Transcript Ch.2 Limits and derivatives

Derivatives of polynomials

d
Derivative of a constant function (c)  0
dx

d n
n 1
(
x
)

nx
We have proved the power rule
dx

We can prove
d 1
1
( ) 2
dx x
x
1
1

1
1
1
x

h
x
( )  lim
  lim
 2
h 0
h 0 x ( x  h)
x
h
x
Rules for derivative

The constant multiple rule:
d
d
(cf ( x))  c
f ( x)
dx
dx

The sum/difference rule:
d
d
d
[ f ( x)  g ( x)] 
f ( x)  g ( x)
dx
dx
dx
Exponential functions

Derivative of
xh
f ( x)  lim
h 0

a
f ( x)  a x
 ax
ah 1
x
 a lim
 a x f (0)
h 0
h
h
The rate of change of any exponential function is
proportional to the function itself.

eh  1
1
e is the number such that lim
h 0
h

Derivative of the natural exponential function
d x
(e )  e x
dx
Product rule for derivative
d
d
d
The product rule: [ f ( x) g ( x)]  f ( x) g ( x)  g ( x) f ( x)
dx
dx
dx
( fg )  f ( x  x) g ( x  x)  f ( x) g ( x)
 [ f ( x  x) g ( x  x)  f ( x) g ( x  x)]  [ f ( x) g ( x  x)  f ( x) g ( x)]
 g ( x  x)f  f ( x)g ,
( fg )
f
g
 g ( x  x)  f ( x) .
x
x
x
g is differentiable, thus continuous, therefore,
( fg )
f
g
lim
 lim g ( x  x) lim  f ( x) lim  g ( x) f ( x)  f ( x) g ( x).
x 0 x
x 0
x 0 x
x 0 x
Remark on product rule

In words, the product rule says that the derivative of a
product of two functions is the first function times the
derivative of the second function plus the second function
times the derivative of the first function.

Derivative of a product of three functions:
( f ( x) g ( x)h( x))  f ( x)(g ( x)h( x))  f ( x)(g ( x)h( x))
 f ( x) g ( x)h( x)  f ( x) g ( x)h( x)  f ( x) g ( x)h( x)
Example
Find f (x ) if f ( x)  x2e x .
Sol. f ( x)  ( x2 )e x  x2 (e x )  2xex  x2e x  ( x 2  2x)e x .
Quotient rule for derivative

The quotient rule:  f ( x)   g ( x) f ( x)  f ( x) g ( x) .
2
 g ( x) 
g
( x)


 f  f ( x  x) f ( x) f ( x)  f f ( x)
  



 g  g ( x  x) g ( x) g ( x)  g g ( x)
f ( x) g ( x)  g ( x)f  f ( x) g ( x)  f ( x)g g ( x)f  f ( x )g


g ( x)( g ( x)  g )
g ( x)( g ( x )  g )
( f / g )
g ( x)
f
f ( x)
g

 
 .
x
g ( x)( g ( x)  g ) x g ( x )( g ( x )  g ) x
Example
Using the quotient rule, we have:

 1 
f ( x)

   2
f ( x)
 f ( x) 

n
 1 
( x  n )   n    n 1  ( n) x (  n ) 1 which means ( x k )  kxk 1
x
x 
is also true for any negative integer k.
Homework 4

Section 2.7: 8, 10

Section 2.8: 16, 17, 22, 24, 36

Section 2.9: 28, 30, 46, 47

Page 181: 13
Example
We can compute the derivative of any rational functions.
x2  x  2
.
Ex. Differentiate y 
3
x 6
3
2
2
3

(
x

6)(
x

x

2)

(
x

x

2)(
x
 6)

Sol. y 
( x3  6)2
( x3  6)(2 x  1)  ( x 2  x  2)(3x 2 )

( x3  6)2
 x 4  2 x3  6 x 2  12 x  6

( x3  6)2
Table of differentiation formulas
d
(c )  0
dx
(cf )  cf 
d n
( x )  nx n 1
dx
d x
(e )  e x
dx
( f  g )  f   g 
( fg )  fg   gf 

 f  gf   fg 
  
2
g
g
An important limit
Prove that

Sol. It is clear that when x  (0, ), sin x  x  tan x
2
thus
Since cos x and sin x are even functions,
x
we have
sin x
cos x 
 1, x  ( / 2, 0) (0,  / 2)
x
Now the squeeze theorem together with
gives the desired result.
Derivative of sine function
Find the derivative of f ( x)  sin x.
Sol. By definition,
f ( x  h)  f ( x )
sin( x  h)  sin x
f ( x)  lim
 lim
h 0
h 0
h
h
2x  h
h
2 cos
sin
2x  h
sin(h / 2)
2
2
 lim
 lim cos
 lim
h 0
h 0
h 0 ( h / 2)
h
2
sin t
 cos x  lim
 cos x
t 0
t
Derivative of cosine function
Ex. Find the derivative of f ( x)  cos x.
Sol. By definition,
f ( x  h)  f ( x )
cos( x  h)  cos x
 lim
h 0
h 0
h
h
2x  h
h
2sin
sin
2x  h
sin(h / 2)
2
2
 lim
  limsin
 lim
h 0
h

0
h 0 ( h / 2)
h
2
sin t
  sin x  lim
  sin x
t 0
t
f ( x)  lim
Derivatives of trigonometric functions
Using the quotient rule, we have:
(sec x)  sec x tan x, (csc x)   csc x cot x
(tan x)  sec2 x,
(cot x)   csc2 x
Change of variable

The technique we use in
sin(h / 2)
sin t
lim
 lim
1
h 0 ( h / 2)
t 0
t
is useful in finding a limit.

The general rule for change of variable is:
g ( x )l ( x a )
lim f ( g ( x))   lim f (u ).
x a
u l
Example
Ex. Evaluate the limit
lim
xa
sin x  sin a
.
xa
xa xa
sin
Sol. Using the formula sin x  sin a  2 cos
2
2
and putting u=(x-a)/2, we derive
xa
xa
2 cos
sin
sin x  sin a
2
2
lim
 lim
x a
x a
xa
xa
xa
2 sin u
 lim cos
 lim
 cos a.
x a
u 0
2
2u
Example
1  cos x
.
Ex. Find the limit lim
2
x 0
x
Sol. Using the trigonometry identity 1  cos x  2 sin 2
and putting u=x/2, we obtain
1  cos x
2 sin 2 ( x / 2)
sin 2 u
lim
 lim
 lim
2
2
x 0
x

0
u 0 2u 2
x
x
2
2
1
1
sin u 
1
 sin u 
 lim 
   lim
  .
x

0
x

0
2
2
u 
2
 u 
x
2
Example
cos x
arcsin x
lim
.
Ex. Find the limits: (a) lim
, (b)


x0
x
x
x
2
2
Sol. (a) Letting u  arcsin x, then x  sin u, and
arcsin x
u
lim
 lim
 1.
x 0
u

0
x
sin u
(b) Letting u   / 2  x,
lim
x

2
cos x

2
x
 lim
x

2
sin(


2
2
then
 x)
x
sin u
 lim
 1.
u 0
u