Objectives: To recognise the number e and be able to differentiate y=e kx

Objectives:

• To recognise the number e and be able to differentiate y=ekx

• To recognise the inverse of y=ekx

xey The function

Let us consider these graphs. We notice that all graphs pass through the same point (0,1) This is because anything to the power zero gives the answer 1. From the graphs we can see that as the value of the base increases, the gradient increases.

The gradient of the graph of at x = 0 is 0.693.

The gradient of the graph of at x = 0 is 1.098 There must then be some value between 2 and 3 which gives us a gradient of 1 when x = 0 . This value is 2.71828...

It continues on irrationally. This number is a special function and is known as the Exponential function and is denoted by the letter e.

y 2x

y 3x

xey

gradient of equalsxa

)3(7182 d.p.e

The value of a where the is an irrational number, written as e, where

xay

xx edx

dyey kxkx ke

dx

dyey

xey

xy

xey

The Inverse of xey

We can sketch the inverse by reflecting in y = x.

is a one-to-one function so has an inverse function.

xexf )(

xy ln

0xN.B. The domain is .

SUMMARY xexf )(• is a growth function.

7182e• (3 d.p.)

xey • At every point on , the gradient equals y:

xx edx

dyey

• The inverse of is

xexf )(

xxf ln)(1

( log with base e )is defined for x > 0

onlyxln

xy

xy lnxey

Calculus, ex and logarithms

Objectives:

To apply the laws of logs to solve equations involving ex and lnx

To be able differentiate and integrate functions involving exponential functions and natural logarithms.

Laws of logarithms

xyln

y

xln

kxln

x

x

e

e

e

ln

)ln(

1ln

ln

Laws of logarithms

xyln

y

xln

kxln

yx lnln

yx lnln

xk ln xe

xe

e

x

x

ln

)ln(

01ln

1ln

Activities:

• Multiple Choice• Natural Log Examples• Tarsia puzzle

HomeworkEx. 4A: 1,2,3 (2 parts of each) & 6

Ex. 4B: 1,2,3 (2 parts of each) & 5

Ex 4C: choose 3 (3 parts of each)

Ex 4E: 1,2,3,6 & one other

Mixed Ex: 2, 3 & choose 1 exam q. [A]