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UNIVERSITI KUALA LUMPUR DIFFERENTIATION – F2-A

1. INTRODUCTION

y y t fig : a fig : b

X

The line of sight of a person sitting in a car traveling along a roller coaster track P will be parallel to the tangent line T

Consider the graph of a function f ( fig : a ) that gives the distance traveled by a car on a straight road t seconds after starting from rest . At the initial stage , the curve rises slowly . Reflecting the slow speed of the car ( rate of change of the position y with respect to time t ). As time passes , the curve rises more rapidly. Showing that the speed of the car in the latter phase is greater than that in the earlier phase .The speed of the car is therefore not constant over the period of time . What is the speed of the car at any given instant of time t ? This is called the instantaneous speed of the car which is the speed of the car at that particular instant of time . How to measure the changes in f for any value of x ? Consider the graph of f ( fig : b ) . Think of the curve as a stretch of roller coaster track . When the car is at the point P on the curve , a passenger sitting in the car , looking straight ahead , will have a line of sight that is parallel to the line T (tangent to the curve at P ) Therefore , the rate at which y is increasing or decreasing with respect to x can be measured by considering the slope of the tangent line to the curve at the point in question .

F2-A - 1 - MATHEMATICS UNIT


1.1 TANGENT LINE Let QX be any point on the graph of f distinct from P (a point on the graph). The straight line that passes through the two points , P and QX is called a secant line If QX move toward P along the curve , the secant line through P and QX rotates about the fixed point P. And the point QX approaches a fixed line through P . This fixed line is the tangent line .

S3 S4

S2

S1

T

Q1

Q2

Q4Q3

P

y

x Consider the point P(2,1) Points Q1 , Q2 , …. , Q6 lie on the graph of f(x) = ¼ x2 . The lines S1 , S2 , …. , S6 are the corresponding secant line through P . By using the formula to find the slope of a straight line that passes through 2 points

12

12

xxyym

−−

=

find the slopes of the secant lines .

point Q1 Q2 Q3 Q4 Q5 Q6 coordinates

(5, 6.25)

(4 , 4)

(3 , 2.25)

(2.5 ,

1.5625)

(2.1 ,

1.1025)

(2.01 ,

1.010025) slope m1 m2 m3 m4 m5 m6

1.75

1.5

1.25

1.125

1.025

1.0025

From the results , we see that the slopes of the secant lines seem to approach the number 1 CONCLUSION : the slope of the tangent line is 1



To find a general expression involving the function f(x) that describes the slopes of the approximating sequence of secant lines , we consider :

f(x)

f(x + h)

x + h x

P( x , f(x) )

Q( x+h , f(x+h) ) y = f(x)

S

y

point coordinate

P

( x , f(x) )

Q

( x+h , f(x+h)

) x

Using the equation for the slope of a line :

h

h

h

h

T

Q

P

y

12

12

xxyym

−−

=

x

Observe that as the number h approaches zero , the point Q ( x+h , f(x+h)) approaches the point P along the graph of f . So the sequence lines through points P & Q approaches the tangent line through P . DEFINITION : The slope of the tangent line to the graph of f at the point P is

given by : ( ) ( )

hxfhxf

h

−+→0

lim if it exists



EXAMPLE 1 Find the slope of the tangent line to the graph of the linear function f(x) = 3x + 5

SOLUTION

53)( += xxf

( )

hxhx

hxfhxf

hh

5353lim)()(lim00

−−++=

−+→→

3

3lim0

=

=→h

therefore the slope of the linear function is 3

EXAMPLE 2 Find the slope of the tangent line to the graph of y = f(x) = ¼ x2 at any point ( x , y ) . What is the slope of the tangent line T to the graph of f at the point (2 , 1), what is the equation of T ? SOLUTION

2

41)( xxf =

( )h

xhx

hxfhxf

h

22

00

41

41

lim)()( −+=

−+→→h

lim

The

The

Slope of the tangent line to the graph of f at any point

2

2

42lim

41

41

42

41

lim

0

22

0

x

hxh

xhxhx

h

h

=

+=

−++=

→

→

slope of the tangent line T to the graph of f at the point ( 2 , 1 ) is 122

2==

x

equation of T , at point ( 2 , 1 ) with slope m = 1

1

)2(111)2(

−=−=−−=−

xyxyyxm



1.2 RATE OF CHANGE Suppose that we are given a function f that describes the relationship between the two quantities x and y :

y = f (x) the number f(x+h) – f(x) measures the change in y that corresponds to a change of h in x .

