1. Differentiation

2. Finding 𝑑𝑦 𝑑𝑥 given an equation
Usually questions will ask you to find 𝑑𝑦 𝑑𝑥 .
Just differentiate both sides of the equation with respect to 𝑥. Then make 𝑑𝑦 𝑑𝑥 the subject.
C4 Jan 2008 Q5
Find 𝑑𝑦 𝑑𝑥 in terms of 𝑥 and 𝑦 when 𝑥 3 +𝑥+ 𝑦 3 +3𝑦=6
𝟑 𝒙 𝟐 +𝟏+𝟑 𝒚 𝟐 𝒅𝒚 𝒅𝒙 +𝟑 𝒅𝒚 𝒅𝒙 =𝟎 𝒅𝒚 𝒅𝒙 𝟑 𝒚 𝟐 +𝟑 =−𝟑 𝒙 𝟐 −𝟏 𝒅𝒚 𝒅𝒙 =− 𝟑 𝒙 𝟐 +𝟏 𝟑 𝟏+ 𝒚 𝟐 ?
−512−4 𝑦 2 =−96𝑦 𝑦 2 −24𝑦+128=0 𝑦−16 𝑦−8 =0 Gives point −8,16 , −8,8
Implicitly differentiating: 3 𝑥 2 −8𝑦 𝑑𝑦 𝑑𝑥 =12𝑥 𝑑𝑦 𝑑𝑥 +12𝑦 𝑑𝑦 𝑑𝑥 not needed, but it’s 3 𝑥 2 −12𝑦 12𝑥+8𝑦
If 𝑥=−8, 𝑦=8, 𝑑𝑦 𝑑𝑥 =−3
If 𝑥=−8, 𝑦=16, 𝑑𝑦 𝑑𝑥 =0 ? ?

3. Differentiation
You can differentiate the general power function ax where a is a constant
Differentiate
𝑦= 𝑎 𝑥
where a is a constant
This is important – as a is a constant it can be treated as a number
𝑦= 𝑎 𝑥
Take natural logs
𝑙𝑛𝑦= 𝑙𝑛𝑎 𝑥
Use the power law
𝑙𝑛𝑦=𝑥𝑙𝑛𝑎
Differentiate each side
 ln a is a constant so can be thought of as just a number
1 𝑦 𝑑𝑦 𝑑𝑥 =𝑙𝑛𝑎
Multiply both sides by y
𝑑𝑦 𝑑𝑥 =𝑦𝑙𝑛𝑎
y = ax as from the first line
𝑑𝑦 𝑑𝑥 = 𝑎 𝑥 𝑙𝑛𝑎 4C

4. Differentiation 4C 𝑑𝑦 𝑑𝑥 = 𝑎 𝑥 𝑙𝑛𝑎
You can differentiate the general power function ax where a is a constant
Differentiate
𝑦= 3 𝑥
𝑦= 3 𝑥
Differentiate
𝑑𝑦 𝑑𝑥 = 3 𝑥 𝑙𝑛3
Differentiate
𝑦= 6𝑟 𝑥
where r is a constant
𝑦= 6𝑟 𝑥
Differentiate
𝑑𝑦 𝑑𝑥 = 6𝑟 𝑥 𝑙𝑛𝑟 4C

5. Exercise 4C

6. Differentiation 4D 𝑑𝐴 𝑑𝑡 = 𝑑𝑟 𝑑𝑡 × 𝑑𝐴 𝑑𝑟
You can relate one rate of change to another. This is useful when a situation involves more than two variables.
Given that the area of a circle A, is related to its radius r by the formula A = πr2, and the rate of change of its radius in cm is given by dr/dt = 5, find dA/dt when r = 3
You are told a formula linking A and r
You are told the radius is increasing by 5 at that moment in time
You are asked to find how much the area is increasing when the radius is 3
This is quite logical – if the radius is increasing over time, the area must also be increasing, but at a different rate…
𝑑𝐴 𝑑𝑡 =
𝑑𝑟 𝑑𝑡 × 𝑑𝐴 𝑑𝑟
𝑑𝐴 𝑑𝑡 =
𝑑𝑟 𝑑𝑡
× 𝑑𝐴 𝑑𝑟
Replace dr/dt and dA/dr
𝑑𝐴 𝑑𝑡 =
5 × 2𝜋𝑟
We are trying to work out dA/dt
We have been told dr/dt in the question
You need to find a derivative that will give you the dA and cancel the dr out
𝑑𝐴 𝑑𝑡 =
10𝜋𝑟
r = 3
𝐴=𝜋 𝑟 2
= 30𝜋
𝑑𝐴 𝑑𝑟 =2𝜋𝑟
Differentiate 4D

7. Differentiation
You can relate one rate of change to another. This is useful when a situation involves more than two variables.
𝑉= 2 3 𝜋 𝑟 3
The volume of a hemisphere is related to its radius by the formula:
The total surface area is given by the formula:
𝑆=3𝜋 𝑟 2
𝑑𝑉 𝑑𝑡 =6
Find the rate of increase of surface area when the rate of increase of volume:
You need to use the information to set up a chain of derivatives that will leave dS/dt
𝑑𝑆 𝑑𝑡 =
𝑑𝑉 𝑑𝑡
𝑑𝑟 𝑑𝑉
𝑑𝑉 𝑑𝑟
× 𝑑𝑆 𝑑𝑟 ×
Rate of change of surface area over time
We are told dV/dt in the question so we will use it
We have a formula linking V and r so can work out dV/dr
We have a formula linking S and r so can work out dS/dr
This isn’t all correct though, as multiplying these will not leave dS/dt
However, if we flip the middle derivative, the sequence will work! 4D

