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Limits and continuity. Start from an example. If the voltage apply to the resistor can be express by the function Determine the maximum voltage across the resistor. We can calculate the maximum or minimum by use derivative. Contents. Concepts of Limits and Continuity - PowerPoint PPT Presentation
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Start from an example If the voltage apply to the resistor can be express
by the functionDetermine the maximum voltage across the resistor.
We can calculate the maximum or minimum by use derivative
296 23 xxxy
dx
dy
3
Contents
Concepts of Limits and Continuity Derivatives of functions Differentiation rules and Higher
Derivatives Applications
5
The idea of limits
Consider a function
The function is well-defined for all real values of x
The following table shows some of the values:
2x)x(f
x 2.9 2.99 2.99 3 3.001 3.01 3.1
F(x) 8.41 8.94 8.994 9 9.006 9.06 9.61
92
33
xlim)x(flim
xx
7
Concept of Continuity
E.g. is continuous at x=3?
The following table shows some of the values:
exists asand => f(x) is continuous at x=3!
2x)x(f
x 2.9 2.99 2.99 3 3.001 3.01 3.1
F(x) 8.41 8.94 8.994 9 9.006 9.06 9.61
93
)x(flimx
93
)x(flimx
)x(flimx 3
933
)x(flim)x(flimxx
)(f)x(flimx
393
9
Derivative (導數 )
Given y=f(x), if variable x is given an increment x from x=x0, then y would change to f(x0+x)
y= f(x0+x) – f(x)
yx is the slope of triangular ABC
x
f(x0+x)
f(x0)
x0+xx0
y
x
y
Y=f(x)
A
B
C
10
Derivative
What happen with yx as x tends to 0? It seems that yx will be close to the slope of the
curve y=f(x) at x0. We defined a new quantity as follows
If the limit exists, we called this new quantity as the derivative of f(x).
The process of finding derivative of f(x) is called differentiation.
x
)x(f)xx(flim
x
ylim
dx
)x(df
dx
dyxx
00
11
Derivative
y
f(x0+x)
f(x0)
x0+xx0x
x
y
Y=f(x)
A
B
C
Derivative of f(x) at Xo = slope of f(x) at Xo
12
Differentiation from first principle
Find the derivative of with respect to (w.r.t.) x
xx)x(fy 32
32
32
32
3332
33
0
2
0
222
0
22
0
0
0
x
)xx(limx
x)x()x(xlim
x
)xxxx)x()x(xxlim
x
)xx()xx()xx(lim
x
)x(f)xx(flim
x
ylim
dx
dy
x
x
x
x
x
x
To obtain the derivative of a function by its definition is called differentiation of the function from first principles
14
Fundamental formulas for differentiation I
Let f(x) and g(x) be differentiable functions and c be a constant.
0dx
)c(d
dx
)x(dg
dx
)x(df
dx
))x(g)x(f(d
dx
)x(dfc
dx
))x(cf(d
1 nn
nxdx
)x(dfor any real number n
15
Examples
Differentiate and w.r.t. x
143 2 xx 9685 23 xxx
46
0423
043
143
143
2
2
2
x
)x(dx
)x(d
dx
)x(d
dx
)(d
dx
)x(d
dx
)x(d
dx
dy
xxy
61615
062835
968
55
9685
2
2
23
23
xx
)x()x(
dx
)(d
dx
)x(d
dx
)x(d)(
dx
)x(d
dx
dy
xxxy
16
Fundamental formulas for differentiation II
Let f(x) and g(x) be differentiable functions
dx
))x(f(d)x(g
dx
))x(g(d)x(f
dx
))x(g)x(f(d
2))x(g(dx)x(dg
)x(fdx)x(df
)x(g
dx
))x(g/)x(f(d
17
Example 1
Differentiate w.r.t. xx
xy
32
32
2
2
2
32
12
32
332332
32
3232
3232
)x(
)x(
))(x())(x(
)x(dx
)x(d)x(
dx)x(d
)x(
dx
dy
x)x(f 32 x)x(g 32
18
Example 2
Differentiate w.r.t. x)xx)(xx(y 523142 22
2222424
4452326142
422523223142
142523
523142
142523
523142
23
22
22
22
22
22
22
xxx
)x)(xx()x)(xx(
))x()(xx())x()(xx(
)dx
)(d
dx
dx
dx
dx)(xx()
dx
)(d
dx
dx
dx
dx)(xx(
dx
)xx(d)xx(
dx
)xx(d)xx(
dx
dy
19
Fundamental formulas for differentiation III
)xsin(dx
)xcos(d )xcos(
dx
)xsin(d
22
1))x(sec(
))x(cos(dx
)xtan(d
22
1))x(csc(
))x(sin(dx
)xcot(d
)xtan()xsec(dx
)xsec(d )xcot()xcsc(
dx
)xcsc(d
!
x
!
x
!
x
!
xxewhere x
54321
5432x
x
edx
de
xdx
)xln(d 1 x)eln(where x
ln(x) is called natural logarithm
20
Differentiation of composite functions
To differentiate w.r.t. x, we may have problems as we don’t have a formula to do so.
