Academic Skills Advice
Differentiation Summary
Differentiation is about gradients. You can find the gradient at any point on a curve by
finding the gradient function. To find the gradient function you need to differentiate.
Differentiation of Polynomials:
Original Function:
Gradient function:
Multiply the
expression by the
power then reduce
the power by 1.
𝒚 = 𝒂𝒙𝒏
𝒅𝒚
𝒅𝒙
= 𝒂𝒏𝒙𝒏−𝟏
For functions with more than one term, differentiate each term separately.
e.g.
𝑦 = 5𝑥 2 − 3𝑥 + 4
𝑑𝑦
𝑑𝑥
= 10𝑥 − 3
Notation and Meaning
If you differentiate something once you have the 1st differential (or 1st derivative), if you
differentiate again you have the 2nd differential (or 2nd derivative).
Differentiation can be denoted in a few ways:
If you have:
𝑦=
1st differential/derivative is written:
2nd differential/derivative is written:
𝑑𝑦
𝑑𝑥
𝑓(𝑥) =
= or 𝑦 ′ =
𝑑2 𝑦
𝑑𝑥 2
𝑓 ′ (𝑥) =
= or 𝑦 ′′ =
𝑓 ′′ (𝑥) =
Remember: the 1st differential is the gradient function and can be used to find the
gradient of a curve at any point.
 Find the gradient of the curve, 𝒚 = 𝟒𝒙𝟑 − 𝟑𝒙𝟐 − 𝟕
when 𝒙 = 𝟐.
𝑦 = 4𝑥 3 − 3𝑥 2 − 7
Function:
𝒅𝒚
𝒅𝒙
= gradient function
Gradient function:
Gradient when 𝑥 = 2:
𝑑𝑦
(substitute 2 into 𝑑𝑥 )
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
© H Jackson 2010/2013 / Academic Skills
= 12𝑥 2 − 6𝑥
= 12(2)2 − 6(2)
= 36
We have found that when 𝑥 = 2, the gradient = 136
Stationary Points
There are 3 types of stationary points: maximum, minimum and point of inflexion.
Maximum turning point
Point of inflexion
Minimum turning point
Where? - Differentiate once (1st differential):
Use the 1st differential to find out where there are any stationary points.
Put
𝑑𝑦
𝑑𝑥
= 0 and solve to find the value of 𝑥.
Then substitute 𝑥 into the original equation to find 𝑦 so that you have the co-ordinates of
the stationary point(s).
Type? - Differentiate again (2nd differential):
Use the 2nd differential to find out the type of stationary point(s) you have found.
Substitute your value(s) of 𝑥 into the 2nd differential and:
If
If
If
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
= 0 and
= 0 and
= 0 and
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
𝑑2 𝑦
𝑑𝑥 2
>0
then you have a minimum (positive is minimum).
<0
then you have a maximum (negative is maximum).
=0
then it could be maximum, minimum or point of inflexion.
In the last case – differentiate again:
If
𝑑3 𝑦
𝑑𝑥 3
≠0
then it is a point of inflexion.
1st differential tells us where, 2nd differential tells us type.
© H Jackson 2010/2013 / Academic Skills
2
Product Rule
𝑑𝑦
If 𝑦 = 𝑢𝑣 then
𝑑𝑥
𝑑𝑣
For both of these rules make sure you
write down 4 bits of information:
𝑑𝑢
= 𝑢 𝑑𝑥 + 𝑣 𝑑𝑥
𝑢=
𝑑𝑢
𝑑𝑥
Quotient Rule
𝑑𝑦
𝑢
If 𝑦 = 𝑣 then
=
𝑑𝑥
𝑣
𝑑𝑢
𝑑𝑣
−𝑢
𝑑𝑥
𝑑𝑥
𝑣2
=
𝑣=
𝑑𝑣
𝑑𝑥
=
Once you have these 4 things you can just
substitute into the correct equation.
