Cambridge Additional Mathematics

Introduction to differential calculus (Chapter 13) 357

We have already shown that if f(x)=bx then f^0 (x)=bx

μ

lim

h! 0

bh¡ 1

h

¶

.

So if f^0 (x)=bx then we require lim

h! 0

bh¡ 1

h

=1.

) lim

h! 0

bh= lim

h! 0

(1 +h)

Letting h=

1

n

, we notice that

1

n

! 0 if n!1

) lim

n!1

b

1

n= lim

n!1

³

1+

1

n

́

) b= lim

n!1

³

1+

1

n

́n

if this limit exists

We found that as n!1,

³

1+^1

n

́n

! 2 :718 281 828 459 045 235::::

and this irrational number is the natural exponentiale.

We now have: If f(x)=ex then f^0 (x)=ex.

THE DERIVATIVE OF ef(x)

The functions e¡x, e^2 x+3, and e¡x

2

all have the form ef(x).

Since ex> 0 for allx, ef(x)> 0 for allx, no matter what the function f(x).

Suppose y=ef(x)=eu where u=f(x).

Now

dy

dx

=

dy

du

du

dx

fchain ruleg

=eu

du

dx

=ef(x)£f^0 (x)

Function Derivative

ex ex

ef(x) ef(x)£f^0 (x)

Example 13 Self Tutor

Find the gradient function foryequal to:

a 2 ex+e¡^3 x b x^2 e¡x c

e^2 x

x

a If y=2ex+e¡^3 x then

dy

dx

=2ex+e¡^3 x(¡3)

=2ex¡ 3 e¡^3 x

b If y=x^2 e¡x then

dy

dx

=2xe¡x+x^2 e¡x(¡1)

=2xe¡x¡x^2 e¡x

fproduct ruleg

e

x

xe

x

1-x

is sometimes written

as. For example,

.

exp( )

exp(1 ) =¡

We have in fact already seen this limit inChapter 4

Discovery 2on page 123.

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_13\357CamAdd_13.cdr Friday, 4 April 2014 5:25:30 PM BRIAN