f(x + h) - f(x)

hf(x)

f(x + h)

x + h x

P( x , f(x) )

Q( x+h , f(x+h) ) y = f(x)

S

y

x

The difference quotient : ( ) ( )

hxfhxf

xhxxfhxf )()( −+=

−+−+

measures the average

rate of change of f with respect to x over the interval [ x , x+h ] As this equation is the same as the equation of the slope , we conclude that the difference quotient also measure the slope of the secant line that passes through the two points , P and Q on the graph f . By taking the limit of the difference quotient as h approaches zero ,

( ) ( )

hxfhxf

h

−+→0

lim

we obtain the rate of change of f at the point x often called the instantaneous rate of change of f at the point x .



1.3 SUMMARY AVERAGE RATE OF CHANGE of f over the interval [ x , x + h ] or slope of the secant line to the graph of f through the point ( x , f(x) ) and ( x+h , f(x+h)) is

( ) ( )

hxfhxf −+

INSTANTANEOUS RATE OF CHANGE of f at the point x or slope of tangent line to the graph of f at ( x , f(x) ) is

( ) ( )

hxfhxf

h

−+→0

lim

2. DERIVATIVE 2.1 THE DERIVATIVE OF A FUNCTION The derivative of a function f with respect to x is the function f ‘ defined by :

( ) ( ) ( )h

xfhxfxfh

−+=

→0lim'

the domain of f ‘ is the set of all x when the limit exists Thus the derivative of a function f is a function f ‘ that gives the slope of the tangent line to the graph of f at any point ( x , f(x) ) and thus the rate of change of f at x . 2.2 NOTATION :

DX f(x) : “ dee sub x of f of x “

dxdy : “ dee y dee x “

y ‘ : “ y prime “ f ‘ : “ f prime “ f ‘ (x) : “ f prime of x “



EXAMPLE 1 : Let f (x) = x2 a: compute the derivative f ‘ of f

h

xhxh

xfhxfhh

22

00

)(lim)()(lim −+=

−+→→

h

xhxhxxfh

222

0

2lim)(' −++=

→

x

hxh

2

2lim0

=

+=→

b: compute f ‘(2) and interpret your result

f

4)2(2)2('

2)('==

= xxfThe slope of the tangent line to the graph of f at the point ( 2 , 4 ) is 4. The function f is changing at the rate of 4 units per unit change in x

EXAMPLE 2 : Let f(x) = x2 - 4x

(a) compute the derivative f ‘ of f (b) find the point on the graph of f where the tangent line to the curve is

horizontal (c) sketch the graph of f and the tangent line to the curve at the point

found in (b) (d) what is the rate of change of f at this point ?

SOLUTION (a) xxxf 4)( 2 −=

h

xxhxhxh

xfhxfhh

)4()(4)(lim)()(lim22

00

−−+−+=

−+→→

hhhxhxf

h

42lim)('2

0

−+=

→

42

42lim0

−=

−+=→

x

hxh



(b) If the tangent line to the curve is horizontal , then the slope is zero.

)2(4)2( 2 −=y

2

0420)('

==−=

xxxf

the point is ( 2 , - 4 ) (c)

-6

-4

-2

0

2

4

6

8

10

12

14

-3 -2 -1 0 1 2

(d) the rate of change of f at EXAMPLE 3 : Let f(x) = 1/x

(a) compute the derivative f ‘ o(b) find the slope of the tangen(c) find an equation of the tang

SOLUTION

(a) x

xf 1)( =

h

xfhxfhhlim)()(lim

00=

−+→→

xfhlim)('

0

=

→

2

0

1

lim

x

h

−=

−=→

F2-AMATHEMA

4−=y

3 4 5 6 7

this

f f t linen

x1+

(xx −

(x +

- 8TIC

Tangent line

point ( 2 , - 4 ) is zero

e T to the graph of f when x = 1 t line T in (b)

hxh1.1

−

( )) hxhhx 1.

