1. Derivative of Logarithm Function
If y = logex then = dy dx 1 x
Proof
Given: y = logex
then x = ey dx dy
= ey
Exercise 4.6 (page 146) dy dx = 1 ey = 1 x

2. Function of a Function Rule
If y = logef(x) then = f’(x) = dy dx 1
f(x)
f’(x)
Proof
Let u = f(x)
then y = logeu du dx
Also = f’(x) dy du = 1 u dy dx du
= x
Exercise 4.6 (page 146)
= f’(x) 1 u
= f’(x) 1
f(x)
f’(x)
f(x) =

3. Function of a Function Rule – Example 1/3
f’(x)
f(x) = dy dx
Differentiate loge(2x2 - x + 1)
y = loge(2x2 - x + 1)
Exercise 4.6 (page 146)
4x – 1 dy dx =
2x2 - x + 1

4. Function of a Function Rule – Example 2/3
f’(x)
f(x) = dy dx
x + 1
x - 1
Differentiate loge =
y = loge(x + 1) - loge(x - 1) 1 1
(x - 1).1-(x + 1).1
(x + 1)(x - 1) = dy dx =
x + 1
x - 1
Exercise 4.6 (page 146)
x – 1 - x - 1
(x + 1)(x - 1) =
- 2
x2 - 1 =

5. Function of a Function Rule – Example 3/3
f’(x)
f(x) = dy dx
Find the gradient of the normal to
loge(x2 - x + x = 1
g(x) = loge(x2 - x + 1)
Gradient of Normal
m2 = -1 m1
2x – 1
g’(x) =
Exercise 4.6 (page 146)
x2 - x + 1
2 – 1
g’(1) = = -1 1 m1
= 1 m2
= -1