Limits and continuity

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Limits and continuity

2

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

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Differential Calculus

Concepts of Limits and Continuity

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

6

The idea of limits

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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

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Derivatives of functions

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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

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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

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Differentiation rules and Higher Derivatives

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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

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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

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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

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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

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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

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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

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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

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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

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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



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

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Example 3



)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

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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

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Applications

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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

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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

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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

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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

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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

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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

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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

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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 +

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Example 2

By first derivative test, x=1 is the local maximum (+ve -> 0 -> -ve) x=3 is the local minimum (-ve -> 0 -> +ve)

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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

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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

Limits and continuity

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