Chapter 8A Full Version - Differentiation Techniques

8/3/2019 Chapter 8A Full Version - Differentiation Techniques

1/15

Raffles Junior CollegeH2 Mathematics 9740JC1 2008 (Year 5) Full Version

__________________________________________________________

______________________________________________________Chapter 8A: Differentiation I Techniques of Differentiation & Limits

Page 1 of 15

Chapter 8ADifferentiation I Techniques of Differentiation & Limits

Syllabus will include:

differentiation of simple functions defined implicitly or parametrically

finding the numerical value of a derivative at a given point using a graphic calculator

PRE-REQUISITES

Understand the idea of a derived function.

Usethederivativesofn

x (foranyrational n ),sinx ,cosx ,tanx ,ex ,lnx ,together with

constant multiples, sums, and composite functions of these. Differentiate products and quotients of functions.

CONTENT

1 Limits and Derivative

1.1 Limits

1.2 Derivative as a Limit

2 Basic Rules of Differentiation

3 Derivatives of Standard Functions

3.1 Power Functions3.2 Exponential and Logarithmic Functions

3.3 Trigonometric Functions

3.4 Inverse Trigonometric Functions

4 Higher Order Derivatives

5 Implicit Differentiation

6 Parametric Differentiation

6.1 Parametric Equations

6.2 Differentiation of Parametric Equations

INTRODUCTION

Differentiation deals with the concept of change. It is useful for understanding phenomena inmany fields of study,including business,economics and social sciences.This topic reinforcesthe ideas of derivatives that students learnt in O-Level Additional Mathematics and furtherdevelops the idea of derivatives as a limit.In this chapter,students will learn more techniquesof differentiation such as implicit differentiation and parametric differentiation. Students willalso learn to use the graphic calculator to calculate derivatives.


2/15

Raffles Junior College H2 Mathematics (9740) JC1 2008______________________________________________


Page 2 of 15

1 LIMITS AND DERIVATIVES

1.1 Limits

Consider the expression2

1x + .

Let x approaches 2 from the left. Let x approaches 2 from the right.

We write 2x . We write 2x + .

From the table above, we see that From the table above, we see that

as 2x ,2

1x + approaches 5, as 2x + ,2

1x + approaches 5,

i.e. the limiting value of2

1x + is 5. i.e. the limiting value of2

1x + is 5.

We write2

2lim ( 1) 5

xx

+ = . We write2

2lim ( 1) 5

xx

+

+ = .

Since2 2

2 2lim( 1) lim ( 1) 5

x xx x

+

+ = + = , the limit of2

1x + as 2x exists and is given by

the value 5. We write2

2lim( 1) 5x

x

+ = .

Definition: If lim f ( ) lim f ( ) x a x a

x x k +

= = , where k is real, then lim f ( )x a

x k

= .

In the above example, we see that2 2

2lim( 1) 5 2 1x

x

+ = = + .

Question: Is it always true that if lim f ( )x a x exists, then limf ( ) f ( )x a x a = ?

Answer: No. Consider the following function1 if 0

f( )2 if 0

xx

x

=

=.

The graph of f is as shown on the right.

Then0

lim f ( )x

x

exists

since0 0

lim f ( ) lim f ( ) 1x x

x x +

= = .

Hence,0

limf ( ) 1x

x

= .

But f(0) 2= .

Hence, 0limf ( ) f (0)x x .

Note: In fact, if the graph of f is continuous at x a= , then lim f ( ) f ( )x a

x a

= .

Example: The graph of2 1y x= + is continuous at 2x = , so

2 2

2lim( 1) 2 1 5x

x

+ = + = .

(Please refer to Appendix A on page 14 for more information and results on Limits)

y

0 x

1

2

Key in Y1 = X2 + 1 andGo To TBLSET and put

TblStart = 1.993 and

=


3/15



Page 3 of 15

1.2 Derivative as a Limit

Consider a curve with equation f ( )y x= .

