑒 𝑟𝑖𝑔𝑡.
3. 𝑦 = 3𝑥 5 − 5𝑥 3
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02. Differentiation 2
𝑎 𝑓𝑜𝑟 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 𝑝𝑢𝑡 𝑦 = 0;
3𝑥 5 − 5𝑥 3 = 0
At the 𝑥-intercept, put 𝑦 = 0 to
𝑥 3 3𝑥 2 − 5 = 0
𝑥 = 0
𝑥2 =
𝑥
𝑝𝑜𝑖𝑛𝑡𝑠
find the values of 𝑥.
3𝑥 2 − 5 = 0
𝑜𝑟
5
,
3
0, 0
We’ve solved the equation by
factorization.
5
3
5
3
=±
0
𝑎𝑛𝑑
−
5
,
3
0
At the 𝑦-intercept, put 𝑥 = 0 to
.
find the value of 𝑦.
𝑏 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 𝑖𝑠 0, 0
𝑐 𝑠𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑟𝑦 𝑝𝑜𝑖𝑛𝑡𝑠 𝑎𝑛𝑑 𝑡𝑒𝑖𝑟 𝑛𝑎𝑡𝑢𝑟𝑒;
𝑑𝑦
= 15𝑥 4 − 15𝑥 2
𝑑𝑥
Equate
𝑑𝑥 to zero to find the 𝑥coordinates of the stationary
points.
15𝑥 4 − 15𝑥 2 = 0
∴
𝑑𝑦
15𝑥 2 𝑥 2 − 1 = 0
𝑥=0
We put the 𝑥-values into the curve
𝑜𝑟 𝑥 = 1 𝑜𝑟 𝑥 = −1
𝐴𝑡 𝑥 = 0; 𝑦 = 0
to find the 𝑦-values.
We then use the table to
𝑝𝑜𝑖𝑛𝑡 0, 0
determine the nature of the points.
𝐴𝑡 𝑥 = 1; 𝑦 = −2 𝑝𝑜𝑖𝑛𝑡 1, − 2
𝑎𝑡 𝑥 = −1; 𝑦 = 2
𝑑𝑦
𝑑𝑥
𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑥 𝐿
−1, 2
= 15𝑥 4 − 15𝑥 2
0
𝑅
𝐿
1
𝑅
𝐿
0
−
−
0
+
+
𝑠𝑖𝑔𝑛 𝑜𝑓
𝑑𝑦
𝑑𝑥
𝑝𝑜𝑖𝑛𝑡𝑠
0, 0 𝑖𝑛𝑓𝑙,
−
𝑝𝑜𝑖𝑛𝑡
1, −2 min 𝑎𝑛𝑑
−1 𝑅
0
−
−1, 2 max .
𝑇𝑒 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠 𝑠𝑘𝑒𝑡𝑐𝑒𝑑 𝑎𝑠 𝑠𝑜𝑤𝑛 𝑜𝑛 𝑡𝑒 𝑟𝑖𝑔𝑡.
The reader may later review 𝑐𝑎𝑝𝑡𝑒𝑟 16 which is dedicated to curve sketching, where more elaborate methods
have been explained.
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02. Differentiation 2
Exercise 2C:
Find where the following curves meet the axes. Find also the coordinates of their stationary points and use these
results to sketch the curves:
1. 𝑦 = 𝑥 2 + 4𝑥
6. 𝑦 = 4𝑥 4 − 3𝑥 3 + 1
2. 𝑦 = 𝑥 2 𝑥 − 2
7. 𝑦 = 2𝑥 4 + 5𝑥 3 − 12𝑥 2 + 7𝑥 − 1
3. 𝑦 = 𝑥 3 − 2𝑥 2 + 5𝑥 − 3
8. 𝑦 = 𝑥 3 𝑥 2 − 1
4. 𝑦 = 2𝑥 − 1
2
9. 𝑦 = 𝑥 5 + 2𝑥 4
𝑥+1
5. 𝑦 = 𝑥 3 𝑥 + 2
10. 𝑦 = 2𝑥 5 − 3𝑥 4 + 12
2.7. Second derivative
𝑊𝑒 𝑎𝑣𝑒 𝑙𝑒𝑎𝑟𝑛𝑡 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒
𝑑𝑦
. 𝐴𝑛 𝑎𝑡𝑡𝑒𝑚𝑝𝑡 𝑡𝑜 𝑓𝑢𝑟𝑡𝑒𝑟
𝑑𝑥
𝑑𝑦
𝑔𝑖𝑣𝑒𝑠 𝑢𝑠:
𝑑𝑥
𝑑 𝑑𝑦
𝑑2 𝑦
𝑎𝑛𝑑 𝑖𝑠 𝑤𝑟𝑖𝑡𝑡𝑒𝑛 𝑎𝑠
; 𝑟𝑒𝑎𝑑 𝑎𝑠 "𝒅 − 𝒕𝒘𝒐 − 𝒚 𝒃𝒚 𝒅 − 𝒙 − 𝒔𝒒𝒖𝒂𝒓𝒆𝒅" .
𝑑𝑥 𝑑𝑥
𝑑𝑥 2
𝐴𝑙𝑠𝑜 𝑖𝑓 𝑦 =
𝑓 𝑥
𝑑𝑦
= 𝑓′ 𝑥
𝑎𝑛𝑑
𝑑𝑥
2
𝑑 𝑦
= 𝑓 ′′ 𝑥 .
𝑑𝑥 2
Differentiating once gives the first derivative;
and twice gives the second derivative.