8. Differentiation
You can relate one rate of change to another. This is useful when a situation involves more than two variables.
𝑉= 2 3 𝜋 𝑟 3
The volume of a hemisphere is related to its radius by the formula:
The total surface area is given by the formula:
𝑆=3𝜋 𝑟 2
𝑑𝑉 𝑑𝑡 =6
Find the rate of increase of surface area when the rate of increase of volume:
You need to use the information to set up a chain of derivatives that will leave dS/dt
𝑑𝑆 𝑑𝑡 =
𝑑𝑉 𝑑𝑡
× 𝑑𝑟 𝑑𝑉
× 𝑑𝑆 𝑑𝑟
𝑉= 2 3 𝜋 𝑟 3
𝑆=3𝜋 𝑟 2
Sub in values
Differentiate
Differentiate
𝑑𝑆 𝑑𝑡 =
× 𝜋𝑟 2 6
× 6𝜋𝑟
𝑑𝑉 𝑑𝑟 =2 𝜋𝑟 2
𝑑𝑆 𝑑𝑟 =6𝜋𝑟
You can write as one fraction
Flip to obtain the derivative we want
𝑑𝑆 𝑑𝑡 =
36𝜋𝑟 2 𝜋𝑟 2
𝑑𝑟 𝑑𝑉 = 1 2 𝜋𝑟 2
Simplify
𝑑𝑆 𝑑𝑡 =
18 𝑟 4D

9. Exercise 4D

10. Differentiation 𝑑𝑁 𝑑𝑡 = −𝑘𝑁 4E
You can set up differential equations from information given in context
Differential equations can arise anywhere when variables change relative to one another
In Mechanics, speed may change over time due to acceleration
The rate of growth of bacteria may change when temperature is changed
As taxes change, the amount of money people spend will change
Setting these up involve the idea of proportion, a topic you met at GCSE…
Radioactive particles decay at a rate proportional to the number of particles remaining. Write an equation for the rate of change of particles…
 Let N be the number of particles and t be time
𝑑𝑁 𝑑𝑡 =
−𝑘𝑁
The rate of change of the number of particles
The number of particles remaining multiplied by the proportional constant, k.
 Negative because the number of particles is decreasing 4E

11. Differentiation 𝑑𝑃 𝑑𝑡 = 𝑘𝑃 4E
You can set up differential equations from information given in context
A population is growing at a rate proportional to its size at a given time. Write an equation for the rate of growth of the population
 Let P be the population and t be time
𝑑𝑃 𝑑𝑡 = 𝑘𝑃
The population multiplied by the proportional constant k
 Leave positive since the population is increasing
The rate of change of the population 4E

12. Differentiation 𝑑𝜃 𝑑𝑡 = −𝑘(𝜃− 𝜃 0 ) 4E
You can set up differential equations from information given in context
Newton’s law of cooling states that the rate of loss of temperature is proportional to the excess temperature the body has over its surroundings. Write an equation for this law…
Let the temperature of the body be θ degrees and time be t
The ‘excess’ temperature will be the difference between the objects temperature and its surroundings – ie) One subtract the other
(θ – θ0) where θ0 is the temperature of the surroundings
𝑑𝜃 𝑑𝑡 =
−𝑘(𝜃− 𝜃 0 )
The rate of change of the objects temperature
The difference between temperatures (θ - θ0) multiplied by the proportional constant k
 Negative since it is a loss of temperature 4E

13. Differentiation
You can set up differential equations from information given in context
The head of a snowman of radius r cm loses volume at a rate proportional to its surface area. Assuming the head is spherical, and that the volume V = 4/3πr3 and the Surface Area A = 4πr2, write down a differential equation for the change of radius of the snowman’s head…
𝑑𝑉 𝑑𝑡 =−𝑘𝐴
The first sentence tells us
The rate of change of volume
Is the Surface Area multiplied by the proportional constant k (negative as the volume is falling)
𝑑𝑟 𝑑𝑡 =
𝑑𝑉 𝑑𝑡
𝑑𝑟 𝑑𝑉
𝑉= 4 3 𝜋 𝑟 3 ×
Differentiate
𝑑𝑉 𝑑𝑟 =4𝜋 𝑟 2
We want to know the rate of change of the radius r
We know dV/dt
We need a derivative that will leave dr/dt when cancelled
Flip over
𝑑𝑟 𝑑𝑉 = 1 4𝜋 𝑟 2 4E

14. Differentiation
You can set up differential equations from information given in context
The head of a snowman of radius r cm loses volume at a rate proportional to its surface area. Assuming the head is spherical, and that the volume V = 4/3πr3 and the Surface Area A = 4πr2, write down a differential equation for the change of radius of the snowman’s head…
𝑑𝑉 𝑑𝑡 =−𝑘𝐴
The first sentence tells us
The rate of change of volume
Is the Surface Area multiplied by the proportional constant k (negative as the volume is falling)
𝑑𝑟 𝑑𝑡 =
𝑑𝑉 𝑑𝑡
𝑑𝑟 𝑑𝑉
𝑉= 4 3 𝜋 𝑟 3 ×
Fill in what we know
Differentiate
𝑑𝑟 𝑑𝑡 =
1 4𝜋 𝑟 2
𝑑𝑉 𝑑𝑟 =4𝜋 𝑟 2
−𝑘𝐴 ×
We know A from the question
Flip over
𝑑𝑟 𝑑𝑡 =
1 4𝜋 𝑟 2
𝑑𝑟 𝑑𝑉 = 1 4𝜋 𝑟 2
−𝑘(4𝜋 𝑟 2 ) ×
Simplify
𝑑𝑟 𝑑𝑡 = −𝑘 4E

15. Exercise 4E and Mixed 4F