The problem can be simplified by considering composite function:
Let so and 12 xy12 xu uy
12 xy
21
Chain Rule
Chain Rule states that :given y=g(u), and u=f(x)
So our problem and
dx
du
du
dy
dx
dy
2
122
1
2
122
1
1
)x(x)x(u
dx
du
du
dy
dx
xd
dx
dy 2
1
2
1 u
du
ud
du
dy
x
dx
xd
dx
du2
12
u)u(gy 12 x)x(fu
22
Example 1
Differentiate w.r.t. x
Simplify y by letting so nowBy chain rule
3))x(cos(y
3uy
dx
du
du
dy
dx
dy
23
3udu
du
du
dy )xsin(
dx
))x(cos(d
dx
du
22 33 ))x)(cos(xsin())xsin((udx
du
du
dy
dx
dy
)xcos(u
23
Example 2
Differentiate w.r.t. x
Simplify y by letting so nowBy chain rule
2xey
2xu uey
dx
du
du
dy
dx
dy
uu
edu
de
du
dy x
dx
dx
dx
du2
2
2
22 xu xexedx
du
du
dy
dx
dy
24
Example 3
Differentiate w.r.t. x
Simplify y by letting so nowBy chain rule
)xln(y 32 2
32 2 xu )uln(y
dx
du
du
dy
dx
dy
udu
)uln(d
du
dy 1 x
dx
)x(d
dx
du4
32 2
32
44
12
x
xx
udx
du
du
dy
dx
dy
25
Higher Derivatives)
If the derivatives of y=f(x) is differentiable function of x, its derivative is called the second derivative of y=f(x) and is denoted by or f ’’(x). That is
Similarly, the third derivative =
the n-th derivative =
2
2
dx
yd
)dx
dy(
dx
d
dx
yd
2
2
)dx
yd(
dx
d
dx
yd)x(f )(
2
2
3
32
)dx
yd(
dx
d
dx
yd)x(f
n
n
n
n)n(
1
1
26
Example
Find if 3
3
2
2
dx
yd,
dx
yd,
dx
dy 68124 234 xxxxy
8243166
8124 23234
xxxdx
)(d
dx
dx
dx
dx
dx
dx
dx
dx
dx
dy
246488
24316 223
2
2
xxdx
)(d
dx
dx
dx
dx
dx
dx)
dx
dy(
dx
d
dx
yd
69624
6482
2
2
3
3
xdx
)(d
dx
dx
dx
dx)
dx
yd(
dx
d
dx
yd
28
Slope of a curve
Recall that the derivative of a curve evaluate at a point is the slope of the curve at that point.
f(x0+x)
f(x0)
x0+xx0x
x
y
Y=f(x)
A
B
C
Derivative of f(x) at Xo = slope of f(x) at Xo
29
Slope of a curve Find the slope of y=2x+3
at x=0 To find the slope of a
curve, we have to compute the derivative of y and then evaluate at a point
The slope of y at x=0 equals 2
(y=mx+c now m=2)
232
dx
)x(d
dx
dy
30
Slope of a curve Find the slope of at
x=0, 2, -2
The slope of y = 2x The slope of y (at x=0) = 2(0) = 0 The slope of y (at x=-2) =2(-2) = -4 The slope of y (at x=2) =2(2) = 4
xdx
)x(d
dx
dy2
12
12 xy
X=0
X=2X=-2
31
Local maximum and minimum point
For a continuous function, the point at which is called a stationary point. This gives the point local maximum or local
minimum of the curve
0dx
dy
A
B
C
D
X1
X2
32
First derivative test (Max)
Given a continuous function y=f(x)
If dy/dx = 0 at x=xo & dy/dx changes from +ve to –ve through x0,
x=x0 is a local maximum point
x=x0
local maximum point
33
First derivative test (Min)
Given a continuous function y=f(x)
If dy/dx = 0 at x=xo & dy/dx changes from -ve to +ve through x0,
x=x0 is a local minimum point
x=x0 local minimum point
34
Example Determine the position of any local maximum and
minimum of the function
First, find all stationary point (i.e. find x such that dy/dx = 0)
, so when x=0
By first derivative test x=0 is a local minimum point
12 xy
xdx
)x(d
dx
dy2
12
0dx
dy
020 xdx
dy,xwhen 020 x
dx
dy,xwhen
35
Example 2
Find the local maximum and minimum of
Find all stationary points first:
296 23 xxxy
296 23 xxxy )x)(x(xxdx
dy3139123 2
310 xandxwhen,dx
dy
x 0 x<1 x=1 1<x<3 x=3 x>3
dy/dx + + 0 - 0 +
36
Example 2
By first derivative test, x=1 is the local maximum (+ve -> 0 -> -ve) x=3 is the local minimum (-ve -> 0 -> +ve)
37
Second derivative test
Second derivative test states:
There is a local maximum point in y=f(x) at x=x0, if at x=x0 and < 0 at x=x0.
There is a local minimum point in y=f(x) at x=x0, if at x=x0 and > 0 at x=x0.
If dy/dx = 0 and =0 both at x=x0, the second derivative test fails and we must return to the first derivative test.
0dx
dy2
2
dx
yd
0dx
dy2
2
dx
yd
2
2
dx
yd
38
Example
Find the local maximum and minimum of
Find all stationary points first:
By second derivative test, x=1 is max and x=3 is min
296 23 xxxy
296 23 xxxy )x)(x(xxdx
dy3139123 2
310 xandxwhen,dx
dyxx
dx
yd126
2
2
0612161
2
2
)(dx
yd
x
0612363
2
2
)(dx
yd
x