Let ( , f ( ))A x x and ( , f ( ))B x x x x + +

be two points on the curve, where x isa small increment in x .

Notice that the gradient of the curve atAcan be approximated by the gradient of

the chordAB. The closerB is toA, thebetter is the approximation.

Hence,

Gradient of the curve atA = gradient of tangent to the curve atA

limB A

= (gradient of chordAB) (See Appendix B)

limB A

BC

AC=

0

f ( ) f ( )limx

x x x

x

+

= (since 0 B A as x )

0limx

y

x

=

d

d

y

x= (by definition)

Definition of derivative: If f ( )y x= , thend

d

y

x=

0

f ( ) f ( )limx

x x x

x

+ .

Example 1

By considering the derivative as a limit, find ( )d

2 ( 4)d

x xx

.

Solution 1

Let f ( ) 2 ( 4) y x x x= = .

( )( )d

2 4d

x xx

=d

d

y

x=

0

f ( ) f ( )limx

x x x

x

+ (by definition)

( )0

2( ) 4 2 ( 4)limx

x x x x x x

x

+ + =

( )22 2

0

2( 4 4 ) 2 8limx

x x x x x x x x x x

x

+ + + +=

( )

22 2

02 4 8 2 8 2 8lim

xx x x x x x x x

x

+ + +=

0

(4 2 8)limx

x x x

x

+ =

0(4 8 2 )lim

xx x

+= (since 0x )

4 8x= .

0

C

y

x

( ), f ( )A x x

( )( ), fB x x x x + +

x

yy = f(x)


4/15



Page 4 of 15

GRAPHING CALCULATOR TASK 1

Using a graphic calculator to find the gradient of f ( ) 2 ( 4) x x x= at the point on the graph

where 5.2x = .

Solution:Note: We know that the answer is 4(5.2) 8 12.8 = . (from Example 1)

Method 1:

In the Y= screen, enter the rule for the function as Y1 = 2X2 - 8X, and choose 6:Zstandard

from the ZOOM menu.Choose 6:dy/dx from the CALC menu. Type 5.2 and press ENTER.

Note: 1. Thepointonthecurveatwhichwewishtofindthegradientatmustbeinthewindow.

2.The graphic calculator may not always give the correct gradient especially at sharppoints.

Method 2:

Alternatively, on the Home Screen, choose 8:nDeriv from MATH menu and press ENTER.

Key in 2X28X, X, 5.2) and press ENTER.

Note: The advantage of this method is that we do not need to know the shape of the graph.


5/15



Page 5 of 15

2 BASIC RULES OF DIFFERENTIATION

Let fand g be two functions ofx. The following are some basic rules of differentiation:

(1) ( ) ( ) ( )d d d

f ( ) g( ) f( ) g( )d d d

x x x x x x x

= .

(2) ( ) ( ) ( )

d d d

f ( )g( ) g( ) f( ) f( ) g( )d d d x x x x x x x x x

= +

. (Product Rule)

(3)

( ) ( )

( )2

d dg( ) f( ) f( ) g( )

d f ( ) d d

d g( ) g( )

x x x xx x x

x x x

=

. (Quotient Rule)

(4) Ifyisafunctionofuanduisafunctionofx,thend d d

d d d

y y u

x u x= . (Chain Rule)

Exercise: Use the chain rule to show thatd 1

dd

d

y

xx

y

= .

Answer: The result follows sinced d d

( ) 1d d d

y xy

y x y= = (by chain rule).

Note :d d

( )d d

yy

x x= and hence

d

d

y

xis not a fraction.

3 DERIVATIVES OF STANDARD FUNCTIONS

3.1 Power Functions

y d

d

y

x

y d

d

y

x

k 0 kx k

nkx

1nnkx [f ( )]

nk x

1[f ( )] f '( )

nnk x x

Example 2 Differentiate (a)5 3( 1)x + ,(b) (2 3) 5x x and (c)

2

1

3

x

x

+

with respect tox.