++

)1xh

- S UNIT


(b) when x = 1 ,

the slope , 2

1)('x

xf −=

1)1(' −=f (c) when x = 1 , y = 1 at point ( 1 , 1 ) , m = -1

2

111

+−=

−−

=−

xy

xy

2.3 FUNCTIONS THAT ARE NOT DIFFERENTIABLE AT A POINT Functions that fail to be differentiable are functions that do not possess derivative at certain values of x . A continuous function f(x) , fails to be differentiable at a point x = a when :

The graph of f(x) makes an abrupt change of direction at that point The tangent line is vertical ( since the slope of a vertical line is undefined )

f is not differentiable at x = a

( a , f(a) )

( a , f(a) )

x a

x a

y y

The graph makes an abrupt

change of direction at x = a The slope at x = a is undefined



2.4 CONTINUITY AND DIFFERENTIABILITY In general , the continuity of a function at a point x = a does not necessarily imply the differentiability of the function at that point .The converse however is true . EXAMPLE 4 : Explain why the function fails to be differentiable at each of the points x = a , x = b , x = c , x = d , x = e , x = f , x = g

y At x = a , b , c : The function is discontinuous at these points At x = d , e , f : The function has a kink at each of these points At x = g : The tangent line is vertical at this point

x

g
a b c d e f


3. RULES IN DIFFERENTIATION

3.1 BASIC RULES These rules are used to simplify the process of finding the derivative of a function

(a)

DERIVATIVE OF CONSTANT

0)( =cdxd

Where c = constant

(b)

POWER RULE

1)( −= nn nxx

dxd

Where n = any real number

(c)

DERIVATIVE OF A

CONSTANT MULTIPLE OF A FUNCTION

[ ] [ ])()( xfdxdcxcf

dxd

=

Where c = constant

(d)

THE SUM /

DIFFERENCE RULE

[ ] [ ] [ ])()()()( xgdxdxf

dxdxgxf

dxd

±=±



EXAMPLES Find the derivative of

Functions

Derivative

(a)

28)( =xf

0)(' =xf

(b)

2)( −=xf

0)(' =xf

(c)

xxf =)(

1)(' 011 === − xxxf

(d)

8)( xxf =

718 88)(' xxxf == −

(e) 2

5

)( xxf = 231

25

25

25)(' xxxf ==

−

(f) 2

1

)( xxxf == 211

21

21

21)('

−−== xxxf

(g) 3

1

3

1)(−

== xx

xf 341

31

31

31)('

−−−−=−= xxxf

(h)

35)( xxf =

( ) 213 1535)(' xxxf == −

(i) 2

1

33)(−

== xx

xf 231

21

23

213)('

−−−−=

−= xxxf

(j)

3834)( 245 ++−+= xxxxxf

1161220)(' 34 +−+= xxxxf

(k)

( ) 323

2

5515

5−+=+= tt

tttg

( ) 41552' −−= tttg

(l)

( ) 1223

3434 −+−=+−

= rrrrrrrf

( ) 2342' −−−= rrrf



3.2 PRODUCT RULE The derivative of the product of two differentiable functions is given by :

( ) ( )[ ] ( )xgdxdxfxgxf

dxdxgxf

dxd )()().(. +=

EXAMPLES

(a) Find the derivative of the function ( ) ( )( )312 32 +−= xxxf ( ) ( ) ( ) 223 31234' xxxxxf −++=

[ ][ ]12310

361243

33

+−=

−++=

xxxxxxx

(b) Find the derivative of the function ( ) ( )13 += xxxf

( ) ( )

++=

−21

32

2113' xxxxxf

22

5

25

225

327

233

xx

xxx

+=

++=



3.3 QUOTIENT RULE The derivative of the quotient of two differentiable functions is given by

( )( ) [ ] 2)(

)()()().(

xg

xgdxdxfxgxf

dxd

xgxf

dxd −

=

EXAMPLE

(a) Find the derivative f ‘ of ( )11

2

2

−+

=xxxf

( ) 22

22

)1()2)(1()1)(2('

−+−−

=x

xxxxxf

22

22

22

)1(4

)1()11(2

−−

=

−−−−

=

xxx

xxx

(b) Find the derivative f ‘ of ( )12 +

=xxxh

( )( )

22

21

221

)1(

2)1(21

'+

−+=

−

x

xxxxxf

22

2

)1(231

+−xxx

=



3.4 CHAIN RULE If the function f is differentiable and h (x) = [ f (x) ] n , where n = real number , then

( ) ( )[ ] ( )[ ] ( )xfxfnxfdxdxh nn '.' 1−==

EXAMPLES

(a) Find the derivative h’ of the function : ( ) ( )22 1++= xxxh . ( ) ( )( )1212' 2 +++= xxxxh

