revision differentiation mat181q
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M O D U L E 2
M O D U L E 9
DIFFERENTIATION CONTENTS Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions. Calculus is a subject, which falls into two parts: differentiation (module 9) and integration (module 10). PAGE UNIT 1 FUNCTIONAL NOTATION 415
1. FUNCTIONS ...................................................................................................... 417 1.1 Dependent and Independent Variables ............................................................... 417 1.2 Functional Notation ............................................................................................ 417 2. RESPONSES TO ACTIVITIES......................................................................... 420 2.1 Response 1 .......................................................................................................... 420
UNIT 2 LIMITS 423
1. INTRODUCTION .............................................................................................. 425 2. THE TANGENT QUESTION............................................................................ 426 3. WHAT IS MEANT BY A LIMIT? .................................................................... 427 4. TECHNIQUES FOR FINDING LIMITS........................................................... 428 4.1 When the limit of the denominator of a quotient is zero .................................... 431 4.2 Limits at Infinity ................................................................................................. 432 5. ONE-SIDED LIMITS AND CURVES .............................................................. 434 6. RESPONSES TO ACTIVITIES......................................................................... 436 6.1 Response 1 .......................................................................................................... 436 6.2 Response 2 .......................................................................................................... 437 6.3 Response 3 .......................................................................................................... 437
UNIT 3 THE DERIVATIVE 439
1. THE TANGENT QUESTION AND THE DERIVATIVE ................................ 441 1.1 Notation............................................................................................................... 441 2. DIFFERENTIATION FROM FIRST PRINCIPLES ......................................... 443 3. RESPONSES TO ACTIVITIES......................................................................... 445 3.1 Response 1 .......................................................................................................... 445
UNIT 4 STANDARD FORMS 449
1. STANDARD FORMS ........................................................................................ 451 2. RESPONSES TO ACTIVITIES......................................................................... 455 2.1 Response 1 .......................................................................................................... 455 2.2 Response 2 .......................................................................................................... 455 2.3 Response 3 .......................................................................................................... 456
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M O D U L E 9
UNIT 5 RULES OF DIFFERENTIATION I 457
1. DERIVATIVES OF COMBINED EXPRESSIONS .......................................... 459 1.1 Constant times a function.................................................................................... 459 1.2 Sums and Differences ......................................................................................... 459 1.3 Product Rule........................................................................................................ 462 1.4 Quotient Rule ...................................................................................................... 463 2. RESPONSES TO ACTIVITIES......................................................................... 464 2.1 Response 1 .......................................................................................................... 464 2.2 Response 2 .......................................................................................................... 464 2.3 Response 3 .......................................................................................................... 465 2.4 Response 4 .......................................................................................................... 466 2.5 Response 5 .......................................................................................................... 466
UNIT 6 RULES OF DIFFERENTIATION II 469
1. DERIVATIVES OF COMPOSITE FUNCTIONS............................................ 471 1.1 Function of-a-Function or Chain Rule.............................................................. 471 1.2 General Standard Forms ..................................................................................... 480 2. RESPONSES TO ACTIVITIES......................................................................... 484 2.1 Response 1 .......................................................................................................... 484 2.2 Response 2 .......................................................................................................... 485 2.3 Response 3 .......................................................................................................... 485 2.4 Response 4 .......................................................................................................... 485 2.5 Response 5 .......................................................................................................... 487 2.6 Response 6 .......................................................................................................... 488 2.7 Response 7 .......................................................................................................... 488 2.8 Response 8 .......................................................................................................... 488
UNIT 7 HIGHER ORDER DERIVATIVES 489
1. HIGHER ORDER DERIVATIVES ................................................................... 491 1.1 Notation for Higher Order Derivatives .............................................................. 492 2. RESPONSES TO ACTIVITIES......................................................................... 497 2.1 Response 1 .......................................................................................................... 497
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M O D U L E 9
UNIT 8 APPLICATIONS I 499
1. L HOSPITAL S RULE .................................................................................... 501 2. THE GRADIENT OF A CURVE....................................................................... 502 2.1 The Equation of a Tangent to a Curve ................................................................ 502 2.2 The Equation of a Normal to a Curve................................................................. 504 3. RATE OF CHANGE .......................................................................................... 505 3.1 Velocity............................................................................................................... 505 3.2 Acceleration ........................................................................................................ 505 4. RESPONSES TO ACTIVITIES......................................................................... 507 4.1 Response 1 .......................................................................................................... 507 4.2 Response 2 .......................................................................................................... 508 4.3 Response 3 .......................................................................................................... 508
UNIT 9 APPLICATIONS II: MAXIMA AND MINIMA 509
1. MAXIMA AND MINIMA ................................................................................. 511 1.1 Definitions........................................................................................................... 511 1.2 Maximum Value ................................................................................................. 512 1.3 Minimum Value .................................................................................................. 512 2. DERIVED CURVES .......................................................................................... 513 3. PRACTICAL APPLICATIONS......................................................................... 521 4. RESPONSES TO ACTIVITIES......................................................................... 525 4.1 Response 1 .......................................................................................................... 525 4.2 Response 2 .......................................................................................................... 529
UNIT 10 GRAPHS: INTERCEPTS, SYMMETRY, ASYMPTOTES AND
CONCAVITY 531 1. INTERCEPTS AND SYMMETRY OF GRAPHS ............................................ 533 1.1 Intercepts............................................................................................................. 533 1.2 Symmetry............................................................................................................ 534 1.21 Symmetry with respect to the y axis ............................................................. 534 1.2.2 Symmetry with respect to the x-axis............................................................... 534 1.2.3 Symmetry with respect to the origin............................................................... 535 1.2.4 Summary of tests for symmetry...................................................................... 536 2. ASYMPTOTES .................................................................................................. 537 3. CONCAVITY AND POINTS OF INFLECTION.............................................. 540 3.1 Concavity ............................................................................................................ 540 1.2 Points of Inflection.............................................................................................. 542 4. RESPONSES TO ACTIVITIES......................................................................... 549 4.1 Response 1 .......................................................................................................... 549 4.2 Response 2 .......................................................................................................... 550
POST-TEST 553 SOLUTIONS POST-TEST 557
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M O D U L E 9
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DIFFERENTIATION Functional Notation
CONTENTS PAGE
1. FUNCTIONS .......................................................................................................... 7 1.1 Dependent and Independent Variables ................................................................... 7 1.2 Functional Notation ................................................................................................ 7 2. RESPONSES TO ACTIVITIES........................................................................... 10 2.1 Response 1 ............................................................................................................ 10
MODULE 9 UNIT 1
OUTCOMES
At the end of this unit you should be able to: Use functional notation Determine the value of a function Interpret functional notation
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1. FUNCTIONS
A function is a special type of relation. It is a relation in which each element of the domain corresponds to only one element of the range. However the reverse is not true, which implies that each element of the range does not necessarily correspond with only one element of the domain.
1.1 Dependent and Independent Variables
An equation involving two variable quantities has the property that when a value is assigned to one of the variable quantities, the other is determined. For example, the letters x and y indicate the variable quantities in 3 23 5y x x= + . If we assign a value to x in the equation we can determine y. Say we assign the value zero to x, we can determine the value of y as 5. Because x is the value which is assigned it is called the independent variable. An arbitrary value can be assigned to x. Because the value of y is dependent on the value assigned to x, y is called the dependent variable.
1.2 Functional Notation
The functional notation is mathematical shorthand which is extremely convenient to use. It gives an expression an identification tag within a certain area of discussion and specifies the variables involved. For example instead of writing out expressions such as 4 2 4 3 2 75 3 4 and 3 5 4x x x x x y + + we can identify them in the following way:
( )( )
4 2
4 3 2 7
5 3 4 and
, = 3 5 4
f x x x
G x y x x x y
= +
+
Now f and G are the tags, which identify the functions in our discussion, and the x and y in brackets indicate the independent variables. Once identified we refer to f(x) and G(x,y) in our discussion or problem solving instead of writing the expressions in full. This notation is also useful when we work with equations. For example if 3 23 5y x x= + and we let ( ) 3 23 5f x x x= + , then we can say
( )y f x= . This clearly indicate x as the independent variable and we say y is a function of x. If we need to discuss more than one expression or equality in the same context, then we must have several tags so that each one is clearly distinguished. Other letters in common use as identification tags are F, G, g and the Greek letters and . We often need to know the value of a function for an assigned value of the independent variable. We can use a graph or calculate the value if the function is known.
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Example 1 If ( )y f x= figure 1 shows the meaning of f (2). The length of the ordinate at the point x = 2 represents the value f (2).
Figure 1
In general f (x) is the length of the ordinate at any point x.
Example 2 If 2( ) 1f x x= + , find the value of a) f (1) b) f (3) c) f (a) a) ( ) ( )21 1 1 1 1 2f = + = + = b) ( ) ( )23 3 1 9 1 10f = + = + = c) ( ) ( )2 1 1f a a a= + = +
Example 3 Given that ( )
2 4 72
x xG xx+ +=
+ find ( ) ( ) ( )0 , 10 and G G G x h+ .
( ) ( ) ( )( )
( ) ( ) ( )( )
2
2
0 4 0 70
0 2
7 3,52
10 4 10 710
10 2
100 40 712
147 12,2512
G
G
+ +=
+
= =
+ +=
+
+ +=
= =
-
( ) ( ) ( )( )
2
2 2
4 72
2 4 4 72
x h x hG x h
x h
x xh h x hx h
+ + + ++ =
+ +
+ + + + +=+ +
Example 4 If angular displacement is given by = wt + at2, where w and a are constants and t is the time in seconds, evaluate and interpret = f(0) and = f(10). Solution: In our shorthand notation, since is a function of the time t we have ( )f t = .