Solution 2

( )( )

( )

( )

35

25 4

24 5

d1

d

3 1 (5 )

15 1 .

xx

x x

x x

+

= +

= +

(a)

( )( )

( )

d2 3 5

d

12 3 2 52 5

2 3 4( 5)

2 5

6 23.

2 5

x xx

x xx

x x

x

x

x

= +

+ =

=

(b)

2

2

2 2

2 2

2 2

2

2 2

d 1

d 3

( 3)(1) ( 1)(2 )

( 3)

3 2 2

( 3)

2 3.

( 3)

x

x x

x x x

x

x x x

x

x x

x

+

+=

=

=

(c)


6/15



Page 6 of 15

3.2 Exponential and Logarithmic Functions

y d

d

y

x y

d

d

y

x

ex ex ln x ,x > 01

x

f ( )e

x

f ( )f '( )e

xx

[ ]ln f ( )x ,

f(x) > 0

f '( )

f ( )

x

x

Example 3 Differentiate the following with respect tox.

(a) ( )23 2

e 1x x , (b) ( )ln ln x , 1>x , (c)1 + e

2 + e

x

x

,

(d)2

ln 1x + , (e)2

5log ( 1)x + (f)32x+

Solution 3

( )( )

( )

( )( )

2

2 2

2

2

3 2

3 2 3

3 2

3 2

de 1

d

e (2 ) 1 e (6 )

2 e 1 3 1

2 e (3 2).

x

x x

x

x

xx

x x x

x x

x x

= +

= +

=

(a)

( )( )d

ln lnd

1 1

ln

1

ln

xx

x x

x x

=

=

(b)

2

2

2

d 1+e

d 2+e

(2+e )( e ) (1 e )(e )

(2+e )

2e 1 e 1

(2+e )

2e e 2.

(2+e )

x

x

x x x x

x

x x

x

x x

x

x

+=

=

=

(c)

( )

( )

( )

2

12 2

2

2 2

dln 1

d

d ln 1d

d 1ln 1

d 2

2.

2( 1) 1

xx

xx

xx

x x

x x

+

= +

= +

= =+ +

(d)

( )252

2

dlog ( 1)

d

d ln( 1)d ln 5

2.

(ln 5)( 1)

xx

xx

x

x

+

+=

=+

(e)

( )

( )

( )

3

3

ln 2

( +3)ln 2

( +3)ln 2

3

d2

d

d ed

de

d

e (ln 2)

(ln 2)2 .

x

x

x

x

x

x

x

x

+

+

+

=

=

=

=

(f)


7/15



Page 7 of 15

3.3 Trigonometric Functions

y d

d

y

x y

d

d

y

x

sinx cosx cosecx cosec cotx x

cosx sin x secx sec tanx x tan x 2sec x cot x 2cosec x

sin f ( )x f '( ) cos f ( )x x cosec f ( )x f '( ) cos ec f ( ) cot f ( ) x x x

cos f ( )x f '( )sin f ( )x x sec f ( )x f '( )sec f ( ) tan f ( ) x x x

tan f ( )x 2f '( )sec f ( )x x cot f ( )x 2

f '( ) cos ec f ( )x x

Note : 1. The above results hold forx measured in radians only.

Ifx is measured in degrees, then covertxo

to180

xradians.

2. The derivatives of secx can be found in the List of Formulae MF15.

Exercise: Use the quotient rule to show that ( ) 2d

tan secd

x xx

= .

Answer: ( ) 2

2 22

2 2

d d sin (cos )(cos ) (sin )( sin )tan

d d cos cos

cos sin 1sec . (shown)

cos cos

x x x x xx

x x x x

x xx

x x

= =

+= = =

Example 4 Differentiate the following with respect tox.

(a) tan2x , (b) 3cos sin2

xx

, (c)22+sec

ex, (d) sin x .

