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