( )
( ) ( )212
Then 0
0 0
0
f
w a
=
= +
=
Thus the displacement is zero when the time is counted zero, which is usually at the beginning of the movement. Further ( ) ( ) ( )21210 10 10
10 50
f w a
w a
= = +
= +
that is, the displacement is 10 50w a+ units when the time has moved 10 seconds from the start of the counting.
ACTIVITY 1: 1. Given 2( ) 5 2 1f x x x= + determine
( ) ( ) ( ) ( ) ( )0 , 5 , 7 , 1 and f f f f f x h + . 2. ( )
( ) ( )( ) ( )( ) ( ) ( )
2If 3 5 8 find
a 1
b 3 1
c 2 1
x x x = +
3. ( )
( ) ( ) ( )( ) ( ) ( )
( ) ( )
3If 6 , find
a 3 1 4 1
b
c
F x x
F F
F x h F x
F hh
=
+
-
4. ( )( ) ( )( ) ( ) ( )
( ) ( ) ( )
2If 2 , find expressions for
a
b
c
g x x
g x h
g x h g x
g x h g xh
=
+
+
+
5. Consider the following two functions: f(x) = 2x2 + 3x and h(x) = 3x3 x + 4. Calculate the value of ( ) ( )23 2f x h x .
6. Consider the functions 2( ) 4 2 and ( ) 3f t t t g s s= + + = + . Find the following values (a) (2)f (b) (9)g (c) 2 (1) 3 (2)f g+
7. Given that ( ) ( )3 3 1 and 2 3f x x x g t t= + = . Determine the following values:
( )( ) ( )( ) ( ) ( )( ) ( )
(a) 3
b 3
c 1
d ( 2)
f
f
g a g a
f g
+
Remember to check the response on page 420.
2. RESPONSES TO ACTIVITIES
2.1 Response 1
1. ( ) ( )2(0) 5 0 2 0 11
f = +
=
( ) ( )
( ) ( )
2
2
(5) 5 5 2 5 1
125 10 1
116
(7) 5 7 2 7 1
245 14 1
232
f
f
= +
= +
=
= +
= +
=
-
( ) ( )2( 1) 5 1 2 1 15 2 1
8
f = +
= + +
=
( ) ( )
( )
2
2 2
2 2
( ) 5 2 1
5 2 2 2 1
5 10 5 2 2 1
f x h x h x h
x xh h x h
x xh h x h
+ = + + +
= + + +
= + + +
( )( )( )
( )( )( )
( )( )( )
2 2 3
2
2 2
2
2. a 10
b 0
c 0
3. a 42
b 18 18 6
c 6
4. a 2 4 2
b 4 2
c 4 2
+ +
+ +
+
+
x h xh h
h
x hx h
hx h
x h
5. ( ) ( )( )
22 2 3
4 3 2 3
4 3 2 3
4 3 2
3[ ( )] - 2[ ( )]=3 2 + 3 2 3 + 4
2 4 12 9 6 2 8
8 24 18 6 2 8
8 18 18 2 8
f x h x x x x x
x x x x x
x x x x x
x x x x
= + + +
= + + +
= + + +
6. ( )a (2) = 2 + 4(2) + 2
= 4 + 8 + 2
= 14
f
( )
( )
b (9) = 9 + 3.
= 12
b 2 (1) + 3 (2) = 2[(1) + 4(1) + 2] + 3[2 + 3]
= 2[7] + 3[5]
= 29
g
f g
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7. ( ) ( ) ( ) ( )3a 3 3 3 3 13 3 3 3 1
1
f = +
= +
=
( ) ( ) ( ) ( )3b 3 3 3 3 13 3 3 3 1
1
f = +
= + +
=
( ) ( ) ( )( ) [ ]
c 1
2 1 3 2 3
2 2 3 2 3
2
g a g a
a a
a a
+
= + = + +
=
( ) ( )( ) ( )
( ) ( )
( ) ( )3
d First calculate 2
2 2 2 3
4 3
7
Thus ( 2) 7
7 3 7 1
321
g
g
f g f
=
=
=
=
= +
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DIFFERENTIATION Limits
CONTENTS PAGE
1. INTRODUCTION .............................................................................................. 425 2. THE TANGENT QUESTION............................................................................ 425 3. WHAT IS MEANT BY A LIMIT? .................................................................... 427 4. TECHNIQUES FOR FINDING LIMITS............................................................. 18 4.1 When the limit of the denominator of a quotient is zero ...................................... 21 4.2 Limits at Infinity ................................................................................................... 22 5. ONE-SIDED LIMITS AND CURVES ................................................................ 24 6. RESPONSES TO ACTIVITIES........................................................................... 26 6.1 Response 1 ............................................................................................................ 26 6.2 Response 2 ............................................................................................................ 27 6.3 Response 3 ............................................................................................................ 27
MODULE 9 UNIT 2
OUTCOMES
At the end of this unit you should be able to: Explain what is meant by a limit. Find the value of a limit. Give examples of one-sided limits.
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1. INTRODUCTION
It is often important to know how quickly a quantity is changing. For example, the rate at which the speed of a car is increasing or decreasing, the rate at which the temperature of a chemical is rising in a tank and the rate at which currency are changing. Differentiation focuses on analysing the rate at which a function is changing in a situation. Graphically differential calculus solves the tangent question.
2. THE TANGENT QUESTION
In module 7 unit 2 we studied the slope of a straight line. We found that if ( )1 1, andx y ( )2 2,x y are two points on a line, then the slope of the line is given by
2 1
2 1
y ym
x x
=
.
Consider the curve in figure 1.
Figure 1
This is not a straight line, but it is the curve of a function. The slope of a straight line is the same at every point on the line. The curve in figure 1 gives the impression that it gets steeper as x increases. We might expect that the slope of a non-linear curve would be different at different points on the curve. We would like a way to measure the steepness or slope of such a curve at any specific point on the curve. The slope of the tangent to a curve at some point can be used for the slope of the curve at that point. Compare the slopes of the two tangents in figure 2. We can see that the slope of the tangent to a point becomes greater as x increases. (Note: In mathematics a straight line can also be referred to as a curve. For example, you may be asked to draw the curve of 2 2y x= + , which is a straight line.)
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Figure 2
Now we need to find a way to determine the slope of a tangent to a curve at any point. Consider figure 3. Points A and B are different points on the curve. The line that passes through points A and B are called the secant line.
Figure 3
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In figure 4 we see that if B approaches A the secant line (dotted lines) becomes the tangent (solid line) to the curve at point A.
Figure 4
Therefore the limiting value of the slope of secant line will be the equal to the slope of the tangent line. To answer our question on how to find the slope of the tangent line we first need to develop the concept of limits.
3. WHAT IS MEANT BY A LIMIT?
Let us first get a feeling for limits by discussing an example. Consider the function ( ) 2 1f x x= when x is near to two but not equal to two. If we tabulate values of x which approach 2 with the corresponding values of the function f(x) we observe that the closer x comes to the value 2 the closer f(x) comes to the value 3.
x < 2 x > 2 f(1,7) = 2,4 f(1,8) = 2,6 f(1,9) = 2,8
f(1,99) = 2,98 f(1,999) = 2,998
f(2,3) = 3,6 f(2,2) = 3,4 f(2,1) = 3,2
f(2,01) = 3,02 f(2,001) = 3,002
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Figure 5 Both the table and figure5 illustrate that the nearer we take the value of x to 2, the nearer the value of f (x) lies to 3. Note also that it does not matter whether x approaches 2 from the left (x < 2) or from the right (x > 2). We summarise our observations by saying that 3 is the limit of f(x) when x approaches 2. We write this as ( )
2lim 2 1 3x
x
= .
In general we write ( )lim
x af x L
= to say that the limit of f (x), as x approaches a, is
L.
Finding limits by using graphs or tables is a tedious process. In many cases it would be a lot easier to use our knowledge of algebra to determine a limit.
4. TECHNIQUES FOR FINDING LIMITS
Rule 1 and 2 for Limits: lim , where and are real numbers
lim , where is a real numberx a
x a
C C a C
x a a
=
=
Examples:
2 3lim 7 7 and lim 3x x
x
= =
In the following rules we will assume that
( ) ( )1 2 1 2lim and lim where and are real numbers.x a x af x L g x L L L = =
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Rule 3: Limit of a Sum or Difference ( ) ( ) ( ) ( )
1 2
lim lim limx a x a x a
f x g x f x g x
L L
=
=
Thus in order to find the sum (or difference) of two functions, you can find the sum (or difference) of their limits. Example: ( ) [ ]
[ ]3 3 3
lim 2 lim lim 2 rule 3
3 2 rule 1 and 2
5
x x xx x
+ = +
= +
=
Rule 4: Limit of a Product
( ) ( ) ( ) ( )
1 2
lim . lim .lim
.x a x a x a
f x g x f x g x
L L
=
=
This rule states that the limit of the product of two functions is the product of their limits.
Example: [ ]
( ) ( ) [ ]1 1 1
lim3 lim3 lim rule 4
3 1 rule 1 and 2
3
x x xx x
=
=
=
Rule 5: Limit of a Quotient
( )( )
( )( )
12
2
limlim ,if 0
limx a
x ax a
f xf x L Lg x g x L
= =
If the limit of the denominator is not zero, then the limit of the quotient of two functions is the quotient of their limits.
