Differentiation
from first principles
Gradient of a straight line
, or
𝑦2 −𝑦1
.
𝑥2 −𝑥1
If 𝑦 = 𝑓 𝑥 then this formula can be written as
𝑦1
𝑓(𝑥2 )−𝑓(𝑥1 )
𝑥2 −𝑥1
𝑥2 − 𝑥1
=ℎ
= 𝑓(𝑥2 ) − 𝑓(𝑥1 )
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑦𝑠
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑥𝑠
𝑦2
𝑦2 − 𝑦1
As you know, the gradient of a straight line is given by
𝑦=𝑓 𝑥
If the difference between 𝑥1 and 𝑥2 is ℎ, then 𝑥2 = 𝑥1 + ℎ
so the formula becomes
𝑓(𝑥1 +ℎ)−𝑓(𝑥1 )
(𝑥1 +ℎ)−𝑥1
𝑥
𝑥1
Since we’re now only dealing with a single 𝑥 value,
we can lose the subscript 1, leaving
𝑓(𝑥+ℎ)−𝑓(𝑥)
,
(𝑥+ℎ)−𝑥
or simply
𝑓(𝑥+ℎ)−𝑓(𝑥)
.
ℎ
Put that formula aside for a minute and let’s look at what happens with
a curve.
𝑥2
= 𝑥1 +ℎ
Gradient of a curve
If we’re dealing with a curve then the gradient at a
given point is the gradient of the tangent at that point.
𝑦=𝑓 𝑥
To find the gradient of the curve at P, we need to find
the gradient of the tangent (blue line) at that point…
… but in order to find a gradient we need the
coordinates of TWO points on the tangent, and we only
have one.
Let’s use a chord (a line that cuts the curve in two
places) as an approximation – i.e. the green line on this
graph.
The gradient of the chord is given by the formula from
𝑓(𝑥+ℎ)−𝑓(𝑥)
the previous slide:
ℎ
… but that’s clearly not the same as the gradient of the
tangent at P.
Px
𝑥
𝑥
𝑥+ℎ
Gradient of a curve
𝑦=𝑓 𝑥
How can we make the gradient of the chord
closer to the gradient of the tangent?
What happens if we move the second point
closer to the first (reducing h)?
You can see that the gradient of the new chord
is closer to the gradient of the tangent.
We can continue reducing h until h approaches
0, at which point the chord and the tangent
become indistinguishable, as do their gradients.
The gradient of the chord when this happens is
the limit of the gradient as h approaches 0, or
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ→0 (𝑥+ℎ)−𝑥
𝑓 ′ 𝑥 = lim
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
ℎ→0
or lim
P
x
𝑥
𝑥
𝑥+ℎ
𝑥+ℎ
The formula is given to you
so you don’t need to
memorise it.
Derivation from first principles
We are going to show that if 𝑓 𝑥 = 𝑥 2
then 𝑓′ 𝑥 = 2𝑥.
′
𝑓 𝑥 =
𝑓(𝑥+ℎ)−𝑓(𝑥)
lim
ℎ
ℎ→0
First, substitute in for 𝑓 𝑥 and 𝑓 𝑥 + ℎ :
𝑓′
𝑥 =
(𝑥+ℎ)2 −𝑥 2
lim
ℎ
ℎ→0
ℎ→0
Now expand the brackets and tidy up:
𝑓′
𝑥 =
=
𝑥 2 +2𝑥ℎ+ ℎ2 −𝑥 2
lim
ℎ
ℎ→0
2𝑥ℎ+ ℎ2
lim
ℎ
ℎ→0
= lim (2𝑥 + ℎ)
ℎ→0
Now consider what will happen
as h approaches zero…
As h approaches zero, any term
with h as a factor will become
insignificant
So 𝑓 ′ 𝑥 = lim 2𝑥 + ℎ = 2𝑥
ℎ(2𝑥+ℎ)
ℎ
ℎ→0
= lim
Hence if 𝒇 𝒙 = 𝒙𝟐 then 𝒇′ 𝒙 = 𝟐𝒙.
You need to be able to
replicate this proof for small
integer powers of x, i.e. for f(x)
= x3, x4, etc.
For higher powers, the binomial
expansion will prove helpful.
In Year 2 you will also have to
apply it to sin x and cos x.
Questions
1. Derive from first principles the expressions of the gradients of the
curves
(a) 𝑦 = 𝑥
(b) 𝑓(𝑥) = 𝑥 3
Prior knowledge required
We will be using the same formula as before:
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
ℎ→0
𝑓 ′ 𝑥 = lim
This time our f(x) is either sin x or cos x
So 𝑓 𝑥 + ℎ = sin(𝑥 + ℎ) or cos(𝑥 + ℎ)
and we’re going to need to use the identities
𝐴+𝐵
𝐴−𝐵
sin
2
2
𝐴+𝐵
𝐴−𝐵
−2 sin
sin
2
2
sin 𝐴 − sin 𝐵 = 2 cos
cos 𝐴 − cos 𝐵 =
(these are in your formulae booklet)
and also the small angle approximation that, as 𝜃 → 0, sin 𝜃 → 𝜃
so
sin 𝜃
𝜃
→ 1, or
sin 𝜃
lim
𝜃→0 𝜃
=1
Angles
MUST be
in
radians!
