Upload
pirawin92
View
221
Download
0
Embed Size (px)
Citation preview
8/12/2019 Derivatives Module 1 September 2011
1/47
8/12/2019 Derivatives Module 1 September 2011
2/47
26/09/2011Derivatives - by definition 2
Learning Objective
At the end of the module, studentsshould be able to
compute derivative of a functionat arbitrary point using definition.
determine whether a functionis differentiable or not.
8/12/2019 Derivatives Module 1 September 2011
3/47
26/09/2011Derivatives - by definition 3
Secant line
x0 x0+ hx
y
P f ( x0 +h)
f ( x0)
0
y = f ( x)Q
h x f h x f
xh x x f h x f
m PQ)()(
)()()( 00
00
00
Slope of the secant line PQ:
8/12/2019 Derivatives Module 1 September 2011
4/47
26/09/2011Derivatives - by definition 4
x0 x0+ hx
y
P f ( x0 + h)
f ( x0)
0
y = f ( x)
Q
, provided the limit exists.
Derivative as slope to a curve y = f ( x)at the point x = xo.
which can be represented as limit below:
h x f h x f
mh
)()(lim 00
0
By moving the red dot closer andcloser to the black dot we canapproximate slope on the curve at point P.
8/12/2019 Derivatives Module 1 September 2011
5/47
26/09/2011Derivatives - by definition 5
Derivative of a function
The derivative (slope) of a function f atwith respect to the variable x,is
is defined by
),( x f
h
x f h x f
x f h)()(
lim)( 0
8/12/2019 Derivatives Module 1 September 2011
6/47
8/12/2019 Derivatives Module 1 September 2011
7/4726/09/2011Derivatives - by definition 7
Example 1 - solution ,)( 2 x x x f
h x f h x f x f
h)()(lim)(
0
(i)
8/12/2019 Derivatives Module 1 September 2011
8/4726/09/2011Derivatives - by definition 8
Example 1 - solution ,)( 2 x x x f
h
x xh xh x
h
)][()]()[(lim
22
0
h
x xh xh xh x
h
)][()]()2[(lim
222
0
(i)
8/12/2019 Derivatives Module 1 September 2011
9/4726/09/2011Derivatives - by definition 9
Example 1 - solution
h
hh xh x f
h
2
0
2lim)(
1212lim)(0
xh x x f h
(i)
hh xh
x f h
)12(lim)(
0
8/12/2019 Derivatives Module 1 September 2011
10/4726/09/2011Derivatives - by definition 10
Example 1 - solution 12)( x x f
(ii) ,)( 2 x x x f
-1/41/2 1
x
y
0
Domain?
8/12/2019 Derivatives Module 1 September 2011
11/4726/09/2011Derivatives - by definition 11
Example 2
Find the derivative of ,1)( x x f
Is there any difference between the domain
of ? and f f
8/12/2019 Derivatives Module 1 September 2011
12/4726/09/2011Derivatives - by definition 12
Example 2- solution ,1)( x x f
h x f h x f
x f h
)()(lim)(
0
h xh x
h
11)(lim
0
8/12/2019 Derivatives Module 1 September 2011
13/4726/09/2011Derivatives - by definition 13
Example 2 - solution
11)(
11)(.