Example: ( )( ) [ ]
[ ]
22
lim 2 42 4lim rule 51 lim 1
2(2) 4 rule 1 to 4(2) 1
xx
x a
xxx x
++ =
+=
Rule 6: Limit of ( ) nf x or ( )n f x If n is a positive integer
-
( ) ( ) ( )
( ) ( )
1
1
lim lim
lim lim
nn n
x a x a
nn nx a x a
f x f x L
f x f x L
= =
= =
( ) ( )3 335 53 33
2 2
lim lim 5 and
lim 4 lim 4 8 2
x x
x x
x x
x x
= =
= = =
Examples 1 ( )
( ) ( )
( ) ( )( ) ( )
( )( )( )
3 3
3 3 3
3
3 2 3 2
3 3 3 3
3 2
22
22
2
2
2 2
2 2
2
a) lim 2 5 lim 2 lim5
2 3 5 3
39
b lim 4 6 lim lim 4 lim 6
3 4 3 6
3
lim 11c lim4 1 lim 4 1
lim lim1
lim 4 lim1
(2) 1(4)2 1
57
x x x
x x x x
xx
x
x x
x x
x x x x
x x x x
xxx x
x
x
=
=
=
+ = +
= +
= +
++ =
+=
+=
=
ACTIVITY 1: Determine: a) ( )2
2limy
y y
+
b) ( )( )3
lim 3 4x
x x
+
c) 25
lim3k
k
d) 2
41
2 5lim3x
x xx
+
-
Remember to check the response on page 436.
4.1 When the limit of the denominator of a quotient is zero
To find the limit of ( )2
0
2 4limx
xx
+ we find that we cannot use rule 5. Checking the
denominator we see that 0
lim 0x
x
= . But this does not mean that the limit does not
exist. If we simplify the fraction we find that ( )
( )
2 2
2
2 4 4 4 4
4
4
4
x x xx x
x xx
x xxx
+ + + =
+=
+=
= +
We are interested in the limits as x approaches 0 and not when x has the value 0. We may thus divide by x that is cancel x in the above manipulation.
Now ( ) ( )
2
0 0
2 4lim lim 4 4x x
xx
x +
= + =
Thus whenever substitution results in 00 , we must do more work to determine whether the limit exists.
Examples 2 ( ) ( )( )
2
2 2
2 24a lim lim2 2x x
x xxx x
+ =
Note that the denominator becomes 0 and start to simplify the fraction.
The factor ( )2x can be cancelled. This factor is sometimes called the vanishing factor.
Thus ( )( )
( )
2
2 2
2
2 24lim lim2 2
lim 2
2 2
4
x x
x
x xxx x
x
+ =
= +
= +
=
-
( ) ( )( )23 3
3
3 3b lim lim3 39
1lim3
16
x x
x
x xx xx
x
= +
=+
=
( ) ( )( )1 1
1
1 1c lim lim Treating 1 as the difference between squares1 1 1
1lim1
11 112
x x
x
x x xx x x
x
= +
=+
=+
=
ACTIVITY 2: Determine:
a) 2
22
3 10lim5 14x
x xx x
+
b) 2
22
8 12lim3 10xx xx x
+
c) 2
30
3 2lim1x
x xx +
+
Remember to check the response on page 437.
4.2 Limits at Infinity
Consider 1lim , 0nx nx> . Now 1 1 1lim lim lim
n n
nx x xx xx = =
Remember that 1yx
= is the standard form of a rectangular hyperbola. Using the
graph of the hyperbola we find that 1lim 0x x
= .
( )1 1 1Thus lim lim lim 0 0n n
nnx x xx xx
= = = =
.
Notice from this example that as 2 3 41 1 1, , , ,...xx x x
all have limits of zero.
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Using this with rule 3 we see that ( ) 1lim lim lim .0 0n
nx x x
c c cxx
= = =
.
Thus we have the following which we will use to find limits at infinity: Limits at Infinity
If c is a constant then lim 0nxcx
=
Examples 3 a) ( )2lim 3 2
xx x
+ Take out the highest power of x as a common factor.
( ) ( )
22
2
3 2lim 1
1 0 0
xx
x x = +
= +
=
b) 2
33 2lim
1xx x
x +
+
We will first divide both the numerator and denominator by the largest power of x in the denominator in this case 3x . This will make each term into a constant or a term with a variable in the denominator and allow us to use the properties for limits at infinity.
2 3
3
2
2 3 3 3
3 3
3 3
31 2
1
3 23 2lim lim
1 1
lim1
x x
x x xx
x
x xx x x x x
x xx x
+ + =+ +
+=
+
Now, according to the properties for limits at infinity, all the terms with a x in the denominator have a limit of 0. So we have
2
33 2 0 0 0lim
1 01010
x
x xx + +=
++
=
=
Hint: When you want to find a limit at infinity, divide both the numerator and denominator by the largest power of the variable in the denominator.
-
ACTIVITY 3: Evaluate:
a) 3 2
33 4 1lim
2 2 1xx x x
x x +
+
b) ( )2 3 4lim 4 5 4x
x x x
+
Remember to check the response on page 437.
5. ONE-SIDED LIMITS AND CURVES
Figure 6 shows the graph of a function f.
Figure 6 Notice that f (x) is not defined when x = 0. As x approaches 0 from the right, f (x) approaches 1. We write this as ( )
0lim 1x
f x+
= .
On the other hand, as x approaches 0 from the left, f (x) approaches 1 and we write
( )0
lim 1x
f x
= .
Limits like these are called one-sided limits. From the last section we know that the limit of a function as x a is independent of the way x approaches a. Thus the limit will exist if both one-sided limits exist and are equal. We therefore conclude that
( )0
limx
f x
does not exist.
( )As a second example of a one-sided limit, consider 3 as approaches 3 (see figure below). Since is defined only when 3, we speak of the limit as
approaches 3 from the right. From the dia
f x x xf x x
=
3gram it is clear that lim 3 0.
xx
+ =
-
Figure 7 A third example:
( ) 21Now let us look at near 0. Figure 4 below shows the graph of the function.
Notice that as 0, both from the left and from the right, ( ) increases without bound. Hence no limit exists a
y f x xx
x f x
= = =
20
t 0. We say that as 0, ( ) becomes positively infinite 1and symbolically we write lim
x
x f x
x
=
Figure 8 A fourth example: The graph of the hyperbola:
( ) 1Consider the graph of for 0. 1As approaches 0 from the right, becomes positively infinite.
y f x xx
xx
= =
-
Figure 9
0 0
1As approached 0 from the left, becomes negatively infinite.
1 1Symbolically we write lim and lim .x x
xx
x x+ = =
Either one of these facts implies that 0
1limx x
does not exist.
6. RESPONSES TO ACTIVITIES
6.1 Response 1
a) ( )2 22 2 2
2
lim lim lim
2 2
6
y y yy y y y
+ = +
= +
=
b) ( )( ) ( ) ( )
( )( )
3 3 3
3 3 3 3
lim 3 4 lim 3 .lim 4
lim lim3 . lim lim 4
3 3 3 4
6
x x x
x x x x
x x x x
x x
+ = +
= +
= +
=
c)
( )
2 2
5 5lim3 3lim
3 25
75
k kk k
=
=
=
-
d) 2
21
4 411
lim 2 52 5lim3 lim 3
22
1
xx
x
x xx xx x
+ + =
=
=
6.2 Response 2
( ) ( )( )( )( )
( )
2
22 2
2
3 5 23 10a lim lim7 25 14
3 5lim7
3 2 52 7
119
x x
x
x xx xx xx x
xx
+ =+ +
+=+
+=
+
=
( ) ( )( )( )( )2
22 2
2
2 68 12b lim lim2 3 53 10
6lim3 5
2 63(2) 5
411
x x
x
x xx xx xx x
xx
+ = +
=+
=+
=
( ) ( ) ( )( )
22
3 30
0 3 0 23 2c lim1 0 1
212
x
x xx
+ + =+ +
=
=
Note in this case we can use rule 5, as the denominator does not become 0.
6.3 Response 3
-
a)
2 3
2 3
3 2
3
4 1 1
2 1
3 4 1lim2 2 1
3lim
2
32
x
x x xx
x x
x x xx x
+ +
+ =
+
=
b) ( )2 3 4lim 4 5 4
xx x x
+
( ) ( )
( ) ( )
2 3 44
4 4 4 4
44 2
4
4
4 5 4lim
4 5 1lim 4
0 0 0 4
4
x
x
x x xxx x x x
xxx x
= +
= +
= +
=
=
-
Differentiation The Derivative
CONTENTS PAGE
1. THE TANGENT QUESTION AND THE DERIVATIVE .................................. 31 1.1 Notation................................................................................................................. 31 2. DIFFERENTIATION FROM FIRST PRINCIPLES ........................................... 33 3. RESPONSES TO ACTIVITIES........................................................................... 35 3.1 Response 1 ............................................................................................................ 35
MODULE 9 UNIT 3
OUTCOMES
At the end of this unit you should be able to: Relate the rate of change of a function to the gradient of the tangent at a point. Differentiate simple expressions from first principles. Recognise the different notations used to denote derivatives.
-
1. THE TANGENT QUESTION AND THE DERIVATIVE
We return to the discussion of the tangent question in unit 2. We want to find an expression for the slope of the curve, or the slope of the tangent at a point on the curve. Consider figure 1.
Figure 1
We can write the slope (gradient) of the secant line through A and B as
( ) ( )change in change in AB
f x h f xy ymx x h
+ = = =
If we move B closer to A, h becomes smaller and is getting closer to 0 and the secant line becomes the tangent. The slope of the tangent line will thus be the limit
( ) ( )0
limh
f x h f xh
+
This formula gives us the slope of the curve at the point A. This special limit is called the derivative of a function. The mathematical process of finding the expression for the gradient of a curve at any point is called differentiation.