Solution 4

( )

2

2 2

dtan2

d

d 2tan

d 180

2 2sec

180 180

sec or sec 2 .90 90 90

xx

x

x

x

xx

=

=

=

(a)

3

3 2

2

d(cos sin )

d 2

1(cos )( cos ) (sin )(3cos )( sin )

2 2 2

1cos (cos cos 6sin sin ).

2 2 2

xx

x

x x x x x

x x x x x

= +

=

(b)

( )2

2

2

(2 sec )

(2 sec )

(2 sec ) 2

de

d

= e (2sec )(sec tan )

= 2e sec tan .

x

x

x

x

x x x

x x

+

+

+

(c)

( )

( )

dsin

d

1cos

2

1cos .

2

xx

xx

xx

=

=

(d)


8/15



Page 8 of 15

3.4 Inverse Trigonometric Functions

y d

d

y

x y

d

d

y

x

1sin x

2

1, 1

1

x

x

.

Solution 7

2 3

2

2

2

2 3

d d2 6 (1) 0

d d

d(6 ) 2

d

d 2.

d 6

x y xy

y y x y x y

x x

y y x y x

x

y y x

x y x

+ = +

+ = + +

=

=

(a)

2cos 3

d d( sin ) cos 6

d d

dcos (6 sin )

d

d cos.

d 6 sin

x y y

y y x y y y

x x

y y y x y

x

y y

x y x y

=

+ =

= +

=+

(b)

( ) 2ln 2cos

1 d d(1 ) 4cos ( sin )d d

d 1 d 14 cos sin

d d

d d4( )cos sin 1

d d

d 1.

d 1 4( )cos sin

x y y

y yy y x y x x

y yy y

x x y x x y

y y x y y y

x x

y

x x y y y

+ =

+ = +

+ = + +

+ + =

= + +

(c)

sin

sin

sin sin

sin

sin

e 2

d de (1 cos )d d

de (1 e cos )

d

d e.

d 1 e cos

x y

x y

x y x y

x y

x y

y

y yyx x

yy

x

y

x y

+

+

+ +

+

+

= +

+ =

=

=

(d)

, 0

ln ln

1 d 1( ) ln

d

d(1 ln ) (1 ln ).

d

x

x

y x x

y x x

yx x

y x x

y y x x x

x

= >

=

= +

= + = +

(e)

ln

ln

Alternatively,

, 0

e

d 1e [ ( ) ln ]

d

d(1 ln ).

d

x

x x

x x

x

y x x

y

yx x

x x

yx x

x

= >

=

= +

= +

(e)


11/15



Page 11 of 15

6 PARAMETRIC DIFFERENTIATION

6.1 Parametric Equations

The coordinates of a point ( , )x y on a curve can usually be expressed in terms of a

third independent variable t, which is called a parameter.

In other words, we can often describe a curve by the equations: f ( )x t= , g( )y t= ,

where t . Such equations are called parametric equations.

Consider the curve described by the parametric equations: 2x t= and2

y t= , t .

Weseethatthecoordinatesof thepointsonthiscurvecanbeexpressedas2

(2 , )t t ,

t and different values of t correspond to different points on the curve.

For example, 1t= corresponds to the point (2,1) on the curve.

If it is possible to eliminate t, then we can obtain the cartesian equation of the curve,which is an equation involving x and y only.

In this case, the cartesian equation is

2

21=

2 4

xy x

=

.

6.2 Differentiation of Parametric Equations

If f ( )x t= and g( )y t= , where t , then by chain rule,

d

d

y

x=

d

d

y

t

d

d

t

x=

d d

d d

y x

t t.

Example 8 Findd

d

y

xfor each of the following parametric equations.

(a)

21

,1 1

tx y

t t= =

+ +, (b) sin 2 , cos 4x y = = , (c)

1ln , sin x t y t

= = .

Solution 8

(a)2

2 2 2

2 2

1 d 1

1 d (1 )

d (1 )(2 ) (1) 2=

1 d (1 ) (1 )

.