(b) Find the derivative H’ of the function : ( ) ( )1002 1++= xxxH

( ) ( ) ( )121100' 992 +++= xxxxH

(c) Differentiate the function : ( ) ( )21

22 11 +=+= xxxG

( ) ( ) ( )xxxG 2121' 2

12 −+=

12 +x

x=

(d) Differentiate the function : ( ) 23 382 xxxf −=

( ) ( ) ( ) ( )xxxxxxf 638212386' 2

1232

122 −−+−=

−

( ) ( )( ) ( )[ ]( )

2

22

2221

22

21

2421

22

38

48638386

386386

x

xxxxxx

xxxx

−

−=

−−−

−−−=

−

−

=



4. THE DERIVATIVE OF CERTAIN FUNCTIONS

4.1 TRIGONOMETRIC FUNCTIONS - THE DERIVATIVE OF THE STANDARD FORM AND THE GENERAL FORM

STANDARD FORM

( ) xxdxd cossin =

( ) xxdxd sincos −=

( ) xxdxd 2sectan =

( ) xxxdxd cotcsccsc −=

( ) xxxdxd tansecsec =

( ) xxdxd 2csccot −=

GENERAL FORM

( ) )(')(cos)(sin xfxfxfdxd

=

( ) )(')(sin)(cos xfxfxfdxd

−=

( ) )(').(sec)(tan 2 xfxfxfdxd

=

( ) )(').(cot)(csc)(csc xfxfxfxfdxd

−=

( ) )(').(tan)(sec)(sec xfxfxfxfdxd

=

( ) )(').(csc)(cot 2 xfxfxfdxd

−=



EXAMPLES Differentiate each of the following functions

Functions

Derivatives

(a)

( ) ( )12sin += xxg

( ) ( )12cos2' += xxg general form

(b)

( ) ( )12cos 2 −= xxf

( ) ( ) ( )12sin4' 2 −−= xxxf general form

(c)

( )35sec)( 2 −= xxf

( ) ( )35tan35sec10)(' 22 −−= xxxxf general

form

(d) ( ) ( )25cot5 xxg =

( ) ( )22 5csc50' xxxg −= general form

(e)

= 2

52csc

32)( xxf

−= 22

52cot

52csc

158)(' xxxxf general form

(f)

( ) xxg 2cos=

( )21

2cos x=

( ) ( ) ( )22sin2cos21' 2

1xxxg −= −

( ) 21

2cos2sin −−= xx

(g)

( ) xxg 2cos=

( )21

2cos x=

( ) ( ) ( )

−= − 22

212sin' 2

121

xxxg

( ) ( )21

21

2sin2 xx −−=

(h)

( ) ( )102sinxxxh +=

( ) ( ) [ ]292 cos21sin10' xxxxxh ++= chain rule

(i)

( ) xxxf sin2=

( ) xxxxxf cossin2' 2+= product rule

( )xxxx cossin2 +=



4.2 THE DERIVATIVE OF THE EXPONENTIAL FUNCTION The derivative of the exponential function with base e is equal to the function itself

STANDARD FORM GENERAL FORM

xx eedxd

= )('.)()( xfeedxd xfxf =

EXAMPLES : Compute the derivative of each of the following :

Functions

Derivatives

(a)

( ) 12 += xexf

( ) ( ) 11 22

22' ++ == xx xexexf general form

(b)

( ) ( ) 23

2+= tetg

( ) ( ) ( tt eetg 123

223' −

+= ) chain rule

( )21

223

+= tt ee

(c)

( ) xexxf 2=

( ) xx exxexf 22' += product rule

( )xxe x += 2

(d)

xxey 2−=

( )xx exey 22 21' −− −+= product rule

( )xe x 212 −= −



4.3 THE DERIVATIVE OF THE LOGARITHMIC FUNCTION The derivative of natural logarithm is ,

STANDARD FORM GENERAL FORM

xx

dxd 1ln = ( )

)()('ln

xfxfxf

dxd

=

EXAMPLES : Compute the derivative of each of the following functions

Functions Derivatives

(a)

( ) ( )1ln 2 += xxf

( ) ( )12' 2 +

=xxxf

(b)

( ) ( )22ln tettg −= ( ) ( )2

22

2

322't

tt

etettetg

−

−− −=

(c)

( ) xxxf ln= ( ) 1ln1.ln1' +=+= x

xxxxf

(d)

( )( )[ ]632 21ln ++= xxy

( ) ( ) ( ) ( )

( )( )( ) [ ]

( )( )[ ]( )1

29102'

21)1(9222'

21326122'

2

3

632

2353

632

253263

+++

=

++

++++=

++

++++=

xxxxy

xxxxxxxy

xxxxxxxy

(e)

