Getting to know the gradient
f(x + h)
f(x)
x
x+h
Differentiation from first principles
400
360
320
P
280
f(x+h)
240
200
160
120
80
A
f(x)40
0
0
0.5
x1
1.5
Gradient of AP =
2
2.5
f( x  h)  f( x)
h
3
x3.5
+h
4
Differentiation from first principles
400
360
320
280
240
200
P
f(x+h)
160
120
80
A
f(x)40
0
0
0.5
x1
1.5
Gradient of AP =
2
2.5
f( x  h)  f( x)
h
x +3 h
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
P
120
f(x+h)
80
A
f(x)40
0
0
0.5
x1
1.5
Gradient of AP =
2
2.5
x+
h
f( x  h)  f( x)
h
3
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
120
P
80
f(x+h)
A
f(x)40
0
0
0.5
x1
1.5
Gradient of AP =
2
x+
h
2.5
f( x  h)  f( x)
h
3
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
120
80
f(x+h)
f(x)40
A
P
0
0
0.5
x1
x 1.5
+h
Gradient of AP =
2
2.5
f( x  h)  f( x)
h
3
3.5
4
Differentiation from first principles
400
360
320
280
240
200
160
120
80
A
40
f(x)
0
0
0.5
x1
1.5
2
Gradient of tangent at A =
2.5
3
Limit f( x  h)  f( x)
h0
h
3.5
4
Finding the gradient using a secant line
x+h
Secant line
y 2  y1
gradient 
x 2  x1
x
𝑦2 = 𝑓 𝑥 + ℎ = (𝑥 + ℎ)2

2 − 𝑥2
𝑥
+
ℎ)
𝑓 ′ (𝑥) =
𝑥+ℎ −𝑥
𝑓 ′ (𝑥)
2ℎ𝑥 + ℎ2
=
ℎ
𝑓 ′ (𝑥) =
ℎ(2𝑥 + ℎ)
ℎ
Expand & simplify
𝑦1 = 𝑓 𝑥 = 𝑥 2
2 + 2ℎ𝑥 + ℎ2 ) − 𝑥 2
𝑥
𝑓 ′ (𝑥) =
ℎ
𝑓 ′ 𝑥 = 2x + h
This is the gradient of the Secant line
As h → 𝟎 the secant line becomes the tangent
x+h
x
𝑓 ′ 𝑥 = lim 2x + h
ℎ→0
𝑓 ′ 𝑥 = 2x
So to find the gradient of 𝑓 𝑥 = 𝑥 2 for any value of x, we use the
gradient function 𝑓 ′ 𝑥 = 2x
Also known as the derivative of f
In general
The derivative of f(x)
𝑓 𝑥 + ℎ − 𝑓(𝑥)
𝑓 𝑥 = lim
ℎ→0
ℎ
′
Hence find the derivative of 𝑓 𝑥 = 2𝑥 2 + 1
Differentiation from first principles
f(x) = x3
( x  h)3  x 3
x 3  3 x 2 h  3 xh 2  h3  x 3
3 x 2 h  3 xh 2  h
f( x  h)  f( x)



h
h
h
h
(3 x 2  3 xh  h 2 )h  3 x 2  3 xh  h 2

h
f '( x) 
Limit f( x  h)  f( x)
 3x 2
h0
h
𝑓 𝑥 + ℎ − 𝑓(𝑥)
𝑓 𝑥 = lim
ℎ→0
ℎ
′
𝑓′ 𝑥
2
Differentiate: 𝑓 𝑥 = 2𝑥 + 1
2(𝑥 + ℎ)2 + 1 − (2𝑥 2 + 1)
𝑓 𝑥 = lim
ℎ→0
ℎ
𝑓 𝑥 + ℎ − 𝑓(𝑥)
= lim
ℎ→0
ℎ
′
𝑓′
2(𝑥 2 + 2ℎ𝑥 + ℎ2 ) + 1 − (2𝑥 2 + 1)
𝑥 = lim
ℎ→0
ℎ
2𝑥 2 + 4ℎ𝑥 + 2ℎ2 + 1 − 2𝑥 2 − 1
𝑓 𝑥 = lim
ℎ→0
ℎ
′
4ℎ𝑥 + 2ℎ2
𝑓 𝑥 = lim
ℎ→0
ℎ
′
ℎ(4𝑥 + 2ℎ)
ℎ→0
ℎ
𝑓 ′ 𝑥 = lim
𝑓 ′ 𝑥 = lim 4𝑥 + 2ℎ
ℎ→0
𝑓 ′ 𝑥 = 4𝑥
Let’s find an easier general rule
Here are some results – what’s the rule
𝑓 𝑥 = 3𝑥 2 + 2𝑥
𝑓 𝑥 = 5𝑥 3 − 7
𝑓′ 𝑥 = 6𝑥 + 2
𝑓′ 𝑥 = 15𝑥 2
𝑓 𝑥 = 𝑎𝑥 𝑛 + 𝐶,
𝑓 𝑥 = 𝑥 12 − 7𝑥 3
𝑓′
𝑥
= 12𝑥 11 − 21𝑥 2
𝑓 ′ 𝑥 = 𝑎𝑛𝑥 𝑛−1
𝑓 𝑥 = 2𝑥 −4
𝑓′
𝑥
= −8𝑥 −5
Differentiate the following … or find the derivative of f(x)
1. 𝑓 𝑥 = 𝑥 5
𝑓′ 𝑥 = 5𝑥 4
𝑥8
𝑓′ 𝑥 = 8𝑥 7
2. 𝑓 𝑥 =
3. 𝑓 𝑥 =
4. 𝑓 𝑥 =
5. 𝑓 𝑥 =
6. 𝑓 𝑥 =
1
𝑥4
3
𝑥
1
𝑥
5
𝑥3
𝑓
′ 𝑥
4
=− 5
𝑥
𝑓′ 𝑥 =
𝑓′ 𝑥
1. 𝑓 𝑥 =
1
3
3 𝑥2
=−
1
2 𝑥3
3
5
5 𝑥2