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Title: Differentiation 1
Description: An introduction into differentiation, aimed at A level maths but also useful for 1st year science degrees. Contains questions at the end with solutions

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Differentiation I
The Definition of Differentiation
...
The derivative is the
instantaneous rate of change of a function with respect to one of its variables
...
For example, we may have a function
like π‘Œ = π‘₯ 3 + 4π‘₯ + 3, Here we can’t just simply state the gradient as we can with equations in the
form π‘Œ = π‘šπ‘₯ + 𝑐
...
e
...
This is simple enough, but it can become more
complicated
...
e
...
So, the chain rule must be applied:
This can be put in the form 𝑓(π‘₯) = 𝑆𝑖𝑛(𝑒) where 𝑒 = π‘₯ 2 from this both 𝑓(π‘₯) π‘Žπ‘›π‘‘ 𝑒 need
to be differentiated then multiplied together
...
e
...
For example, 𝑓(π‘₯) = π‘₯𝑠𝑖𝑛(π‘₯)
to solve this requires 3 steps (This is the
product rule)
Step1 – split into two variables usually u= and v=
Step2 – differentiate both u and v with respect to x
Robert Airey

Step 3- multiply diagonally and add
If we take the example of 𝑓(π‘₯) = π‘₯𝑠𝑖𝑛(π‘₯)
Step 1𝑒 = π‘₯ π‘Žπ‘›π‘‘ 𝑣 = sin(π‘₯)

the order doesn’t matter

Step 2 and 3 –

𝑒=π‘₯

𝑣 = sin(π‘₯)

𝑒′ = 1

𝑣′ = cos(π‘₯)

The product rule is 𝑓(π‘₯) = 𝑒(π‘₯)𝑣(π‘₯) π‘‘β„Žπ‘’π‘› 𝑓 β€² (π‘₯) = 𝑣𝑒′ + 𝑒𝑣 β€²
Again, the order of this doesn’t matter
𝑒(π‘₯)

Now let’s look at fractions in the form 𝑓(π‘₯) = 𝑣(π‘₯)

the order of this does matter, the

numerator should always be u(x) and denominator v(x), this is the quotient rule
Again 3 steps
Step1- state 𝒖(𝒙) =

𝒗(𝒙) =

Step2- differentiate 𝒖(𝒙) 𝒂𝒏𝒅 𝒗(𝒙)
Step3- diagonally multiply (like for the product rule) except subtract and divide by (𝒗(𝒙))𝟐
3π‘₯ 2

For example, let’s take 𝑓(π‘₯) = cos(π‘₯)
𝑒(π‘₯)

Put this in the form 𝑓(π‘₯) = 𝑣(π‘₯)

𝑒(π‘₯ ) = 3π‘₯ 2

𝑣(π‘₯ ) = cos(π‘₯)

Find the derivatives

𝑒(π‘₯ ) = 3π‘₯ 2

𝑣(π‘₯ ) = cos(π‘₯)

𝑒′ (π‘₯ ) = 6π‘₯

𝑣 β€² (π‘₯) = βˆ’sin(π‘₯)

So 𝑓(π‘₯) =
So 𝑓(π‘₯) =

𝑒(π‘₯)
𝑣(π‘₯)

3π‘₯ 2
cos(π‘₯)

Robert Airey

𝑓 β€² (π‘₯) =

𝑣𝑒′ βˆ’π‘’π‘£ β€²
𝑣2

𝑓 β€² (π‘₯) =

(6π‘₯π‘π‘œπ‘ (π‘₯)βˆ’βˆ’3π‘₯ 2 sin(π‘₯))
cos2(π‘₯)

=

6π‘₯π‘π‘œπ‘ (π‘₯)+3π‘₯ 2 sin(π‘₯)
cos2 (π‘₯)

Natural logs and exponentials are easier to differentiate if we remember the rules,
𝑓 β€² (π‘₯) =

𝑓(π‘₯) = ln(π‘₯)
𝑓(π‘₯) = 𝑒 π‘₯

1
π‘‘π‘–π‘“π‘“π‘’π‘Ÿπ‘’π‘›π‘‘π‘–π‘Žπ‘™ π‘œπ‘“ π‘π‘Ÿπ‘Žπ‘π‘˜π‘’π‘‘
𝑖𝑛 π‘™π‘Žπ‘¦π‘šπ‘Žπ‘›π‘  π‘‘π‘’π‘Ÿπ‘šπ‘  𝑖𝑑 β€² 𝑠
π‘₯
π‘π‘Ÿπ‘Žπ‘π‘˜π‘’π‘‘
2

𝑓 β€² (π‘₯) = 𝑒 π‘₯ , 𝑖𝑓 𝑖𝑑 π‘€π‘’π‘Ÿπ‘’ 𝑓(π‘₯) = 𝑒 π‘₯ π‘‘β„Žπ‘’π‘› 𝑓 β€² (π‘₯) = 2π‘₯𝑒 π‘₯

exponentials it’s easier to generalise as 𝑓(π‘₯) = 𝑒 𝑒(π‘₯) π‘‘β„Žπ‘’π‘› 𝑓 β€² (π‘₯) =

𝑑𝑒(π‘₯)
𝑑π‘₯

2

so, for

𝑒 𝑒(π‘₯)