( )xxxf ln

=

( ) 22

ln11.ln.1

'xx

x

xxxxf −

=−

=



4.4 LOGARITHMIC DIFFERENTIATION The task of finding the derivative of a given function can be made easier by first applying the laws of logarithms to simplify the function. This process is called logarithmic differentiation. 4.4.1 LOGARITHMIC LAW

• NMMN lnlnln +=

• NMNM lnlnln −=

• kMMk lnln = 4.4.2 EXAMPLES

Differentiate ( )( )32 11 +−= xxxy

( )( )( )( )[ ]

( ) ( )( ) ( )

( ) ( )

( )

+

+−

+=

++

−+=

++−+=

++−+=

+−=

+−=

16

111

1231

1111

1ln31lnlnln1ln1lnlnln

11lnln

11

2

2

2

32

32

32

xx

xxy

dxdy

dxdx

xx

dxdx

xdxdx

xdxdy

y

xxxyxxxy

xxxy

xxxy



5. HIGHER ORDER DERIVATIVES 5.1 HIGHER-ORDER DERIVATIVES The derivative f ‘ of a function f is also a function . Thus , the function f ‘ has a derivative f “ at a point x in the domain of f ‘ The function f “ obtained is called the second derivative of the function f

FIRST

DERIVATIVE

SECOND

DERIVATIVE

THIRD

DERIVATIVE

………………

nth

DERIVATIVE

f ‘ (x)

f “ (x)

f ‘’’ (x)

………………

f (n)(x)

D1 f(x)

D2 f(x)

D3 f(x)

………………

Dn f(x)

y ‘

y “

y ‘’’

………………

y n

dxdy

2

2

dxyd 3

3

dxyd

………………n

n

dxyd

D1y

D2y

D3y

………………

Dn y

REMARK : y = f = f(x) 5.2 EXAMPLES

(a) Find the third derivative of the function 32xy =

the first derivative : 311

32

32

32'

−−== xxy

the second derivative : 341

31

92

31

32''

−−−−=

−= xxy

the third derivative : 371

34

278

34

92'''

−−−=

−−= xxy



(b) Find the derivatives of all orders of the polynomial function

( ) 8243 2345 −+−+−= xxxxxxf the function : ( ) 8243 2345 −+−+−= xxxxxxf the first derivative : ( ) 1412125' 234 +−+−= xxxxxf the second derivative : ( ) 4243620'' 23 −+−= xxxxf the third derivative : ( ) 247260 23 +−= xxxf the fourth derivative : ( ) 721204 −= xxf the fifth derivative : ( ) 1205 =xf the sixth derivative : ( ) 06 =xf (c) Find the second derivative of the function ( ) 2

32 32 += xy

: ( ) ( )xxy 43223' 1

23

2 −+=

( )21

2 326 += xx

the first derivative ( applying the chain rule )

: ( ) ( ) ( )

+++=

− xxxxy 432216326'' 2

122

12

11

−

the second derivative ( applying the product rule )

( ) ( ) 22222 321232 +++ xxx6=



6. IMPLICIT DIFFERENTIATION 6.1 DIFFERENTIATING IMPLICITLY Some functions are expressed in the form y = f(x) , where the dependant variable y is expressed explicitly in terms of the independent variable x . But there are functions that express y implicitly as a function of x . When restrictions are placed on x and y , to compute dy/dx , we use a method called implicit differentiation . 6

FINDING dy/dx BY IMPLICIT DIFFERENTIATION Suppose that we are given an equation in x and y and wish to compute dy/dx Differentiate both sides of the equation with respect to x . Make sure that the derivative of any item involving y includes the factor dy/dx

.2 EXAMPLES

(a) Find dxdy

and dydx

given the equation xy =2

ydxdy

dxdx

dxdyy

xy

21

12

2

=

=

=

ydydx

dydx

dydy

y

xy

2

12

2

=

=

=



(b) Find dxdy

given the equation 82 334 =−+−− xxyyy

( )

13461

61134

016134

82

23

2

223

223

334

−−−

=

−=−−

=−+−−

=−+−−

yyx

dxdy

xyydxdy

dxdx

dxdxx

dxdy

dxdyy

dxdyy

xxyyy

(c) Consider the equation . Find 422 =+ yxdxdy

by implicit

differentiation

yx

dxdy

xdxdyy

dxdyy

dxdxx

−=

−=

=+

22

022

(d) Find dxdy

given that x and y are related by the equation

426 2232 ++=+ yyxyx

( )( )

14362

122143

0141232

426

22

3

322

223

2232

−−+−

=

−−=−−

++=+

+

++=+

yyxxxy

dxdy

xxyyyxdxdy

dxdy

dxdyy

dxdxx

dxdyyxy

dxdxx

yyxyx



7. RELATED RATES In a related rates problem the idea is to compute the rate of change of one quantity in terms of the rate of change of another quantity ( which may be more easily measured ) . The procedure is to find an equation that relates the two quantities and then use the Chain Rule to differentiate both sides with respect to time .

ate .

