1. Differentiation

2. Introduction
In this chapter you will learn how to differentiate equations that are given parametrically
You will also learn to differentiate implicit functions, which aren’t necessarily written in the form ‘y = …’
These will be combined with calculating normals and tangents as you have so far
We will also look at creating differential equations from information we are given

3. Teachings for Exercise 4A

4. You can find the gradient of a curve given in parametric coordinates
Differentiation
You can find the gradient of a curve given in parametric coordinates
When a curve is given in parametric equations:
Differentiate y and x with respect to the parameter t
We can then use the following calculation, a variation of the chain rule
𝑑𝑦 𝑑𝑥
= 𝑑𝑦 𝑑𝑡
÷ 𝑑𝑥 𝑑𝑡
𝑑𝑦 𝑑𝑡
÷ 𝑑𝑥 𝑑𝑡
Rewrite as a multiplication
Written differently
𝑑𝑦 𝑑𝑡
× 𝑑𝑡 𝑑𝑥
= 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡
𝑑𝑦 𝑑𝑥
Cancel dts
= 𝑑𝑦 𝑑𝑥 4A

5. You can find the gradient of a curve given in parametric coordinates
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡
Differentiation
You can find the gradient of a curve given in parametric coordinates
𝑥= 𝑡 3 +𝑡
Find the gradient at the point P where t = 2, on the curve given by parametric equations:
𝑦= 𝑡 2 +1
𝑥= 𝑡 3 +𝑡
𝑦= 𝑡 2 +1
Differentiate
Differentiate
𝑑𝑥 𝑑𝑡 =3 𝑡 2 +1
𝑑𝑦 𝑑𝑡 =2𝑡
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡
Write in terms of t using dy/dt and dx/dt
 This is the gradient function for the curve, but in terms of t
𝑑𝑦 𝑑𝑥 = 2𝑡
3 𝑡 2 +1
Sub in t = 2
𝑑𝑦 𝑑𝑥 =
2(2)
3 (2) 2 +1
Work out
𝑑𝑦 𝑑𝑥 =
4 13 4A

6. You can find the gradient of a curve given in parametric coordinates
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡
Differentiation
You can find the gradient of a curve given in parametric coordinates
Find the equation of the normal at the point P where θ = π/6, on the curve with parametric equations:
𝑥=3𝑠𝑖𝑛𝜃
𝑦=5𝑐𝑜𝑠𝜃
We need to find the gradient at point P by differentiating and substituting θ in
We then need to find the gradient of the normal
We also need the coordinates of x and y at point P
Then we can use the formula y – y1 = m(x – x1) to obtain the equation of the line
𝑥=3𝑠𝑖𝑛𝜃
𝑦=5𝑐𝑜𝑠𝜃
Differentiate
Differentiate
𝑑𝑥 𝑑𝜃 =3𝑐𝑜𝑠𝜃
𝑑𝑦 𝑑𝜃 =−5𝑠𝑖𝑛𝜃
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝜃 𝑑𝑥 𝑑𝜃
−5 3 3
Gradient of the tangent at P:
Sub in dx/dθ and dy/dθ
𝑑𝑦 𝑑𝑥 = −5𝑠𝑖𝑛𝜃 3𝑐𝑜𝑠𝜃
Gradient of the normal at P:
Sub θ = π/6 in to obtain the gradient at P
𝑑𝑦 𝑑𝑥 = −5𝑠𝑖𝑛 𝜋 6 3𝑐𝑜𝑠 𝜋 6
Work out the fraction
𝑑𝑦 𝑑𝑥 = −5 3 3 4A

7. You can find the gradient of a curve given in parametric coordinates
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡
Differentiation
You can find the gradient of a curve given in parametric coordinates
Find the equation of the normal at the point P where θ = π/6, on the curve with parametric equations:
𝑥=3𝑠𝑖𝑛𝜃
𝑦=5𝑐𝑜𝑠𝜃
Gradient of the normal at P: =
We need to find the gradient at point P by differentiating and substituting θ in
We then need to find the gradient of the normal
We also need the coordinates of x and y at point P
Then we can use the formula y – y1 = m(x – x1) to obtain the equation of the line
𝑥=3𝑠𝑖𝑛𝜃
𝑦=5𝑐𝑜𝑠𝜃
Sub in P = π/6
Sub in P = π/6
𝑥=3𝑠𝑖𝑛 𝜋 6
𝑦=5𝑐𝑜𝑠 𝜋 6
Work out
Work out 𝑦=
𝑥= 3 2
3 2 ,
The coordinates of P are: 4A

8. You can find the gradient of a curve given in parametric coordinates
𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡
Differentiation
You can find the gradient of a curve given in parametric coordinates
Find the equation of the normal at the point P where θ = π/6, on the curve with parametric equations:
𝑥=3𝑠𝑖𝑛𝜃
𝑦=5𝑐𝑜𝑠𝜃
Gradient of the normal at P: =
We need to find the gradient at point P by differentiating and substituting θ in
We then need to find the gradient of the normal
We also need the coordinates of x and y at point P
Then we can use the formula y – y1 = m(x – x1) to obtain the equation of the line
3 2 ,
The coordinates of P are:
𝑦 − 𝑦 1 =𝑚(𝑥 − 𝑥 1 )
Sub in y1, m and x1 −
𝑥 − 3 2 𝑦 =
Multiply by 2
𝑥 − 3 2 2𝑦
− 5 3 =
Multiply by 5
(You could multiply by 10 in one step if confident)
𝑥 − 3 2
10𝑦 − =
6 3
Multiply the bracket out
10𝑦 − = 6𝑥 3 − 9 3
Add 25√3
10𝑦 = 6𝑥
Divide by 2
5𝑦 =3𝑥 4A