Example 13:
Find the second derivative of the following:
𝑎 𝑦 = 2𝑥 3 + 4𝑥 2 − 12
𝑑𝑦
= 6𝑥 2 + 8𝑥
𝑑𝑥
∴
𝑑2 𝑦
= 12𝑥 + 8
𝑑𝑥 2
We simply differentiate once and then
twice to obtain the second derivative.
𝑏 𝑓 𝑥 = 4𝑥 4 + 2𝑥 3 − 3𝑥 2 + 5𝑥 − 12
𝑓 ′ 𝑥 = 16𝑥 3 + 6𝑥 2 − 6𝑥 + 5
𝑓 ′′ 𝑥 = 48𝑥 2 + 12𝑥 − 6
𝑓 ′ 𝑥 and 𝑓 ′′ 𝑥 are the first and
second derivatives respectively.
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02. Differentiation 2
𝑥2 − 3 𝑥2 + 4
𝑐 𝑓 𝑥 =
First, we expand the expression.
𝑓 𝑥 = 𝑥 4 + 4𝑥 2 − 3𝑥 2 − 12
= 𝑥 4 + 𝑥 2 − 12
𝑓 ′ 𝑥 = 4𝑥 3 + 2𝑥
Then we differentiate once and then
twice to obtain the second derivative.
𝑓 ′′ 𝑥 = 12𝑥 2 + 2.
Example 14:
Find the second derivative at the following points on the given curves:
𝑎
𝑦 = 𝑥 3 + 2𝑥 − 17;
−1, −20
𝑑𝑦
= 3𝑥 2 + 2
𝑑𝑥
We differentiate once and then twice
to obtain the second derivative.
𝑑2 𝑦
= 6𝑥
𝑑𝑥 2
𝐴𝑡 −1, −20 , 𝑥 = −1;
∴
We then substitute the value of 𝑥 at
𝑑2 𝑦
= 6 −1
𝑑𝑥 2
the given point.
= −6
𝑏
𝑓 𝑥 = 𝑥 2 2𝑥 − 4 3𝑥 − 1 ;
0.5, −0.375
𝑓 𝑥 = 𝑥 2 6𝑥 2 − 14𝑥 + 4
= 6𝑥 4 − 14𝑥 3 + 4𝑥 2
We expand, and then differentiate once
and twice.
𝑓 ′ 𝑥 = 24𝑥 3 − 42𝑥 2 + 8𝑥
𝑓 ′′ 𝑥 = 72𝑥 2 − 84𝑥 + 8
𝐴𝑡 0.5, −0.375 ; 𝑓 ′′ 𝑥 = 72 0.5
2
− 84 0.5 + 8
We then substitute the value of 𝑥 at
the given point.
= −16 .
𝑐
𝑦 = 3𝑥 4 + 16𝑥 3 + 24𝑥 2 + 3;
𝑑𝑦
= 12𝑥 3 + 48𝑥 2 + 48𝑥
𝑑𝑥
2
𝑑 𝑦
= 36𝑥 2 + 96𝑥 + 48
𝑑𝑥 2
1 170
, 27
3
We differentiate once and then twice
to obtain the second derivative.
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02. Differentiation 2
𝐴𝑡
1 170
, 27
3
;
𝑑2 𝑦
= 36
𝑑𝑥 2
1 2
3
+ 96
1
3
+ 48
We then substitute the value of 𝑥 at
the given point.
= 84 .
2.8. Stationary points and Second derivative test
𝑑𝑦
𝑑𝑥 = 0.
Their nature, (whether a maximum, minimum or a point of inflexion) can then be determined by examining the
sign of the gradient at either sides of the stationary point.
From the previous sections, the stationary points of a curve can be found by solving the equation
Another useful method of examining stationary points is using the second derivative test.
4
𝐼𝑓 𝑎 𝑐𝑢𝑟𝑣𝑒 𝑖𝑠;
3
second
derivative to
2
𝑦 = 𝑥 + 2𝑥 − 3𝑥 ;
determine if a
𝑑𝑦
= 4𝑥 3 + 6𝑥 2 − 6𝑥;
𝑑𝑥
𝑖𝑡𝑠 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠
We can use the
𝒅𝒚
𝑡𝑒 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑜𝑓 𝒕𝒉𝒆 𝒈𝒓𝒂𝒅𝒊𝒆𝒏𝒕 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
𝑖𝑠
𝒅𝒙
point is a
𝑑2 𝑦
= 12𝑥 2 + 12𝑥 − 6.
𝑑𝑥 2
maximum or
minimum.
Note the following points from fig.2.4:
1.
2.
Around a maximum point,
𝑑𝑥 passes from positive to negative; so it is
𝑑𝑦
𝑑2 𝑦
decreasing. Hence the gradient of
𝑑𝑥 , i.e.
𝑑𝑥 2 is negative at that point.
𝑑𝑦
Around a minimum point,
𝑑𝑥 passes from negative to positive so it is increasing.
Hence the gradient of
3.
𝑑𝑦
When
𝑑𝑦
𝑑𝑥 i.e.
𝑑2 𝑦
𝑑𝑥 2 is positive at that point.
𝑑2 𝑦
𝑑𝑥 2 is zero, the point may be an inflexion, maximum or minimum. So
𝑑𝑦
the results are inconclusive and the method of finding the sign of
𝑑𝑥 should be
used.
Fig. 2.4
Example 15:
Find the stationary points on the following curves. Using the second derivative test, find their nature.
𝑎 𝑦 = 2𝑥 2 + 4𝑥
𝑦 = 2𝑥 2 + 4𝑥
𝑑𝑦
= 4𝑥 + 4
𝑑𝑥
We obtain
𝑑𝑦
𝑑𝑥
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