Chain Rule
If 𝑦 = 𝑓(𝑔(𝑥)), 𝑙𝑒𝑡 𝑢 = 𝑔(𝑥). 𝑇ℎ𝑒𝑛 𝑦 = 𝑓(𝑢)𝑎𝑛𝑑
𝑑𝑦
𝑑𝑥
𝑑𝑦
= 𝑑𝑢 ×
𝑑𝑢
𝑑𝑥
An easy way to do the chain rule is:
differential of ‘outside’ function X differential of ‘inside’ function
Parametric
Sometimes 𝑥 and 𝑦 are expressed separately in terms of a third variable (usually 𝑡).
If 𝑥 = 𝑓(𝑡) 𝑎𝑛𝑑 𝑦 = 𝑔(𝑡)
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
÷
Differentiate both 𝒙 and 𝒚 with
respect to 𝒕 then do
𝑑𝑥
𝒅𝒚
𝒅𝒕
÷
𝒅𝒙
𝒅𝒕
𝑑𝑡
The 2nd differential is not quite as straightforward as we cannot differentiate a function of 𝑡
with respect to 𝑥.
𝑑𝑦
𝑑2 𝑦
𝑑𝑥 2
=
𝑑
You cannot differentiate 𝑑𝑥 with respect to 𝑥
because it is a function of 𝑡
𝑑𝑦
( )
𝑑𝑥 𝑑𝑥
Therefore use:
𝑑2 𝑦
𝑑𝑥 2
=
𝑑
𝑑𝑦
( ) ×
𝑑𝑡 𝑑𝑥
𝒅𝒚
𝑑𝑡
Notice that the 𝑑𝑡′𝑠 cancel
𝑑𝑥
out leaving you with
𝑑2 𝑦
𝑑𝑥 2
𝒅𝒕
Differentiate 𝒅𝒙 with respect to 𝒕 then multiply by 𝒅𝒙
© H Jackson 2010/2013 / Academic Skills
3
Implicit Differentiation
This is used to differentiate a function which consists of 𝑥′𝑠 and 𝑦′𝑠 together – that cannot
easily be rearranged to make 𝑦 the subject.
Method: Differentiate each term and when you come to a 𝒚-term just differentiate
𝒅𝒚
with respect to 𝒚 then multiply by 𝒅𝒙.
e.g.
3𝑥 2 − 𝑦 2 = 18
becomes:
6𝑥 − 2𝑦 𝑑𝑥 = 0
𝑑𝑦
Which rearranges to give:
𝑑𝑦
𝑑𝑥
=
Don’t forget to use the
product rule for expressions
such as 2𝑦 3 𝑥 4
3𝑥
𝑦
Partial Differentiation
To show that something has been partially (instead of fully) differentiated we use the
symbol 𝜕.
𝝏𝒛
To find 𝝏𝒙 differentiate 𝑧 with respect to 𝑥 just taking 𝒚 to be a ‘constant’
𝝏𝒛
To find 𝝏𝒚 differentiate 𝑧 with respect to 𝑦 just taking 𝒙 to be a ‘constant’.
e.g. 𝑧 = 5𝑥 2 − 𝑥𝑦 2 − 3𝑦 4
𝜕𝑧
𝜕𝑥
𝜕𝑧
𝜕𝑦
= 10𝑥 − 𝑦 2 (−0)
(𝑦 is a constant)
= (0) − 2𝑥𝑦 − 12𝑦 3
(𝑥 is a constant)
𝜕2 𝑧
𝜕𝑥 2
𝜕2 𝑧
𝜕𝑦 2
= 10
(differentiate the 1st one again w.r.t 𝑥)
= −2𝑥 − 36𝑦 2
(differentiate the 2nd one again w.r.t 𝑦)
𝜕2 𝑧
𝜕𝑥𝜕𝑦
= −2𝑦
© H Jackson 2010/2013 / Academic Skills
𝜕𝑧
𝜕𝑧
(EITHER differentiate 𝝏𝒙 w.r.t 𝒚 OR 𝝏𝒚 w.r.t 𝒙)
4