Cambridge Additional Mathematics

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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.

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

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