9. Teachings for Exercise 4B

10. Differentiation 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 4B
You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny
Implicit effectively means all the terms are mixed up, not necessarily in the form ‘y = …’
 This technique is useful as some equations are difficult to arrange into this form…
𝑑 𝑑𝑥 𝑦 𝑛
= 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥
Write dy/dx after differentiating the y term
This is written differently
 The reason is that as the equation is not written as ‘y =…’, y is not a function of x
Differentiate the y term as you would for an x term
6𝑦 𝑑𝑦 𝑑𝑥
For example:
3 𝑦 2
8 𝑑𝑦 𝑑𝑥 8𝑦
This is what happens when you differentiate an equation which starts ‘y =…’
𝑑𝑦 𝑑𝑥 𝑦 4B

11. Differentiation 4B 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥
𝑑 𝑑𝑥 𝑦 𝑛
= 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥
You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny
Find dy/dx in terms of x and y for the following equation:
𝑥 3 + 𝑥 + 𝑦 𝑦 = 6
𝑥 3 + 𝑥 + 𝑦 𝑦 = 6
Differentiate each part one at a time
It would be very difficult to rearrange this into the form ‘y = …’
3𝑦 2 𝑑𝑦 𝑑𝑥
3 𝑑𝑦 𝑑𝑥
3𝑥 2 + 1 + + =
Isolate the terms with dy/dx in
3𝑦 2 𝑑𝑦 𝑑𝑥
3 𝑑𝑦 𝑑𝑥 + =
− 3𝑥 2 − 1
Factorise by taking out dy/dx
𝑑𝑦 𝑑𝑥 3𝑦 =
− 3𝑥 2 − 1
Divide by (3y2 + 3)
𝑑𝑦 𝑑𝑥
−3 𝑥 2 −1 3𝑦 =
You now have a formula for the gradient, but in terms of x AND y, not just x 4B

12. Differentiation 4B 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥
𝑑 𝑑𝑥 𝑦 𝑛
= 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥
You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny
Find the value of dy/dx at the point (1,1) where:
4𝑥𝑦 2 −5𝑦=10
4𝑥𝑦 2 −5𝑥=10
Differentiate each one at a time – remember the product rule for the first term!!!!
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
𝑢=4𝑥
𝑣= 𝑦 2
8𝑥𝑦 𝑑𝑦 𝑑𝑥
+ 4𝑦 2
− 5
= 0
𝑑𝑦 𝑑𝑥 =
8𝑥𝑦 𝑑𝑦 𝑑𝑥
𝑑𝑢 𝑑𝑥 =4
𝑑𝑣 𝑑𝑥 =2𝑦 𝑑𝑦 𝑑𝑥
+ 4 𝑦 2
Add 5, subtract 4y2
8𝑥𝑦 𝑑𝑦 𝑑𝑥 =5−4 𝑦 2
You can substitute x = 1 and y = 1 now (or rearrange first)
 Do not replace the terms in dy/dx
8 𝑑𝑦 𝑑𝑥 =5 − 4
Divide by 8
𝑑𝑦 𝑑𝑥 = 1 8
Keep your workings tidy – you may need to use the product rule on multiple terms as well as factorise – show everything you’re doing! 4B

13. Differentiation 4B 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥
𝑑 𝑑𝑥 𝑦 𝑛
= 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥
You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny
Find the value of dy/dx at the point (1,1) where:
𝑒 2𝑥 𝑙𝑛𝑦 = 𝑥+𝑦−2
Give your answer in terms of e.
𝑒 2𝑥 𝑙𝑛𝑦 = 𝑥+𝑦−2
𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥
𝑢= 𝑒 2𝑥
𝑣=𝑙𝑛𝑦
Differentiate (again watch out for the product rule!
𝑒 2𝑥 𝑦 𝑑𝑦 𝑑𝑥 + 2𝑙𝑛𝑦 𝑒 2𝑥
=1+ 𝑑𝑦 𝑑𝑥
𝑑𝑦 𝑑𝑥 =
𝑒 2𝑥 𝑦 𝑑𝑦 𝑑𝑥
𝑑𝑢 𝑑𝑥 =2 𝑒 2𝑥
𝑑𝑣 𝑑𝑥 = 1 𝑦 𝑑𝑦 𝑑𝑥
+ 2𝑙𝑛𝑦 𝑒 2𝑥
Sub in x = 1 and y = 1
𝑒 𝑑𝑦 𝑑𝑥
=1+ 𝑑𝑦 𝑑𝑥
+ 2𝑙𝑛1 𝑒 2
ln1 = 0 since e0 = 1, cancelling the term out
𝑒 2 𝑑𝑦 𝑑𝑥
=1+ 𝑑𝑦 𝑑𝑥
Subtract dy/dx
𝑒 2 𝑑𝑦 𝑑𝑥 − 𝑑𝑦 𝑑𝑥 =1
Factorise
𝑑𝑦 𝑑𝑥 ( 𝑒 2 −1)=1
Divide by (e2 – 1)
𝑑𝑦 𝑑𝑥 = 1 𝑒 2 −1 4B

14. Teachings for Exercise 4C

15. 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

16. 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 𝑙𝑛6𝑟 4C

17. Teachings for Exercise 4D

18. 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

19. 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

20. 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

21. Teachings for Exercise 4E

22. 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

23. 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

24. 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

25. 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

26. 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

27. Summary
You have learnt how to differentiate equations given parametrically
You can also differentiate implicit equations
You can differentiate the general power function ax
You can also set up differential equations based on information you are given