1.1 Notation
The symbol ddx
considered by on its own is called the differentiating operator, and
indicates that any function written after it is to be differentiated with respect to x. We
-
always differentiate with respect to the independent variable, that is ( )
( )dependent variable
independent variabled
d.
Note: dy ydx x
yx
is the average rate of change over an interval (slope of the secant) while
dydx
is the limiting value equal to the instantaneous rate of change at a point
(slope of the tangent). The following symbols are often used to indicate derivatives instead of
( ) ( )0
limh
f x h f xh
+ :
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
i spoken as by ii ' spoken as prime
iii iv '
v vix
dy dy dx f x f xdxd f x ydxD y Df x
dydx
has various interpretations:
1. the change in y due to the change in x, 2. the differentiation of y with respect to x, 3. the first differential, 4. the first derivative, 5. the slope or gradient, 6. tan where is the angle of inclination of the tangent to the horizontal. See
figure 2.
Figure 2
-
2. DIFFERENTIATION FROM FIRST PRINCIPLES
Differentiation from first principles means to calculate ( ) ( )
0limh
f x h f xh
+ . We will
follow four steps to calculate this special limit: 1. Find ( )f x h+ 2. Find ( ) ( )f x h f x+ 3. Divide by h 4. Let 0h to find the limit
Examples 1 a) Find the expression for the derivative of y = x2. Solution: We follow the steps as set out above:
( )( ) ( )
( ) ( ) ( ) ( )
2
2
2 2
2 2 2
2
Step 1: Thus
2
Step 2: 2
2
f x x
f x h x h
x xh h
f x h f x x xh h x
xh h
=
+ = +
= + +
+ = + +
= +
( ) ( )
( )
22Step 3:
2
2
f x h f x xh hh h
h x hh
x h
+ +=
+=
= +
( ) ( ) ( )
( ) ( )
0 0
0
Step 4: lim lim 2
2
Thus the derivative = = lim 2
h h
h
f x h f xx h
hx
f x h f xdy xdx h
+ = +
=
+ =
This example can be taken a step further: b) Determine the derivative of y = x2 at x = 3.
Solution: Substitute the given value into the expression obtained in (a):
-
32(3) 6
x
dydx =
= = .
Where 3x
dydx =
is a shorthand notation for the derivative at x = 3.
We can say that the slope or gradient of the curve at x = 3 is 6.
Example 2
( ) ( )12If 2 find ' from first principles.f x x f x= Solution:
( ) ( )( )
12
12
3221 11 1
2 212 22 2
1 1 322 2 2
Step 1: 2
The factor may be expanded by using the Binomial Theorem
2 .....
1 12 ....higher order terms in 2 2
St
x h
f x h x h
x h
x x h
x x h x h h
+ = +
+
= + + +
= + + +
( ) ( )
( ) ( )
1 1 3 122 2 2 2
1 322 2
1 32 2
1 1ep 2: 2 ..... 22 2
1 1 ....higher order terms in 2 2
1 1 ....higher order terms in 2 2
Step 3:
f x h f x x x h x h x
x h x h h
h x x h hf x h f x
h h
+ = + + +
= + +
+ + + =
( ) ( )
1 32 2
1 32 2
0 0
12
1 1 ....higher order terms in 2 2
1 1Step 4: lim lim .. ....higher order terms in 2 2
All the terms containing a 0 become 0.
h h
x x h h
f x h f xx x h h
h
x
= + +
+ = + +
=
-
ACTIVITY 1: 1. Determine the derivatives of the following from first principles: e) 23 at 2y x x= = f) ( ) ( )32 2f x x=
g) ( ) 1f x xx
=
h) 8y x= (Hint: Use the binomial theorem.) Remember to check the response on page 445.
3. RESPONSES TO ACTIVITIES
3.1 Response 1
1. a) ( ) 2Put 3y f x x= =
( ) ( )( )
( ) ( ) ( ) ( )
( ) ( )
( )
( ) ( ) ( )
( )
2
2 2
2 2
2 2 2
2
2
0 0
2
Step 1: 3
3 2
3 6 3
Step 2: 3 6 3 3
6 3
6 3Step 3:
6 3
6
Step 4: lim lim 6
6
6 2 12
h h
x
f x h x h
x xh h
x xh h
f x h f x x xh h x
xh h
f x h f x xh hh h
h x hh
x h
f x h f xx h
hx
dydx
=
+ = +
= + +
= + +
+ = + +
= +
+ +=
+=
= +
+ = +
=
= =
-
b) ( ) ( )33 2
2 2
8 24 24 8
f x x
x x x
=
= +
( ) ( )( ) ( ) ( )
( ) ( ) ( )( )
3
3 2
3 2 2 3 2 2
3 2 2 3 2 2
3 2
2 2 3 2
Step 1: 2 2
8 24 24 8
8 24 24 8 24 48 24 24 24 8
Step 2: 8 24 24 8 24 48 24 24 24 8
8 24 24 8
24 24 8 48 24 24
Step
f x h x h
x h x h x h
x x h xh h x xh h x h
f x h f x x x h xh h x xh h x h
x x x
x h xh h xh h h
+ = +
= + + + +
= + + + + +
+ = + + + + +
+
= + + +
( ) ( ) ( )
( ) ( ) ( )
( )
( ) ( )
2 2
2 2
2 2
0 0
2
2
22
24 24 8 48 24 243:
24 24 8 48 24 24
Step 4: lim lim 24 24 8 48 24 24
24 48 24
6 2 2
Thus ' 24 48 24 6 2 2
h h
h x xh h x hf x h f xh h
x xh h x h
f x h f xx xh h x h
hx x
x
f x x x x
+ + ++ =
= + + +
+ = + + +
= +
=
= + = c)
( )
( ) ( ) ( )
( ) ( ) ( )
( )( )
( )2 2
1
1Step 1:
1 1Step 2:
1 1
f x xx
f x h x hx h
f x h f x x h xx h x
hx h x
xh x h x x hx x h
x h xh hx x h
=
+ = + +
+ = + ++
= +++ + +
=+
+ +=+
-
( ) ( )( )
( )( )
( )
2 2
2
2
Step 3:
1 1
1
f x h f x x h xh h hh x x h
h x xh
x x h h
x xhx x h
+ + += +
+ +=
+
+ +=+
( ) ( )( )
( )
2
0 0
2
2
2
2
1Step 4: lim lim
1
11
1' 1
h h
f x h f x x xhh x x h
xx
x
f xx
+ + +=+
+=
= +
= +
d) ( ) 8Put y f x x= =
( ) ( )
( ) ( )
( ) ( ) ( )
8
8 7 6 2
8 7 6 2 8
7 6 2
7 6
7 6
Step 1:
8 28 .....
Step 2: 8 28 .....
8 28 ....higher order terms in
8 28 ....higher order terms in Step 3:
8 28 ....higher orde
f x h x h
x x h x h
f x h f x x x h x h x
x h x h h
h x x h hf x h f xh h
x x h
+ = +
= + + +
+ = + + +
= + +
+ ++ =
= + +
( ) ( ) ( )7 60 0
7
r terms in
Step 4: lim lim 8 28 ....higher order terms in
8h h
h
f x h f xx x h h
hx
+ = + +
=
7Thus 8dy xdx
= .
-
DIFFERENTIATION Standard Forms
CONTENTS PAGE
1. STANDARD FORMS .......................................................................................... 40 2. RESPONSES TO ACTIVITIES........................................................................... 44 2.1 Response 1 ............................................................................................................ 44 2.2 Response 2 ............................................................................................................ 44 2.3 Response 3 ............................................................................................................ 45
MODULE 9 UNIT 4
OUTCOMES
At the end of this unit you should be able to: Differentiate a function in the standard form without referring to your notes.
-
1. STANDARD FORMS
The process of finding the derivative is called differentiation and deriving it via the limit is called differentiation from first principles. It is quite an involved process and to avoid having to evaluate limits every time we differentiate, we use established derivatives called standard forms The standard derivatives summarised below may be proved theoretically and are true for all real values of x. Notice that the list includes derivatives of the trigonometric functions. In the forms in which the derivatives are given it is essential that the angles be measured in radians. You must memorise this list. ( ) or y f x ( ) or 'dy f x
dx
1 constant, k 0
2 x 1
3 nx 1nnx is a constantn
4 nax 1nanx and are constantsa n
5 xe xe
6 kxe kxke is a constantk
7 xa nxa a is a constanta
8 n x 1x
9 loga x 1nx a
is a constanta
10 sin kx cosk kx is a constantk
11 cos kx sink kx is a constantk
12 tan kx 2seck kx is a constantk
13 cot kx 2cos eck kx is a constantk
14 sec kx sec tank kx kx is a constantk
15 cos ec kx cos ec .cotk kx kx is a constantk
-
Examples 1 Differentiate: a) y = 6
b) y = 6x a) 6 is a constant. Use standard form 1.
0dydx
= .
b) Use standard form 4 with a = 6 and n = 1.
( )( ) 1 1 06 1 6 6dy x xdx
= = =
Examples 2 Find the derivatives of a) 312y x=
b) 312yx
=
c) 3y x=
Use standard form 4: If ny ax= then 1ndy anxdx
= .
a) a = 12 and n = 3 thus ( )( ) 3 1 212 3 36dy x xdx
= = .
b) 312yx
= is rewritten in the standard form nax as 312y x= . So a = 12 and n =
3, thus ( )( ) 3 1 4 43612 3 36dy x x
dx x = = =
c) 3y x= is rewritten in the standard form nax as 123y x= . So a = 3 and n
= 12
, thus ( )1 112 2
12
1 3 3 332 2 2
2
dy x xdx x
x
= = = =
.