1. Read the problem carefully . 2. Draw a diagram if possible . 3. Introduce notation . Assign symbols to all quantities that are

functions of time . 4. Express the given information and the required rate in terms of

derivatives 5. Write an equation that relates the various quantities of the problem .

If necessary , use the geometry of the situation to eliminate one of the variables by substitution .

6. Use the Chain Rule to differentiate both sides of the equation with

respect to t . 7. Substitute the given information into the resulting equation and

solve for the unknown r

WARNING !!!! The substitution of the given numerical information should be done only AFTER the differentiation .



EXAMPLE 1 The radius of a circle is increasing at a rate of 4 cm/sec. How fast is the area of the circle increasing when the radius is 20 cm? SOLUTION :

Area of the circle : A

r

Radius of the circle : cmr 20=

Rate of change of the area of the circle : dtdA

Rate of change of the radius : sec/4cmdtdr

=

Relation giving the area of a circle : 2rA π=

Differentiate :

sec/65.502

)4)(20(2

2

2

2cmdtdAdtdA

dtdrr

dtdA

dtdrr

dtdA

=

=

=

=

π

π

π



EXAMPLE 2 The area of a rectangle is increasing at a rate of 5 cm2 / sec while the length is increasing at a rate of 10 m / sec . If the length is 20 m and the width is 16 m , how fast is the width changing ? SOLUTION : L

Area of a rectangle : A

Length : mL 20=Width : mW 16=

Rate of change of the area : 5dtdA

=

W

Rate of change of the length : 1dtdL

=

Relation giving the area of a rectangle : Differentiate :

sec/4

31

(2016)5(

201

1

mdtdWdtdW

dd

LW

dtdA

LdtdW

dtdLW

dtdA

dtdWL

dtdWL

dtdLW

dtdA

−=

−=

−=

−=

+=

F2-A - 2MATHEMATI

sec/2m

sec/0m

WLA ×=

sec/75.7

)10

m

tL

−=

7 - CS UNIT


8. EXERCISES SLOPE OF A TANGENT LINE , RATE OF CHANGE 1. Let f be the function defined by f(x) = x2 (a) Verify by direct substitution that the points Q1(3,9) , Q2(2,4) , Q3(1.5,2.25) ,

Q4(1.1,1.21) and Q5(1.01,1.0201) lie on the graph of f . (b) (i) Compute the slopes of the secant lines S1 , S2 , S3 , S4 and S5 through

P (1,1) and each of the points Q1 , Q2 , Q3 , Q4 and Q5 respectively . (ii) Use these values to guess the slope of the tangent line T to the graph of f at the point P (1,1)

(c) Let P( x,x2 ) be a point that lies on the graph of f and let Q be a neighboring

point on the graph with x-coordinate x+h

(i) Find the y-coordinate of the point Q by evaluating f(x+h) (ii) Show that the slope of the secant line through P and Q is 2x + h (iii) Find the slope of the tangent line T at P by evaluating ( )hx

h+

→2lim

0

(iv)Use this result to find the slope of the tangent line at the point P(1,1) How does this compare with the result obtained in b (ii) ?

2. Find the slope of the tangent line to the graph of the given function at any

point . (a) (b) f (13)( =xf xx 48) −=

(c) 2

21)( xxf −= (d) f xxx 52)( 2 +=

3. Find the slope of the tangent line to the graph of each function at the given

point and determine an equation of the tangent line (a) at ( 2 , 11 ) (b) at ( -2 , -10 ) 72)( += xxf 23)( xxxf −=

4. Find the instantaneous rate of change of y with respect to x at the given

value of x . (a) at x = 2 xxf 2)( =(b) at x = 1 43)( −= xxf

(c) xxxf 221)( 2 += at x = 2



DERIVATIVE OF A FUNCTION 5. Use the definition of the derivative to find the derivative of each of the

following functions (a) f(x) = - 4x (b) f(x) = 2 – 3x

(c) f(x) = 3x2 – 4x + 7

(d) f(x) = - x1

(e) f(x) = 3 / ( x – 5 )