.

xx

t t t

t y t t t t t y

t t t t

= = + +

+ += =

+ + +

22

2

d d d

d d d

2= [ (1 ) = (2 ).

(1 )]

y y t

x t x

t tt t t

t

=

+ + +

+

(b)d

sin 2 2 cos 2 .d

dcos 4 4 sin 4 .

d

xx

yy

= =

= =

d d d 4 sin 4

d d d 2 cos 2

2(2 sin 2 cos 2 )4sin 2 .cos 2

y y

x x

= =

= =

(c)

1

2

d 1ln

d

d 1sin

d

.

.1

xx t

t t

yy

tt

t

= =

= =

2

d d d=

d d d.

1

y y t

x t x

t

t=


12/15



Page 12 of 15

GRAPHING CALCULATOR TASK 2

In Example 8(a), use a graphic calculator to find the gradient of the curve at the point where

3t= .

Change the mode to PAR(i.e. the parametric mode). In the Y= screen, enter the functions.

Set the Window to be:

and press Graph.

Press 2nd [calc] and select 2:dy/dx.

Type 3 and press Enter.

Example 9

A curve has parametric equations1

2x tt

= ,4

y tt

= + , 0.t

Show that2

d 1 91

d 2 2 1

y

x t=

+

.

Deduce thatd 1

d 2

y

x< for all real values of texcept 0.


13/15


14/15



Page 14 of 15

The calculus was the first achievement of modern mathematics and it is difficult tooverestimate its importance. I think it defines more unequivocally than anything else theinception of modern mathematics; and the system of mathematical analysis, which is itslogical development, still constitutes the greatest technical advance in exact thinking.

John von Neumann (1903 1957) - one of the great mathematicians of the century

CHECKLIST

At the end of this chapter, you should know:

the derivatived

d

y

xcan be interpreted as the gradient of a curve;

the idea of a limit and find the derivatives of2

x ,x

1, xsin and ex ;

how to find and use the first derivative of a simple function defined implicitly orparametrically.

At the end of this chapter, you should be able to:

use a graphing calculator to evaluate the derivative at given points.

use a graphing calculator to verify or determine limits such asx

x

x

sinlim

0and

0

e 1lim

x

x x

(to be done in Tutorial).

find and use the first derivative of a simple function defined implicitly or parametrically.

Appendix A Limits (For Your Information)

Useful Results

1. limx a

c c

= , where c is a constant.

2. lim n nx a

x a

= for n +Z .

3. ( )lim f f ( )x a

x a

= if f ( )x is a polynomial.

Theorems on Limits

1. ( )( ) ( )lim f limf x a x a

c x c x

= , where c is a constant.

2. ( ) ( )( ) ( ) ( )lim f g lim f limg x a x a x a

x x x x

= .

3. ( ) ( )( ) ( ) ( )lim f g limf lim g x a x a x a

x x x x

= .

4.( ) ( )limff

limg( ) lim g( )

x a

x a

x a

xx

x x

= , provided ( )limg 0x a

x

.

Useful Results for Limits Involving Infinity

1.1

lim 0nx x

= , .n+

R 2.1

lim 0nx x

= , .n+

R


15/15



Page 15 of 15

APPENDIX B GRAPHING CALCULATOR TASK 0 (In Teachers' Copy only)

To illustrate the limiting process of a secant line approaching the tangent to a curve.

1. Activate the Transfrm Apps

2. Key in Y1 = X2.

3. The equation of the secant line through 2 points (A, Y1(A)) and (B, Y1(B)) isY2 = ((Y1(A) Y1(B))/(A-B))(X A) + Y1(A).

4. Key in Y3 = { }1 2Y , Y , this is the active equation for the Transfrm Apps.

5. Set the WINDOW and SETTINGS

6. Press the Graph key to see the animationQuestion: What is the line that the family of secants is approaching?

Documents

Chapter 8A Full Version - Differentiation Techniques