Examples 3 Find ( )'f x if : a) ( ) 2xf x = b) ( ) nf x x= c) ( ) 3xf x e= d) ( ) 2logf x x= a) Refer to standard form 7 with a = 2. Thus ( )' 2 n 2xf x = .
b) Refer to standard form 8. Thus ( ) 1'x
f x = .
c) Refer to standard form 6. Thus ( ) 3' 3 xf x e= .
-
d) Refer to standard form 9. Thus ( ) 1'n 2
f xx
= .
Examples 4 Determine dy
dx if: a) siny x=
b) sec3y x=
c) cos4xy =
a) Refer to standard form 10. 1k = Thus cosdy xdx
=
b) Refer to standard form 14. 3k = Thus 3sec3 tan 3dy x xdx
=
c) Refer to standard form 11. 14k = Thus 1 sin4 4
dy xdx
=
ACTIVITY 1: Determine the derivatives to x of the following:
5
3
105 6
33
104
a) b) sin 3
c) cot 2 d)e) log f) n 3
g) h) 5
i) tan 3 j) 10k) cos ec l) log
m) n) 6
x
x
x x
x xx x
e x
xx x
x x
4
p) q) sec4 2x x
Remember to check the response on page 455. The table is written with independent variable x. However it can still be used when other independent variables are involved as shown in the next example.
Examples 5 Find the derivatives of: a) ( ) sin 5y t t= b) ( ) n 3y t t= c) ( ) 5f s s=
-
( )
( )
( ) ( )1 4
5 5 54 5 45
a) ' 5cos5(3) 1b) '3
1 1 1c) Thus '5 55
=
= =
= = = = =
y t t
y tt t
f s s s f s sss
ACTIVITY 2: Determine the derivatives of the following:
( )( )( )( )( )
3
3
a) cos ec 5
b) 6
c)
d) n
e) log
y t t
g t t
f s s
u w w
V h h
=
=
=
=
=
Remember to check the response on page 455
ACTIVITY 3: Differentiate the following:
( )
( )( )
32
7
2
2
a) n 2
b)
c)1
1e)
f)
g)
d)
y
f x x
r h
t
wu
f t x
f x x
v
=
=
=
=
=
=
=
Remember to check the response on page 456
-
2. RESPONSES TO ACTIVITIES
2.1 Response 1
4
2
23
2 3 23
5
5
2
a) 5
b) 3 cos 3
c) 2cosec 2
1 1 1d)3 33
1e)n10
1f)3
g) 5
h) 30
i) 3sec 3
j) 10 n10k) cos ec .cot
1l)n 3
x
x
x
x
x
xxx
x
xe
x
x
x x
x
= =
( )
194
3 3
3m) n) 6041 1p) 4 q) sec tan4 2 2 2
x x
x xx x
=
2.2 Response 2
( )( )
( )
( )
( )
2
12
a) ' 5cos ec 5 .cot 5
b) ' 18
1 1c) '2 21d) '
1e) 'n 3
y t t t
g t t
f s ss
u ww
V hh
=
=
= =
=
=
-
2.3 Response 3
a) 0dydx
= , because n 2 is a constant. You can find the value with your
calculator. b) ( ) 6' 7f x x=
c) 32
1213 3
2 2dr h hdh
= = . In this case r is the dependent variable and h the
independent variable.
d) 1 1 312 2 21 1Thus
2 2dv t tdt
v t
= = = .
e) 2 32 31 2. Thus 2dww u u
duu u = = = = .
f) ( )f t x= . Careful the independent variable is given as t, thus x must be
regarded as a constant in this case. ( )' 0f t =
g) ( ) 2 1' 2f x x = .
-
DIFFERENTIATION Rules of Differentiation I
CONTENTS PAGE
1. DERIVATIVES OF COMBINED EXPRESSIONS ............................................ 48 1.1 Constant times a function...................................................................................... 48 1.2 Sums and Differences ........................................................................................... 48 1.3 Product Rule.......................................................................................................... 51 1.4 Quotient Rule ........................................................................................................ 52 2. RESPONSES TO ACTIVITIES........................................................................... 53 2.1 Response 1 ............................................................................................................ 53 2.2 Response 2 ............................................................................................................ 54 2.3 Response 3 ............................................................................................................ 55 2.4 Response 4 ............................................................................................................ 55 2.5 Response 5 ............................................................................................................ 56
MODULE 9 UNIT 5
OUTCOMES
At the end of this unit you should be able to: Use the following rules of differentiation to differentiate combined expressions: Constant times a function. Rule for sums and differences. Product rule. Quotient rule.
-
1. DERIVATIVES OF COMBINED EXPRESSIONS
The rules of differentiation exist for finding the derivatives of functions that have been combined under various operations. The mathematics involved in proving these rules can become complicated. For our purposes it will suffice to present the rules without proof. Although there are many functions for which the derivative does not exist, our concern will be with functions that are differentiable.
1.1 Constant times a function
When the derivative of a function is already known it is a simple matter to find the derivative of a constant multiple of that function. This involves using the rule
( ) ( ) ( ) ( )If then ' ' where is a constant.F x k f x F x k f x k= = That is, the derivative of k times a function is simply k times the derivative of the function.
Examples 1 ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )
3 2 2
5 6 6
5 5 312 2 2
a If 9 then ' 9 3 27
b If 7 then ' 7 5 35
2 2 5 5c If then 3 3 2 3
f x x f x x x
g x x g x x x
dyy x x xdx
= = =
= = =
= = =
ACTIVITY 1: Find the derivatives of: ( ) ( )( )( )
2
4
3
a 3
b 12
c 2
f x x
y x
V y
=
=
=
Remember to check the response on page 464.
1.2 Sums and Differences
( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )
If then ' ' '
If then ' ' '
F x f x g x F x f x g x
F x f x g x F x f x g x
= + = +
= =
According to these rules the derivative of a sum (or difference) is the sum (or difference) of the individual derivatives. To use these rules we differentiate each term.
-
Examples 2 Find the derivatives of:
( )( )
2
5 3
2
a) sin
b) 7 4 2 8
c) nx
F x x x
f x x x x
y e x x
= +
= +
= +
Solutions: Differentiate the given functions term by term using the correct standard forms:
( )( ) 4 2
12 22
a) ' 2 cos
b) 35 12 2
1 1 1 1c) 2 22 2
x x
F x x x
f x x x
y e x ex x x
= +
= +
= + = +
ACTIVITY 2: Determine:
( )
( ) ( ) ( )
( )
2
2
2
a if 3 5 6
4b ' if 2
3c2
dy y x xdx
f x f x xx
d x xdx x
= +
= +
+
Remember to check the response on page 464. In some cases a quotient or product can be simplified to the sum or difference of functions:
Examples 3 Find the first derivative of:
( )
( ) ( )
2 5
3
2
3 4a
b 3 2
t tSt
y x
+=
= +
Solution:
( )2 5 2 5
3 3 3
1 2
3 4 3 4a
3 4
t t t tSt t t
t t
+= = +
= +
-
( )( )( ) ( )( )( )1 1 2 12
2
3
2
3
2
3 1 4 2
3 83 8
13 8
8 3
ds t tdt
t tt
tt
ttt
= +
= +
= +
+=
=
If possible we simplify answers to the same format as the original question. ( ) ( )
( )( ) ( )
( )
2
2
2 1 1 1
0
b 3 2
9 12 4
9 2 12 1 0
18 12 (Remember 1)6 3 2
y x
x xdy x xdx
x xx
= +
= + +
= + +
= + == +
ACTIVITY 3: Differentiate with respect to the independent variable:
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
3 234
3 14 4
2 72
2
a 4 3 2 b2 7
2 3c d 2 3
2e 2 4 f2
d d z zx xdx dz
y f t t tx x
xy x x yx
+
= =
= + =
( )
( )
2
Find in the following cases:
g
h
dydx
a bx cxyx
ay axax
+ +=
= +
Remember to check the response on page 464.
-
1.3 Product Rule
Note that the derivative of a product do not behave as nicely as other products we have encountered thus far. The derivative of a product is not the product of the derivatives. The product rule is given by the following formula.
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )If then ' ' 'F x f x g x F x f x g x f x g x= = + In words, we have the derivative of product equals the first function times the derivative of the second function plus the second function times the derivative of the first function.
Example 4 ( )( )Determine if 3 1 2 1dy y x x
dx= + +
Solution: ( ) ( )( ) ( )
( ) ( ) ( ) ( )
( )( ) ( )( )
Let 3 1 and 2 1
' 3 and ' 2
' '
3 1 2 3 2 16 2 6 312 5
f x x g x x
f x g xdy f x g x f x g xdx
x xx xx
= + = +
= =
= +
= + + += + + += +
Example 5 If 2 sin find .dyy x x
dx=
Solution: ( ) ( )
( ) ( )( )
2
2
2
Let and sin
cos 2 sin
cos 2 sin
f x x g x xdy x x x xdx
x x x x
= =
= +
= +
ACTIVITY 4: Determine the derivatives of: ( ) ( )( )( ) ( )( )
2
2
a 3 5 2
b cos
c x
y x x x
f t t t
y xe
= +
=
=
Remember to check the response on page 465.
-
1.4 Quotient Rule
The quotient rule is given by the following formula:
( ) ( )( ) ( )( ) ( ) ( ) ( )
( ) 2' '
If then 'f x g x f x f x g x
F x F xg x g x
= =
Consider the following examples.