6. Find the slope of the tangent line to the graph of each function at the given

point and determine an equation of the tangent line at that point . (a) f(x) = - 2x at ( 1 , -2 )

(b) f(x) = 3x2 – 4x + 7 at ( 2 , 11 )

(c) f(x) = - x1 at ( 3 , - 1/3 )

7. Let f(x) = 2x2 + 1 (a) Find the derivative f ‘ of f

(b) sketch the graph of f and the tangent line to the curve at the point (1,3)

(c) write an equation of this tangent line 8. Let f(x) = x2 – 2x + 1

(a) find the derivative f ‘ of f (b) find the point on the graph of f where the tangent line to the curve is

horizontal

(c) sketch the graph of f and the tangent line to the curve at the point found in (b)

(d) what is the rate of change of f at this point ?



BASIC RULES 9. Find the derivative of the function f by using the rules of differentiation (a) (f (b) g 3) −=x 5)( rr =

(c) f rr π2) =( (d) 3

34)( rr π=h

(e) (f (f) 127) −= xx 54

45)( mm =k

(g) u

u 2) =f ( (h) r 735)( 2 +−= sss

(i) (f (j) 62) 23 −+−= xxx 543)( 3 ++=ww

wd

(k) ( )63

31) ccc −−= −(f

(l) (f 201.005.0002.0) 23 −+−= xxxx

(m) 4 356)( xxx −=h (n) 312)(

23 −++=

xxxxg

PRODUCT RULES AND QUOTIENT RULES

10. Find the derivative of each of the given functions (a) ( )( )213)( 2 −+= xxxf

(b) ( )( )21)( 223 +−+−= wwwwwf

(c) ( )( ) 281151)( 225 +−−++= xxxxxf

(d) ( )( )1215)( 2 −+= xxxf

(e) ( )( )( )321)( 32 −++= xxxxf

(f) ( )

+++= 2

3 1212)(x

xxxf

(g) 1

)(2 +

=uuuf (h)

1)(

2 +=xxxf



(i) 12)( 2

3

+−

=xxxf (j)

12)(

2

2

+++

=xx

xxf

(k)

−=

xxxf 11)( 3

(l) ( )( )

211)(

2

−++

=xxxxf

CHAIN RULE 11. Find the derivative of the given function

(a) (b) ( )52 2)( += xxf ( )1023 1234)( −+−= xxxxf

(c) ( )42 1

2)(−

=x

xf (d) ( )( )3421)( ++= ttth

(e) 23)( −= xxf (f) ( )23

2 123)( +−= tttf

(g) 42)( 2 −= xxxh (h) 12

1)2 −

=x

x(f

(i) ( )2

344

1

xxy

+= (j) ( ) 323 425)( −

+−+= ttttf

(k) ( ) ( )44 1212)( ++−= tttf

(l) ( ) ( )8252 211)( uuug −+=

(m) (n) ( 522 13)( += xxxf ) ( )( )4

2

2312)(

+−

=tttg

(o) ( )

11)( 3

32

−−+

=uuuuf (p) ( )321)( −− −= tttf

(q) 112)( 2 −

+=xxxf (r)

1

1)(2 −

+=t

ttg

(s)

3

2 11)(

++

=uuug (t)

23

12)(

+=

ttts



DERIVATIVE OF TRIGONOMETRIC FUNCTIONS 12. Find the derivative of the function f . (a) f ( (b) fxx 3cos) = xx πcos2)( =

(c) (f (d) f ( 22tan) xx = xxx 2cos) 2=

(e) 1sin) 2 −= xx(f (f) ( )1csc)( 2 += xxf

(g) xxx3cos1

2sin)+

=f ( (h) f ( xx 2tan) =

(i) xxx 2sec)( +=f (j) ( )52cot1)( xx +=f

(k) f ( ( )xx tansin) =

DERIVATIVE OF EXPONENTIAL FUNCTIONS 13. Find the derivative of the given function (a) f ( (b) g xex 3) = tet −=)(

(c) h (d) (f 2

)( xex −= 22) xex x −=

(e) f ( (f) ueuu −= 2) ( )xx eex −+= 3)f (

(g) w

w

eew 1) +

=f ( (h) (f 132) −= xex

(i) xex1

3)(−

=f (j) 1)( −= xexf

(k) (f (l) ( 334) xex −−= ) xex =)(f

(m) (f (n) ( ) 231) +−= xexx ( ) 2

1) 2 sess −+=(f

(o) 21)

xxex

x

+=

−

(f (p) ( )23

2)( xx exex += −g



DERIVATIVE OF LOGARITHMIC FUNCTIONS 14. Find the derivative of the given function (a) f ( (b) (xx ln5) = ( )1ln) += xxf