11)(lim)(
0
xh x
xh x
h
xh x x f
h
11)(
)1(1)(lim
0
xh xh
xh xh
8/12/2019 Derivatives Module 1 September 2011
14/4726/09/2011Derivatives - by definition 14
Example 2 - solution
)11)((lim)(
0 xh xh
h x f
h
12
1
11)(
1lim
0
x xh xh
8/12/2019 Derivatives Module 1 September 2011
15/47
26/09/2011Derivatives - by definition 15
Example 2 - solution
12
1)(
x x f ,1)( x x f
Domain Domain
8/12/2019 Derivatives Module 1 September 2011
16/47
26/09/2011Derivatives - by definition 16
Example 3
Find .22
)(if )(t t
t g t g
What is the difference between thedomain of ?and g g
8/12/2019 Derivatives Module 1 September 2011
17/47
26/09/2011Derivatives - by definition 17
Example 3 - solution .22
)(if )(t t
t g t g
h
t g ht g t g
h
)()(lim)(
0
ht t
ht ht
t g h
]22
[])(2)(2
[
lim)( 0
8/12/2019 Derivatives Module 1 September 2011
18/47
8/12/2019 Derivatives Module 1 September 2011
19/47
26/09/2011Derivatives - by definition 19
Example 3 - solution
)2)(2()2224()2224(
lim)(22
0 t ht hht t t ht ht ht t t
t g h
)2)(2(4
lim)(0 t ht h
ht g
h
8/12/2019 Derivatives Module 1 September 2011
20/47
26/09/2011Derivatives - by definition 20
Example 3 - solution
2
0
)2(4
)(
)2)(2(
4lim)(
t t g
t ht t g
h
8/12/2019 Derivatives Module 1 September 2011
21/47
26/09/2011Derivatives - by definition 21
Example 3 - solution
2)2(4
)(t
t g
2exceptnumberrealalldomain22
)( t t t t
t g
2exceptnumberrealalldomain t t
No difference in their domain.
8/12/2019 Derivatives Module 1 September 2011
22/47
26/09/2011Derivatives - by definition 22
Example 4
Find the derivative of the function
,13)(2
x x x f
at the number x = 2.
8/12/2019 Derivatives Module 1 September 2011
23/47
26/09/2011Derivatives - by definition 23
Example 4 - solution ,13)( 2 x x x f
By
h x f h x f x f
h)()(lim)( 00
00
8/12/2019 Derivatives Module 1 September 2011
24/47
26/09/2011Derivatives - by definition 24
Example 4 - solution
h f h f
f h
)2()2(lim)2(
0
hhh
f h
]12)2(3[]1)2()2(3[lim)2(
22
0
8/12/2019 Derivatives Module 1 September 2011
25/47
26/09/2011Derivatives - by definition 25
Example 4 - solution
hhh
f h]12)2(3[]1)2()2(3[
lim)2(
22
0
h
hhh f
h
]11[]1)2()44(3[lim)2(
2
0
11113lim113
lim)2(0
2
0h
hhh
f hh
8/12/2019 Derivatives Module 1 September 2011
26/47
26/09/2011Derivatives - by definition 26
Example 5
Given that f ( x) = x3 + 1
(a) Find the average rate of change of f with respect to x over the interval[ 2, 6].
(b) Find the instantaneous rate ofchange of f with respect to x at
x = - 2.
8/12/2019 Derivatives Module 1 September 2011
27/47
26/09/2011Derivatives - by definition 27
Example 5 - solution
(a)
4)12()16(
26)2()6( 33 f f
r ave
524
20826
9217
h x f h x f )()(
changeof rateaverage 00
f ( x) = x3 + 1
8/12/2019 Derivatives Module 1 September 2011
28/47
26/09/2011Derivatives - by definition 28
Example 5 - solution
h f h f
h x f h x f
r
h
hinst
)2()2(lim
)()(lim
0
00
0
(b)
h
hh
)1)2[(]1)2[(
lim
33
0
f ( x) = x3 + 1
8/12/2019 Derivatives Module 1 September 2011
29/47
8/12/2019 Derivatives Module 1 September 2011
30/47
26/09/2011Derivatives - by definition 30
Example 5 - solution
(b)
hhhh
h
)2)2)(2()2)((22(lim
22
0
hhhhh
h
)42444)((lim
2
0
121200)126(lim2
0 hhh
8/12/2019 Derivatives Module 1 September 2011
31/47
26/09/2011Derivatives - by definition 31
PracticeFind the derivative of the given function usingdefinition and state the domain of the functionthe domain of its derivative.
1)( x x f (i)
2
1)( x x f (ii)
8/12/2019 Derivatives Module 1 September 2011
32/47
26/09/2011Derivatives - by definition 32
Some other notations for derivative
If y = f ( x), then the derivative can also be
denoted as
dxdf
dxdy
y x f ),(
)()()( x f D x Df x f dxd
x
8/12/2019 Derivatives Module 1 September 2011
33/47
26/09/2011Derivatives - by definition 33
Definition
A function f is differentiable at c ifexists.