Example 6 ( ) sinIf find .x dyy x
x dx=
Solution: ( ) ( )( ) ( )
( ) ( ) ( ) ( )( )
( )( ) ( )( )
2
2
2
Let sin and
Then ' cos and ' 1
' 'Thus
cos sin (1)
cos sin
f x x g x x
f x x g x
g x f x f x g xdydx g x
x x x
xx x x
x
= =
= =
=
=
=
Example 7 1Find ' if 1
xy yx
+=
Solution:
( ) ( )
( ) ( )
1 12 2
1 12 2
Let 1 +1 and -1
1 1Then ' and ' 2 2
f x x x g x x
f x x g x x
= + = =
= =
-
( ) ( ) ( ) ( )( )
( )( ) ( )( )( )
( ) ( )( )
1 1 1 12 2 2 2
12
1 1 1 1 1 12 2 2 2 2 2
12
2
1 12 2
2
1 1 1 12 2 2 2
2
' 'Thus '
1 1
1
1
g x f x f x g xy
g x
x x x x
x
x x x x
x
=
+=
+=
( )
( )
( )
1 12 2
12
12
12
1 1 1 12 2 2 2
2
2
2
1
1
1
1
=
=
=
x x
x
x
x
x x
ACTIVITY 5: Determine the derivatives of:
( )
( ) ( )
( )
( )
2
2
2a2 3
b2
tanc2
d ( )n
t
xyxef t
txy
xxf x
x
=+
=+
=
=
Remember to check the response on page 466.
2. RESPONSES TO ACTIVITIES
2.1 Response 1
( ) ( ) ( )( ) ( )( ) ( )
3
2
a ' 3 2
b 12 4
c 2 3
f x x
y x
V y
=
=
=
-
2.2 Response 2
( )
( )( ) ( )( )
( ) ( )1 1
2 2
2
2 1 1 1
a 3 5 6
3 2 5 1 0
6 5
4b 2
2 4
y x xdy x xdx
x
f x xx
x x
= +
= + +
= +
= +
= +
( ) ( ) ( )1 12 231
2 2
312 2
32
1 1
3
1 1' 2 42 2
21 2
2
2
2
f x x x
x x
x xxx
x
xxx x
= +
=
=
=
=
=
( )
( ) ( )
2 22
2
2 12 1 1 1
3
3
3c 32 2
2 3 2 12
6 16 1
d x d xx x xdx dxx
x x x
x x
xx
+ = +
= +
= + +
= + +
-
2.3 Response 3
( )( )
( )
( ) ( )
( )
( )
( )
( )
1134
514 4
6
2 3
2
a 3 4
b2 6c
3d ' 22
3e2
1 1f4
g
h2 2
x
z zdydx x x
f t t t
dy x xdxdydx x x xdy acdx xdy a adx ax x ax
= +
=
=
= +
=
=
2.4 Response 4
(a) ( ) ( ) ( )22 2
2
3 5 . 2 2 3. 2
6 4 10 3 6
9 2 10
dy x x x xdx
x x x x
x x
= + + +
= + +
= +
( ) ( ) ( )121b ' sin .cos
2cos.sin2
= +
= +
f t t t t t
tt tt
( )
( )
2 2
2 2
2
c (2 ) (1)( )
2
2 1
x x
x x
x
dy x e edx
xe e
e x
= +
= +
= +
-
2.5 Response 5
( )
( ) ( )
( )( ) ( )( )( )
( )
( )
( )
2
2
2
2
2 2
2
2
2
2
2
2a2 3
Let ( ) 2 and ( ) 2 3
' 2 ' 3
2 3 2 2 3
2 3
4 6 6 32 3
3 4 62 3
3 4 62 3
xyx
f x x g x x
f x x g x
x x xdydx x
x x xx
x xx
x xx
=+
= = +
= =
+ =
+
+=+
=+
+ += +
( ) ( )
( ) ( )( ) ( )( )( )
( )( )
2
2
22
2
22
b22 2
'2
2 2
2
t
t t
t
ef ttt e e t
f tt
e t t
t
=++
=+
+ +=
+
( )
( )( ) ( )( )( )
( )
2
2
2
2
2
2
tanc22 sec tan 2
2
2 sec tan
4sec tan
2
xyxx x xdy
dx x
x x x
xx x x
x
=
=
=
=
-
( )
( )( )( )
( )
( )
2
2
d ( )n
1n 1'
nn 1n
xf xx
x xxf x
xxx
=
=
=
-
DIFFERENTIATION Rules of Differentiation II
CONTENTS PAGE
1. DERIVATIVES OF COMPOSITE FUNCTIONS.............................................. 60 1.1 Function of-a-Function or Chain Rule................................................................ 60 1.2 General Standard Forms ....................................................................................... 69 2. RESPONSES TO ACTIVITIES........................................................................... 73 2.1 Response 1 ............................................................................................................ 73 2.2 Response 2 ............................................................................................................ 74 2.3 Response 3 ............................................................................................................ 74 2.4 Response 4 ............................................................................................................ 74 2.5 Response 5 ............................................................................................................ 76 2.6 Response 6 ............................................................................................................ 77 2.7 Response 7 ............................................................................................................ 77 2.8 Response 8 ............................................................................................................ 77
MODULE 9 UNIT 6
OUTCOMES
At the end of this unit you should be able to: Differentiate composite functions using the chain rule together with the rules for
combined expressions: - Constant times a function - Rule for sums and differences - Product rule - Quotient rule
-
1. DERIVATIVES OF COMPOSITE FUNCTIONS
Suppose we are given a function y(x), where the variable x is itself a function of another variable, t say. We say that y is a function of a function. For example suppose that ( ) ( )3 and sin ,y x x x t t= = we can write ( )3siny t= . There are two functions working on the variable t. When we differentiate both functions must be considered.
1.1 Function of-a-Function or Chain Rule
The chain rule states
( ) ( ) ( ) [ ]If ( ) then ' ' ( ) . '( ).F x f g x F x f g x g x= =
It is often easier to make a substitution before differentiating and using the ddx
notation. The chain rule is then stated as follows:
Given a function ( ) where ( ) then y y x x x t
dy dy dxdt dx dt
= =
=
Note: We have stated the rule for two functions but any number of functions can be involved. We will first look at examples where powers of a function are involved.
Example 1 Find dy
dx if ( )43 1y x= + .
Solution:
[ ]
( ) ( )
( )
3
33 2
32 3
Let ( ) 1
Then from the chain rule
' ( ) . '( )
4 1 3
This answer can be simplified.
=12 1
g x x
dy f g x g xdx
x x
x x
= +
=
= +
+
-
Example 2 If ( )3( ) siny t t= find dy
dt.
Solution: Make the substitution ( ) sinx t t= Thus 3y x= . Now differentiate each of the functions:
23 and cos= =dy dxx tdx dt
Then using the chain rule we can write:
( )23 cos=
=
dy dy dxdt dx dt
x t
Remove the substitution by substituting sinx t= . ( )2
2
3 sin cos
3sin cos
t t
t t
=
=
Example 3 Differentiate 21y x= . Solution: We note that 21y x= is a function, namely the square root of another function,
21 x . We can make a substitution:
( )1
2 2Let 1 then u x y u u= = =
To use the chain rule we require and du dydx du
.
121 and = 2
2dy duu xdu dx
=
Thus
( )12
12
1 22
dy dy dudx du dx
u x
xu
=
=
=
Substitute ( )21u x=
-
( )
( )
12 2
12 2
2
1
1
1
x x
x
x
x
x
=
=
=
In the following examples we will not write out the substitution. If you battle to follow this approach, write out the substitutions and compare your answer to the answer in the notes.
Example 4 Find the derivative of ( )
32 24 1y x= .
( )
( ) ( )
( )
3 12 22
12 2
12 2
4 3. 1 11 2
6 1 2
12 1
dy dx xdx dx
x x
x x
=
=
=
Example 5 21Find if dy y
d
=
.
Solution:
( )( )
221
2 11 1
11 2
2
3
3
1
2
2 1 1
1 12 1
1 1 12
12
S
dy dd d
= =
=
=
= +
= +
=
-
ACTIVITY 1: Differentiate with respect to the independent variable:
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( ) ( )
( )
83 2
2
32
33 2
3
2
1a b1
c 1 d 2 3
1e 4 9 f3
g and are constants
s f y y yt
s x f t t
y x yx
by a a bx
= =
= + =
= =
= +
Remember to check the response on page 484. In the following example we apply the product rule as well as the function-of-a-function rule:
Example 6 ( )2Find if 3 2 5 4.ds s t tdt = + +
Solution:
( )( )( ) 12
2
2
3 2 5 4
3 2 5 4
s t t
t t
= + +
= + +
We will use a substitution for the product rule but not for the function-of-a-function part.
( )
[ ] ( )
( ) ( )
( )
12
12
12
12
Function of a function
2
1
Let 3 2 and 5 4
16 and 5 4 5 421 5 4 525 5 42
u t v t
du dv dt t tdt dt dt
t
t
= + = +
= = + +
= +
= +
Writing the product rule in ddt
notation we have
( ) ( ) ( ) ( )1 12 22. .
53 2 . 5 4 5 4 . 62
ds dv duu vdt dt dt
t t t t
= +
= + + + +
-
There is various methods to simplify we will illustrate two methods: Method 1:
( ) ( ) ( ) ( )1 12 22 53 2 . 5 4 5 4 . 62ds t t t tdt
= + + + +
Take out the highest common factor (H.C.F.) in both terms:
( ) ( ) ( )
( ) ( )
1 1 12 2 2
12
2
H.C.F.