(c) ( )24ln)( tt −=h (d) xx ln) =f (

(e)

= 2

1ln)x

x(f (f) ( )364ln) 2 +−= xxx(f

(g)

+=

12ln)xxx(f (h) f ( xxx ln) 2=

(i) f (j) ( 32ln)( −= uu ) xx ln)( =f

(k) (f (l) f ( )3ln) xx = xex x ln)( =

(m) (f (n) ( 1ln) 2 += tet t ) ( )1ln) 3 += xx(f

(o) xxx ln) =f ( (p) f ( ( )xx lnln) =

LOGARITHMIC DIFFERENTIATION 15. Use log differentiation to find the derivative of each of the given functions (a) y = (b) y = x3 xx ln

(c) ( 43253 −+ xx )=y (d) ( ) 223 2 ++= xxy

(e) ( ) ( ) ( )432 311 ++− xxx=y

HIGHER ORDER DERIVATIVES 16. Find the first and the second derivatives of each function (a) (f 124) 2 +−= xxx (b) h 10362)( 234 +−+−= ttttt

(c) (f (d) ( 102 2) += xx ) 27

) xx =(f



(e) 11)

+−

=(sssf (f) 1

1)( 2

2

+−

=uuuF

(g) 3 2 4)( += ssh (h) 42) 22 −= xxx(f

17. If ( )f , compute f ’’’(x) 128263 245 +−+−= xxxxx18. If ( ) 23 −= ssg compute g ‘’’(s) IMPLICIT DIFFERENTIATION

19. Find the derivative dxdy

:

(i) by solving each of the given implicit equations for y explicitly in terms of x

(ii) by differentiating each of the given equations implicitly Show that in each case the results are equivalent.

(a) xy (b) x 1= 0122 =−+− yxy (c) x (d) 423 =−− xyx 12332 −=+ xyyx

20. Find dx by implicit differentiation. dy

(a) x (b) x 162 22 =− y 042 323 =−++− yyx

(c) x (d) 105 22 =++ yxy 11222 ++= xxy

(e) 1112 =+y2x (f) x

yxyx 3=

−+

(g) ( ) 0333 =+++ yxyx 21. Find the equation of the tangent line to the graph of the function f defined

by the given equation at the indicated point. (a) 4 at ( 0 , 2 ) 369 22 =+ yx (b) x at ( 1 , 1 ) 01232 =−+− xyyy




22. Find the second derivative 2

2

dxyd

of each of the functions defined

implicitly by the given equation

(a) xy (b) y (c) 1= 82 =− xy 131

31

=+ yx RELATED RATES 23. Use the implicit differentiation to solve the following related rates

problems. (a) The area of a circle is decreasing at a rate of 2 cm2 / min . How fast is the

radius of the circle changing when the area is 100 cm2 ? (b) At a certain instant the length of a rectangle is 16 m and the width is 12 m.

The width is increasing at 3 m / sec . How fast is the length changing if the area of the rectangle is remaining constant ?

(c) The volume of a right circular cylinder is 60 cm3 and is increasing at 2 cm3

/ min at a time when the radius is 5 cm and is increasing at 1 cm / min . How fast is the height of the cylinder changing at that time ?

(d) How fast is the volume of a rectangular box changing when the length is 6

cm , the width is 5 cm and the depth is 4 cm , if the length and depth are both increasing at a rate of 1 cm / sec and the width is decreasing at a rate of 2 cm / sec ?

(e) A point moves on the curve y = x2 . How fast is y changing when x = -2

and x is decreasing at a rate 3 ? (f) A point is moving to the right along the first-quadrant portion of the curve

x2y3 = 72 . when the point has coordinates ( 3 , 2 ) its horizontal velocity is 2 units / sec . What is its vertical velocity ?

(g) The top of a ladder 5 m long rests against a vertical wall . If the base of

the ladder is being pulled away from the base of the wall at a rate of 1/3 m / sec , how fast is the top of the ladder slipping down the wall when it is 3 m above the base of the wall ?

(h) Ink is dropped on to blotting paper forming a circular stain which increases

in area at the rate of 5 cm2 / sec . Find the rate of change of the radius when the area is 30 cm2.

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