)(c f
8/12/2019 Derivatives Module 1 September 2011
34/47
26/09/2011Derivatives - by definition 34
Definition
It is differentiable on an open interval ( a ,b)
[ or or or ] if it is
differentiable at EVERY NUMBER IN THEINTERVAL .
),(a ),( a ),(
8/12/2019 Derivatives Module 1 September 2011
35/47
26/09/2011Derivatives - by definition 35
Example
Determine, where the function
x x f )( is differentiable?
8/12/2019 Derivatives Module 1 September 2011
36/47
26/09/2011Derivatives - by definition 36
Example
Recall
0,0,)(
x x x x x x f
0 x
y
8/12/2019 Derivatives Module 1 September 2011
37/47
26/09/2011Derivatives - by definition 37
Example - solution
Case 1: If x > 0
h x f h x f x f
h)()(lim)(
0
11lim)(
lim00 hh
h
xh x
Therefore, f ( x) is differentiable for all x when x > 0 .
0,
0,)(
x x
x x x x f
8/12/2019 Derivatives Module 1 September 2011
38/47
26/09/2011Derivatives - by definition 38
Example - solution
Case 2: If x < 0
h x f h x f x f
h)()(lim)(
0
1)1(lim)()(
lim00 hh
h
xh x
Therefore, f ( x) is differentiable for all x when x < 0 .
0,
0,)(
x x
x x x x f
8/12/2019 Derivatives Module 1 September 2011
39/47
26/09/2011Derivatives - by definition 39
Example - solution
x
y
0
x x f )( x x f )(
Case 3: If x = 0
h x f h x f x f
h)()(lim)(
0
h
hh
00lim0 does not exists.
8/12/2019 Derivatives Module 1 September 2011
40/47
26/09/2011Derivatives - by definition 40
Example - solution
x
y
0
x x f )( x x f )(
Case 3: If x = 0
100
lim0 h
hh
100
lim0 h
hh
and
Therefore, is not differentiable at x = 0 . x x f )(
The domain where is differentiable is x x f )(),0()0,(
8/12/2019 Derivatives Module 1 September 2011
41/47
26/09/2011Derivatives - by definition 41
Example - solution
x
y
0
x x f )( x x f )(
The domain where is differentiable is x x f )(
),0()0,(
8/12/2019 Derivatives Module 1 September 2011
42/47
26/09/2011Derivatives - by definition 42
Theorem
If f is differentiable at x = a , then f iscontinuous at x = a .
Note that the opposite (or converse) of the theoremis not always true (See previous example).
8/12/2019 Derivatives Module 1 September 2011
43/47
26/09/2011Derivatives - by definition 43
3 possibilities a function is notdifferentiable.
y
x0
If the graph of the function
has a sharp corner init, the the graph has notangent at this point and thereforethe function is not differentiable at that point.
1.
8/12/2019 Derivatives Module 1 September 2011
44/47
26/09/2011Derivatives - by definition 44
3 possibilities a function is notdifferentiable.
y
x0
If the graph of the function
has jump discontinuity (i.e thefunction discontinuous), thenthe function is not differentiable.
2.
8/12/2019 Derivatives Module 1 September 2011
45/47
26/09/2011Derivatives - by definition 45
3 possibilities a function is notdifferentiable.
y
x0
If the graph of the function f has vertical tangent at x = a ,that is, f is continuous at x = a ,however
3. (Infinite discontinuities)
.)(lim x f a x
a
This means that the tangent lines becomes steeper and steeper as .a x
8/12/2019 Derivatives Module 1 September 2011
46/47
26/09/2011Derivatives - by definition 46
PracticeShow that
1,2
1,2)(
2
x x
x x x f
is continuous but not differentiable at x = 1 .Sketch the graph of f and . f
8/12/2019 Derivatives Module 1 September 2011
47/47
END OF THE MODULE 1