2 2
2
1 5 4 5 3 2 12 5 42
1 5 4 15 10 60 482
75 48 102. 5 4
t t t t
t t t t
t tt
+
= + + + +
= + + + +
+ +=+
Method 2:
( ) ( ) ( ) ( )1 12 22 53 2 5 4 5 4 62ds t t t tdt
= + + + +
Write first term as a fraction. Add terms using rules for handling fractions.
( )( )
( ) ( )
( ) ( )( )
( )
12
12
1 12 2
12
12
2
2
2 2
2
53 2 5 4 62 5 4
5 3 2 12 5 4
2 5 4
15 10 60 482 5 4
75 48 102. 5 4
t t tt
t t t
t
t t tt
t tt
+
= + + + +
+ + +=
+
+ + +=+
+ +=+
ACTIVITY 2: Differentiate with respect to x and simplify: ( ) ( ) ( )( ) ( ) ( )
( ) ( ) ( )34
5 2
2
2 2 4 3
a 1 2 b
c 2 1 1 d 3 4
e f 2 7
y x x r x ax b
s x x r
x x a y x x
= = +
= + =
= + = + +
Remember to check the response on page 485. In the following examples we will use the quotient rule and the chain rule.
-
Example 7
( )3
4If find 1x dyy
dxx=
+ and simplify.
Solution: We recognise the given function as a quotient and proceed to use the quotient rule. [Note we could handle this question as a product rule by rewriting y as
( ) 43 1y x x = + .]
( ) ( ) ( ) ( )( )
( ) ( )( )
( ) ( )( )
( )( )
Function of a function
4 32 3
24
H.C.F.
32
8
32
8
2
5
1 3 . 4 1 1
1
1 3 1 4
1
1 3 3 4
1
3
1
x x x xdydx x
x x x x
x
x x x x
x
x x
x
+ +=
+
+ + =+
+ + =
+
=
+
Example 8
( ) ( ) ( )2
2
3 2 1Determine ' if
1
xf x f x
x
=
and simplify the answer.
Solution:
( ) ( )
( )( )
( ) ( ) ( ) ( ) ( ) ( )( )
12
1 12 2
12
2
2
2
2
Function of a function
2 2 212
22
3 2 1
1
3 2 1
1
1 .3 4 3 2 1 . 1 2'
1
xf x
x
x
x
x x x x xf x
x
=
=
=
-
There are many different ways to simplify. All of them yield the same answer. We will use the following method: Multiply the top and bottom with the Lowest Common Multiple (L.C.M.) of all the denominators in the top and bottom. L.C.M. in this example is ( )12 21 x .
( )
( ) ( )( )
( )( ) ( ) ( )
( )( )
( ) ( )
( ) ( )( )
( )( )( )
12 22
12 2
1 1 12 2 2 22 2 2
12 2
12 2 2
12
32
12 1 3 2 11 1
2
12 1 1 3 2 1 11 1 1
1
1 11 1
12 2
12
2 2
2
2
32
'1
12 1 3 2 1
1
3 4 4 2 1
1
3 3 2
1
x x x x
x
x x x x x x
x
x x
f xx
x x x x
x
x x x
x
x x
x
+
+ =
+ =
+ =
+ =
=
Example 9 Find the derivative of
21 xyx= using
a) quotient rule b) product rule Solution:
a) ( ) ( )122 212
2
. 1 . 2 1. 1x x x xdydx x
=
2
2
2 2
2
21
2
11
2
2 2
1
1
. 1
xx
x xx
x
x
x
x x
+
=
=
=
-
b) ( )( ) 122
1 21 1xy x xx
= =
( ) ( ) ( ) ( ) ( )
( )( )
( )( )
( )
1 12 2
12
12
1 12 2
12
12
Function of a function
1 2 2 2
2
22
2 2
2 2
2 2
2 2
2 2
1 1 2 12
11
1
1
1
1
1
1
1
dy x x x x xdx
x
xx
x x
x x
x x
x x
x x
+
= +
=
=
+=
=
ACTIVITY 3: Differentiate with respect to the independent variable and simplify:
( ) ( )
( ) ( )
( )
( )
2
2
3
5
43
3
2 1 3a b2 13 1
2 2 3c d5
e Differentiate by using (i) the quotient rule and (ii) the product rule1
1f2 1
x xy rsx x
x x xy yxx
xyx
xyx
+= =+
+ = =+
= +
= +
Remember to check the response on page 485.
Examples 10 By making a substitution 2 3u x x= + , use the chain rule to find the derivative of
( )2n 3y x x= + . Solution:
2Let 3 then nu x x y u= + =
-
In this example the chain rule becomes dy dy dudx du dx
= .
Now 12 3 and du dyxdx du u
= + =
Thus
( )
2
1 2 3
2 33
dy dy dudx du dx
xu
xx x
=
= +
+=+
Note that the numerator is the derivative of the denominator. The result of the previous example can be generalised to any function of the form
( )ny f x= . We have
( ) ( )( )'
If n then .f xdyy f x
dx f x= =
Examples 11 By making a substitution 2 3u x x= + , use the chain rule to find the derivative of
2 3x xy e += . Solution:
2Let 3 then uu x x y e= + =
In this example the chain rule becomes dy dy dudx du dx
= .
Now 2 3 and udu dyx edx du
= + =
Thus
( )
( ) 2 32 3
2 3
u
x x
dy dy dudx du dx
e x
x e +
=
= +
= +
The result of the previous example can be generalised to any function of the form ( )f xy e= . We have
( ) ( )If then '( ) .f x f xdyy e f x e
dx= =
-
Examples 12 By making a substitution 2 3u x x= + , use the chain rule to find the derivative of
( )2sin 3y x x= + . Solution
2Let 3 then sinu x x y u= + =
Now 2 3 and cosdu dyx udx du
= + =
Thus ( )2 3 cosdy dy du x udx du dx
= = +
The result of the previous example can be generalised to any function of the form
( )siny f x= . In fact using the chain rule we can generalise all the standard forms given in unit 4 as follows:
1.2 General Standard Forms
y
dydx
1 ( ) nf x ( ) ( )1. '
nn f x f x
2 (f xe ( ) ( )' f xf x e 3 ( )f xa ( ) ( )' nf xf x a a 4 ( )n f x ( )
( )'f x
f x
5 ( )loga f x ( )( )'1 .
nf x
a f x
6 ( )sin f x ( ) ( )' cosf x f x
7 ( )cos f x ( ) ( )' sinf x f x
8 ( )tan f x ( ) ( )2' secf x f x
9 ( )cot f x ( ) ( )2' cos ecf x f x
10 ( )sec f x ( ) ( ) ( )' sec tanf x f x f x
11 ( )cos ec f x ( ) ( ) ( )' cos ec .cotf x f x f x
-
ACTIVITY 4 Differentiate the following: ( ) ( ) ( ) ( )( ) ( ) ( ) 2
32
3 2
a 1 b 2 5
c d
= + = +
= = x
y n x y n x
y n x y e
( ) ( ) ( )
( ) ( )
( ) ( )
( )
2
2
2 3
2 5 sin 3
5 42
2 2
e n 3 f log 2 1
g 4 h
i 10 j log 3
k 1 .
x
x x x
x
x
y e e y x
y y e
y y x
y x e
+
= + + = +
= =
= =
=
Remember to check the response on page 485. When working with logarithmic or trigonometric functions it is sometimes better to first simplify the given functions before differentiating.
Example 13 2
21Find if .1
dy xy ndx x
= +
Solution:
( ) ( )
( ) ( )( )( )
22 2
2
2 2
2 2
2 2
3 3
4
4
1 a 1 1 log log logb1
2 21 1
2 1 2 1
1 1
2 2 2 21
41
xy n n x n x a bx
dy x xdx x x
x x x x
x x
x x x xx
xx
= = + = + = +
+ =
+
+=
=
Example 14 Find the first derivative if tan 1
secxy
x= .
Solution:
Simplify first:
sin 1tan 1 cos1sec
cos
xx xy
xx
= =
-
sin cos 1cos cos
sin cos coscos 1
sin cos
x xx x
x x xx
x x
=
=
=
The differentiation becomes very easy:
( )cos sin
cos sin
dy x xdx
x x
=
= +
What would happen if we do not simplify first?
( ) ( )( )
( )
2
3 2
2
2
2
3 2
2
sin sin1 1 1cos cos coscot cos1
cos
23
2
tan 1sec
sec sec tan 1 sec tan
sec
sec tan .sec sec .tansec
. .
1 sin sin .cos Multiply top and bottom with coscos
scos sin coscos
x xx x xx x
x
xyx
x x x x xdydx x
x x x x xx
x x x xx
x x xx
=
=
+=
+=
+=
+=
( )
2 2
2 2
in cos 1
cos 1 sin
cos cos sincos
cos sin
x x
x x
x x xx
x x
+ = =
+=
= +
Problems like examples 12 and 13 do not occur frequently but it is nice to ask as it combines different fields. The next activities is aimed at revising what you have learned thus far in differentiation.
ACTIVITY 5 Find the 'y , simplify your answers to the same format as the original function, if possible:
( ) ( ) ( ) ( )( ) ( )
64 2
2 2
2 2 2
a 2 b 2 4 5
1 4c d2
y x y x x
a xy yx a xx
= = +
+= + =
-
( ) ( )2 2
22 2e f 3 2
a xy y x xa x
+= =
( ) ( )
( ) ( ) ( ) ( ) 11 3 22 22
3 2
g h 11 4
i 4 3 j 1
wy y xw
y x x y x x
= = +
= = +
Remember to check the response on page 487.