Differentiation of sin x from first principles
We are going to show that if 𝑓 𝑥 = sin 𝑥
2 cos
′
𝑓 𝑥 = lim
then 𝑓′ 𝑥 = cos 𝑥.
ℎ→0
𝑓′
𝑥 =
𝑓(𝑥+ℎ)−𝑓(𝑥)
lim
ℎ
ℎ→0
First, substitute in for 𝑓 𝑥 and 𝑓 𝑥 + ℎ :
sin 𝑥+ℎ −sin 𝑥
ℎ
ℎ→0
𝑓 ′ 𝑥 = lim
Now use the trig identity for sin A – sin B:
𝐴+𝐵
𝑓′ 𝑥 =
= lim
(𝑥+ℎ)+𝑥
𝑥+ℎ −𝑥
sin
2
2
2 cos
ℎ
ℎ→0
2𝑥+ℎ
=
𝐴−𝐵
2 cos 2 sin 2
lim
ℎ
ℎ→0
2 cos 2
lim
ℎ
ℎ→0
ℎ
sin2
and since
2𝑎𝑏
𝑐
ℎ
2
𝑥+
sin
ℎ
2
ℎ
𝑏
is the same as 𝑎 × 𝑐 ,
ℎ
2
𝑓 ′ 𝑥 = lim cos(𝑥 + ) ×
ℎ→0
= lim cos 𝑥 +
ℎ→0
ℎ
2
2
ℎ
sin
2
ℎ
2
ℎ
since lim
ℎ→0
sin 2
ℎ
2
=1
As h approaches zero, any term with
h as a factor will become insignificant
So 𝑓 ′ 𝑥 = lim cos 𝑥 +
ℎ→0
ℎ
2
= cos 𝑥
Hence if 𝒇 𝒙 = 𝐬𝐢𝐧 𝒙 then 𝒇′ 𝒙 = 𝐜𝐨𝐬 𝒙
Now you try it for cos x.
Differentiation of cos x from first principles
We are going to show that if 𝑓 𝑥 =
cos 𝑥 then 𝑓 ′ 𝑥 = − sin 𝑥.
𝑓′ 𝑥 =
𝑓 ′ 𝑥 = lim
and
𝑓(𝑥+ℎ)−𝑓(𝑥)
lim
ℎ
ℎ→0
First, substitute in for 𝑓 𝑥 and 𝑓 𝑥 + ℎ :
𝐴+𝐵
𝑓′ 𝑥 =
= lim
−2 sin
(𝑥+ℎ)+𝑥
𝑥+ℎ −𝑥
sin
2
2
ℎ
ℎ→0
2𝑥+ℎ
=
𝐴−𝐵
−2 sin 2
lim
ℎ
ℎ→0
ℎ
sin2
ℎ
2
ℎ
𝑏
is the same as 𝑎 × 𝑐 ,
ℎ
2
ℎ→0
cos 𝑥+ℎ −cos 𝑥
ℎ
ℎ→0
−2 sin 2 sin 2
lim
ℎ
ℎ→0
ℎ→0
2𝑎𝑏
since
𝑐
sin
𝑓 ′ 𝑥 = lim −sin(𝑥 + ) ×
𝑓 ′ 𝑥 = lim
Now use the trig identity for sin A – sin B:
ℎ
2
−2 sin 𝑥+
= lim −sin 𝑥 +
ℎ→0
ℎ
2
2
ℎ
sin
2
ℎ
2
ℎ
since lim
ℎ→0
sin 2
ℎ
2
=1
As h approaches zero, any term with
h as a factor will become insignificant
So
𝑓′
𝑥 = lim −sin 𝑥 +
ℎ→0
ℎ
2
= −sin 𝑥
Hence if 𝒇 𝒙 = 𝐜𝐨𝐬 𝒙 then 𝒇′ 𝒙 =
−𝐬𝐢𝐧 𝒙
Worksheet
Question 1
𝑦 = 2𝑥 2 − 5𝑥 + 3
Question 2
5
3
𝑦 = 2 + 𝑥2
𝑥
Question 3
𝑦=
2
𝑥+1
Worksheet
Question 4
𝑦 = cos 2𝑥
Question 5
𝑦 = sin(2 − 𝑥)
Question 6
𝑦 = tan 𝑥