Logs are especially useful to differentiate when equations like this come along𝑦 = 3π‘₯ 2 Γ— sin(π‘₯) Γ— cos2 (π‘₯)
It is possible to product rule this twice however by natural logging both sides we get
ln(𝑦) = ln(3π‘₯ 2 ) Γ— ln(sin(π‘₯)) Γ— ln(cos2 (π‘₯))
Simplifying using log laws we get
ln(𝑦) = 2 ln(3π‘₯) + ln(𝑠𝑖𝑛(π‘₯)) + 2ln(π‘π‘œπ‘ (π‘₯))
Differentiation gives us
1 𝑑𝑦
6 cos(π‘₯) 2 sin(π‘₯)
=
+
βˆ’
𝑦 𝑑π‘₯ 3π‘₯ sin(π‘₯)
cos(π‘₯)
𝑑𝑦

2

cos(π‘₯)

Getting 𝑑π‘₯ = (𝑦) Γ— (π‘₯ + sin(π‘₯) βˆ’ 2 tan(π‘₯))
Substitution for y if needed in terms of x only

Second derivatives are also useful
...

For example, take 𝑦 = 3π‘₯ 2 + 4π‘₯
𝑑𝑦

We differentiate as normal therefore giving us 𝑑π‘₯ = 6π‘₯ + 4
As we are looking for where the gradient is zero, we set
𝑑𝑦
𝑑π‘₯

= 0 π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ π‘‘β„Žπ‘–π‘  π‘π‘’π‘π‘œπ‘šπ‘’π‘  6π‘₯ + 4 = 0
4

4

Therefore at π‘₯ = βˆ’ 6 π‘Žπ‘›π‘‘ 𝑓𝑖𝑛𝑑𝑖𝑛𝑔 𝑦, 𝑦 = βˆ’ 3 at this point there is either a maximum or a
minimum point
To determine this points nature, we must differentiate
𝑑𝑦
𝑑π‘₯

𝑑2 𝑦

= 6π‘₯ + 4 π‘Žπ‘”π‘Žπ‘–π‘› π‘‘β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’ 𝑑π‘₯ 2 = 6

Robert Airey

𝑑2 𝑦

𝑑2 𝑦

If 𝑑π‘₯ 2 > 0 π‘‘β„Žπ‘–π‘  𝑖𝑠 π‘Ž π‘šπ‘–π‘›π‘–π‘šπ‘’π‘š, 𝑖𝑓 𝑑π‘₯ 2 < 0 π‘šπ‘Žπ‘₯π‘–π‘šπ‘’π‘š π‘π‘œπ‘–π‘›π‘‘ think of this as, a minimum
point as having a gradient that starts to increase hence positive and a maximum point
having a gradient that starts to decrease hence negative

Standard differentials

𝑓(x) = tan(kx) βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ 𝑓′(π‘₯) = π‘˜π‘ π‘’π‘ 2 (π‘˜π‘₯)
𝑓(π‘₯ ) = sec(π‘₯ ) βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ 𝑓 β€² (π‘₯) = sec(π‘₯ ) tan(π‘₯ )
𝑓(π‘₯ ) = cot(π‘₯ ) βˆ’ βˆ’ βˆ’ βˆ’ βˆ’ 𝑓 β€² (π‘₯ ) = βˆ’π‘π‘œπ‘ π‘’π‘ 2 (π‘₯ )
𝑓 (π‘₯ ) = π‘π‘œπ‘ π‘’π‘(π‘₯ ) βˆ’ βˆ’ βˆ’ βˆ’π‘“ β€² (π‘₯ ) = βˆ’π‘π‘œπ‘ π‘’π‘(π‘₯ )cot(π‘₯)

Robert Airey

Questions
Q1- if 𝑓(π‘₯) = π‘₯ 3 + 2π‘₯ + ln(π‘₯) ,state the derivative at x=2
Q2- Find the derivative of 𝑓(π‘₯) = cos(π‘₯ 2 )
Q3- Find the derivative of 𝑓(π‘₯) = sin(π‘₯) cos(π‘₯)
Q4- Find the derivative of 𝑓(π‘₯) =

ln(π‘₯)
4π‘₯ 2

Q5- Find the coordinates of the stationary point for 𝑦 = βˆ’π‘’ 2π‘₯ + 4π‘₯ and
determine its nature
Q6- Find the derivative of 𝑦 = sin2 (π‘₯) cos(x) e2x
Q7- Find the derivative of 𝑓(π‘₯) =

sin(π‘₯ 2 )
ln(π‘₯)

Q8- Find the derivative of 𝑦 = π‘₯ 3 + 2π‘₯ + 4 at x=10
Q9- Find the derivative of 𝑓(π‘₯) = ln(π‘₯) 𝑒 π‘₯

2

Q10- 𝑓(π‘₯) = π‘₯ 3 + 3π‘₯ 2 + 1 find the derivative and any coordinates of
stationary points

Robert Airey

Solutions
Q1) apply general power rule and differential rule of natural logs
𝑑𝑦
𝑑π‘₯

= 3π‘₯ 2 + 2 +

1
π‘₯

substitute x=2 then

𝑑𝑦
𝑑π‘₯

= 14
Title: Differentiation 1
Description: An introduction into differentiation, aimed at A level maths but also useful for 1st year science degrees. Contains questions at the end with solutions