ACTIVITY 6 Differentiate with respect to the independent variable and simplify the answer: ( ) ( ) ( ) ( )
( ) ( )
( ) ( ) ( )( )
( ) ( )
( ) ( )
( ) ( )
2
2
10
3
2 sin
a log 4 1 b cos
c cot d
In problems e to g find the value of ' for the given
value of in radians .
e tan ;4
f 5 sin ; 22
g 10 sin 3 ; 1
x
x
x x
R x p n x
y e s e
f x
x
f x n x x
xf x e x
f x e x x
= + =
= =
= =
= =
= =
Remember to check the response on page 488.
ACTIVITY 7 Find the first derivative of: ( ) ( )( ) ( ) ( ) ( )
( ) ( ) ( )( ) ( )( ) ( ) ( )
4
2
3 2
2 2
2 4 2
a sin b n cos
c sin 2 d .sin
1 cose cos f1 cos
g tan h tan
i tan 3 j sin
y x y x
f x x u x x
xy x yx
y x y x
y x f x x x
= =
= + =
= =+
= =
= =
Remember to check the response on page 488.
-
ACTIVITY 8 Differentiate with respect to the independent variable, and simplify where possible: ( ) ( ) ( ) ( )( ) ( )
( ) ( )
2 3
3 2
a 3sin 4 b sin 3 3
c sin d sin
1e cos f tan .sin 22
y x y x
y x x
x y x x
= + = +
= =
= =
Remember to check the response on page 488.
2. RESPONSES TO ACTIVITIES
2.1 Response 1
(a) 1
22
2
1 11
S tt
= =
( )
( ) ( )
( )
( )
12
32
32
12 2
2
2
32
1 1 121 1 22
1
1
ds dt tdt dt
t t
t
t
t
t
=
=
=
=
Only the answers to the rest of the questions are supplied.
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( )
( )( )
( )
73 2 2
22
23
2
3 2423
1b ' 8 3 2 c2 1
3d ' 18 2 3 e4 9
2 6f g3 3
dsf y y y y ydx xdyf t t tdx x
dy x dy b badx dx x xx
= =+= =
= = +
-
2.2 Response 2
( ) ( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
24
2
3 22 2
42 2 4 3
5 4a 1 2 1 12 b2
2 3 53 6c d2 1 3 4
3 4 62e f4. 2 7
dy dr ax bxx xdx dx ax b
ds x drdx dx
x xd x a dydx dxx a x x
+= =+
= =
++= =+ + +
2.3 Response 3
In question (a) to (c) it is not necessary to use the chain rule.
( ) ( )( )
( )( )
( ) ( ) ( )( )
( )( )
( ) ( )( )
2 22
2
3 2
32 34
6 53
5 2 6a b2 13 1
3 3 45 20c d2 5 2 3
36 15e f1 2 1
= =+
+ += =+
= =+ +
x xdy drdx ds sx x
x xdy dy x xdx dxx x x
x xdy x dydx dxx x
2.4 Response 4
a) ( ) ( ) 2This function has the form with 1y n f x f x x= = +
22Therefore
1dy xdx x
=+
b) ( )
( )
3Given 2 5
Note that 3 2 5
y n x
y n x
= +
= +
( ) ( )
( )
232 5
6 2 5
dydx x
x
=+
=+
(c) ( ) ( ) ( ) is of the form n where ny n n x y f x f x x= = =
-
( )( )'
Therefor
1
1 1
1
f xdydx f x
xn x
x n x
x n x
=
=
=
=
(d) 23 26 xdy xe
dx=
(e) 3 30 3 0 3x xdy e edx
= + + = (Since 2 and n 3e are constants.)
(f) Use the table of standard forms page 480 number 5 with a = 10 and
( ) 2 1f x x= + . 1 2.
log10 2 1
2 (Since log 10 = 1)2 1
dydx x
x
=+
=+
(g) Use the table of standard forms page 480 number 3 with a = 4 and
( ) 22 5f x x x= + .
( ) ( )
( )( )( )
2
2
2 5
2 5
4 5 4 . n 4
n 4 4 5 4
x x
x x
dy xdx
x
+
+
= +
= +
It is important to use brackets in your answers to prevent confusion. (h) Use the table of standard forms page 480 number 6.
( )( )sin 33cos3 xdy x edx =
(i) ( )( )25(10 ) 10 n10xdy xdx = Using a calculator this can be simplified to ( )2523,03 10 xx . Note we usually do not simplify using a calculator when finding derivatives except if the questions asks us to find the value of the derivative.
-
(j) 3
41 12.n 2 3
4n 2
dy xdx x
x
=
=
(k) ( ) 11 22 22 2 1. 1 . xxy x e x e= =
( ) ( ) ( )( ) ( )( )
( )( )
( )( )( )
( )
1 12 2
2 2
12
22
12
2 2
12
2
12
2
2 21 12 2
2
2
12
2
2
2
2
2
' 1 . 1 2
1
2 1
1 2
2 1
1 2
2 1
2 1
2 1
x x
xx
x x
x
x
y x e e x x
e x xe
x
e x xe
x
e x x
x
e x x
x
= +
= +
+=
+=
+ =
2.5 Response 5
( ) ( ) ( ) ( )( )( ) ( )
( )( )
( )( )
( )( )
( )
( ) ( ) ( ) ( ) ( ) ( )11 1 22 2
53 2
2
3 23 2 2
2 2
2 2 4 4 2
32
2 2 212
a ' 4 2 b ' 24 1 2 4 5
1 2 4c ' d '
2 3 4e ' f '3 2
1 1g ' h '41 4
i ' 2 3 7 18 j ' 1 5 3
= = + +
= =
= =
= =+
= = + +
y x y x x x
a xy yx x a x
a x xy ya x a x x
y yx x xw
y x x x y x x x
-
2.6 Response 6
( ) ( ) ( )
( ) ( )
( ) ( ) ( )
( ) ( )
2
3
2 2 2 2 sin
sin4 loga b4 1
c 2 cosec d 2 cos
e ' 0,5 f ' 2 21,354
g ' 1 27
= =
+
= =
= =
=
x x x
n xedR dpdx x dx xdy dse e x x edx dx
f f
f
2.7 Response 7
( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( )
( )
( ) ( )
( ) ( ) ( ) ( ) ( )
3
2
33 2 22
22 2
2 2 2
3 4 2 4 2
a 4cos .sin b tan
c ' 2cos 2 d .cos sin
2sine cos cos f1 cos
6 sin cos
g 2 sec h 2 tan sec
i 24 tan 3 sec 3 j ' cos 2 sin
dy dyx x xdx dx
duf x x x x xdxdy xy x xdx x
dy x x xdxdy dyx x x xdx dxdy x x x f x x x x xdx
= =
= + = +
= = =+
=
= =
= = +
2.8 Response 8
( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
2 2
2 2 2
a 6 cos 4 b 9 sin 2 3 cos 2 3
cos cosc d2 2 sin
e 6 cos .sin f 2sin cos sin 2
= + = + +
= =
= = =
dy dyx x x xdx dx
dy x d xdx dxx xd dyx x x x x xdx dx
-
DIFFERENTIATION Applications I
CONTENTS PAGE
1. L HOSPITAL S RULE ...................................................................................... 80 2. THE GRADIENT OF A CURVE......................................................................... 81 2.1 The Equation of a Tangent to a Curve .................................................................. 81 2.2 The Equation of a Normal to a Curve................................................................... 83 3. RATE OF CHANGE ............................................................................................ 84 3.1 Velocity................................................................................................................. 84 3.2 Acceleration .......................................................................................................... 85 4. RESPONSES TO ACTIVITIES........................................................................... 86 4.1 Response 1 ............................................................................................................ 86 4.2 Response 2 ............................................................................................................ 87 4.3 Response 3 ............................................................................................................ 88
MODULE 9 UNIT 8
OUTCOMES
At the end of this unit you should be able to: Determine limits of the form 00 with ' 'Hospital s rule. Determine the gradient to a curve. Find the equations of the tangent and the normal to a curve.
-
1. L HOSPITAL S RULE
Sometimes when we calculate the limit of a quotient we obtain the form 00 . In cases where it is difficult to simplify the fraction we can use LHospitals rule.
LHospitals rule states:
( )( )
( )( )
( )( )
( )( )
'0If lim is of the form then lim lim0 '
'if lim is meaningful.
'
x a x a x a
x a
f x f x f xg x g x g x
f xg x
=
In some cases LHospitals rule must be applied more than once until the denominator
is not zero.
Example 1
0
0
tan 2Evaluate limtan 2
tan 2 0It is clear that the lim yield the indeterminate form .tan 2 0
x
x
x xx x
x xx x
+
+
2
20 0
Therefore we apply L'Hospital's rule:
tan 2 1 2sec 2lim limtan 2 1 2sec 2
1 21 2
3
x x
x x xx x x
+ +=
+=
=
Example 2 2 2
20 0
00
2
0
1 2Evaluate lim lim2
which is still in the form apply rule again
4lim2
32
t t t t
t t
t t
t
e e e ett
e e
=
=
=
-
ACTIVITY 1: Determine:
i) 2
1
1lim1x
xx
+
j) 4 2
2
7 12lim2x
x xx
+
k) 2
20
sinlim1 cos
Remember to check the response on page 507.
2. THE GRADIENT OF A CURVE
From the definition of the derivative we know that the gradient (slope) of a curve ( )y f x= at any point ( ),P a b is given by ( )'f a .
Example 3 Determine the slope of the curve 2y x= at the point 3x = .
3Slope = 2 and 6.
x
dy dyxdx dx =
= =
Examples 4 At which point on the curve 2y
x= is the gradient to the curve equal to 2?
The gradient