MNO2601 Operations performance

Dierentiation

Solutions to a tivities
in MO001

, DSC1520/MO001




Contents


1 The power rule 3
1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.2 A tivity: PE 6.1 (MO p 41) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 A tivity: PE 6.3 (MO p 42) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 A tivity: PE 6.6 (MO p 42) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.5 A tivity: PE 6.7 (MO p 42) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.6 A tivity: PE 6.8 (MO p 42) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.7 A tivity: PE 6.9 (MO p 43) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13


2 Derivatives of exponentials and logarithms 16
2.1 A tivity: PE 6.12 (MO p 43) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16


3 The hain rule 17
3.1 A tivity: PE 6.13 (MO p 44) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18


4 The produ t rule 19
4.1 A tivity: PE 6.14 (MO p 45) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19


5 The quotient rule 20
5.1 A tivity: PE 6.15 (MO p 45) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20


6 Further appli ations, using hain, produ t and quotient rules 21
6.1 A tivity: PE 6.16 (MO p 46) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21


7 Elasti ity and the derivative 24
7.1 A tivity: PE 6.17 (MO p 46) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24




2

, DSC1520/MO001




1 The power rule


The power rule states that if we have the fun tion y = f (x) = xn , then the derivative of the fun tion
is
dy
= f ′ (x) = nxn−1 .
dx
The power rule in words: Multiply by the power of the variable and subtra t one from the power.




1.1 Examples

The derivative of ea h of the following fun tions is found by using the power rule:

Function (simplify) Derivative
dy
y = x3 dx = 3 × x3−1 = 3x2

f (x) = 5x4 f ′ (x) = 5 × 4x4−1 = 20x3

g(x) = x3 + 5x4 g ′ (x) = 3x2 + 20x3

1
f (x) = x = x−1 f ′ (x) = −1 × x−1−1 = −x−2 = − x12
4 dy
y= x3
= 4x−3 dx = 4 × −3x−3−1 = −12x−4 = − x124

1 4
G′ (x) = − x12 − − x124 = − x12 + 12


G(x) = x − x3 x4

√ 1 1 2
f (x) = x = x3
3
f ′ (x) = 31 x 3 −1 = 13 x− 3 = 1
2 = √1
3 2
3x 3 3 x
√ 1 1 3
f (x) = 8 4 x = 8x 4 f ′ (x) = 8 × 41 x 4 −1 = 2x− 4 = √2
4 3
x
√ √ 1 2
g(x) = 3
x+84x g ′ (x) = √
3 2 + √
4 3
3 x x



1.2 A tivity: PE 6.1 (MO p 41)

1. Q3( )

1
Given the fun tion y = 10 + 5x + x2
. After simpli
ation,


y = 10 + 5x + x−2 .

Now,


dy
= 0 + 5x1−1 + (−2)x−2−1
dx
= 5x0 − 2x−3
2
= 5 − 3.
x

2. Q3(e)




3

, DSC1520/MO001



Q3
Dierentiating P = 3 + 70Q − 15Q2 = 13 Q3 + 70Q − 15Q2 gives

dP 1
= × 3Q3−1 + 70Q1−1 − 15 × 2Q2−1
dQ 3
= Q2 + 70 − 30Q
= Q2 − 30Q + 70. (in standard form)

3. Q5(b)

The derivative of
1 1 1
y = √ = 1 = x− 2
x x2
is

dy 1 1
= − x− 2 −1
dx 2
1 3
= − x− 2
2
1
= − √ .
2 x3

4. Q6(b)

We
rst simplify the given fun tion to get
√
Q
P = 4
Q2
1
= 4Q 2 × Q−2
1
= 4Q 2 −2
3
= 4Q− 2 .

Dierentiating this fun tion gives

dP d 3
= 4Q− 2
dQ dQ
3 3
= 4 × − Q− 2 −1
2
5
= −6Q− 2
6
= −p .
Q5

5. Q8( )

We are given the fun tion P (Q) = 10Q + Q0,5 .
The
rst derivative of P is


P ′ (Q) = 10Q1−1 + 0,5Q0,5−1
1 1
= 10 + Q− 2
2
1
= 10 + √ ,
2 Q

4