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LFSE012 Sci/Eng Mathematics A Differentiation
Page 1 of 30
CHAPTER 4 DIFFERENTIATION
4.1 INTRODUCTION
Differential calculus involves the study of finding the rate of change of one quantity with respect
to another.
We can represent the motion of a car with a distance –time graph.
When a car travels at a constant speed (50
km.h–1
) , it has a straight line-graph. Its rate of
change of distance with respect to time is
constant. The gradient of the graph represents
the speed of the car.
dist (km)
200
150
100
50
0
0 1 2 3 4 time (hour)
When the car’s rate of travel with respect to
time is not constant, the graph is not linear.
The gradient of the graph is constantly
changing. The gradient of the curve at each
point gives the speed of the car at that point.
dist (km)
200
150
100
50
0
0 1 2 3 4 time (hour)
The process of differentiation is used to find the rate of change of one quantity with respect to
another, especially when the rate of change is not constant.
4.1.1 FUNCTION NOTATION
Mathematical equations can be written in Cartesian form or function notation.
A function is a rule for which each x-value has a unique y-value.
Function notation is often used in differential calculus.
Cartesian form Function notation
2
2
2
2
2
When = 4, = 4 =16
When = 0, = 0 =0
When = , =
When = , =
y x
x y
x y
x h y h
x x h y x h
2
2
2
2
2
4 4 16
0 0 0
( )
( )
( )
( )
( )
f x x
f
f
f h h
f x h x h
4.1.2 LIMITS
We will consider what happens to the value of a function ( )f x as the value of x approaches
some value. Such a value is called a limiting value. The notation we use is:
lim x a
f x
This is read as “ the limit , as x approaches a, of ( )f x ”
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Example 1
Evaluate each of the following limits (where they exist) :
2
4 3
9a lim 2 (b) lim
3x x
xx
x
( )
Solution:
(a) The function ( )f x = 2 x is a continuous function. So the limit can be evaluated by
substitution.
4
lim 2 =2 4=8 x
x
(b) The function ( )f x = 2 9
3
x
x
is not a continuous function. It does not exist at the point
where x = 3 (we cannot evaluate the expression 0
0, because we cannot divide by 0)
However, if 3x : 2 3 39
3
x xx
x
3x 3x
So 2
3 3
9lim lim 3 6
3x x
xx
x
*The limit exists at x = 3, but the function does not exist.
Theorems of Limits
* In the following, and are different functions of lim and lim( ) ( ) . ( ) , ( )x a x a
f x g x x f x L g x M
1. If ( )f x = c , lim x a
f x c
*If the function is independent of x, the limit is a
constant.
e.g. If ( )f x = 5 , 2
lim 5x
f x
2. lim limx a x a
f x g x L M
( ) ( )
*We can find the limit of a polynomial term by
term.
e.g. If ( )f x = x2 + 3x ,
2
1 1 1lim lim lim 3 1 3 4x x x
f x x x
3. lim lim x a x a
f x g x L M
( ) . ( ) .
*We can split the limit of a product into 2 limits.
e.g. If ( )f x = (x+3) (2x-1) ,
4 4 4lim lim 3 lim 2 1
7 7 49
.x x x
f x x x
4. limlim 0
lim
x a
x a
x a
f xf x LM
g x g x M
( )( )
,( ) ( )
*We can split the limit of a quotient into 2 limits.
e.g. If 3
2 1( )
xf x
x
,
4
4
4
lim 3lim
lim 2 1
7 1
7
x
x
x
xf x
x
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Exercise 4.1
Evaluate each of the following:
1. 4
lim 6x
x
2. 2
1lim 3x
x x
3. 0
lim 5x
4.
2
3 2lim
2x
x x
x
5. 3
0
3lim x
x x
x
6. 2
3
25lim
5x
x
x
7.
2
0
5 25lim h
h
h
8. 2
0
5 4lim h
x h xh
h
9. 2
0lim , where
( ) ( )( )
h
f x h f xf x x
h
4.2 DERIVATIVE OF A FUNCTION
4.2.1 GRADIENT OF A STRAIGHT LINE
The gradient of a straight line is given by:
m = rise
run
For points 2 11 1 2 2
2 1
P and Q m, , ,y y
x y x yx x
The gradient is constant.
y
y = ( )f x
Q(x2,y2)
P(x1,y1)
x
4.2.2 GRADIENT OF A CURVE
The gradient of a curve is constant changing. We find the gradient at any point by drawing a
tangent line at that point.
How do we find the gradient at a point mathematically?
On the curve ( )y f x :
P is the point ,x f x
Q is the point ,x h f x h
We draw a straight line (called a secant)
to join P and Q.
The gradient of this straight line is :
rise
run
f x h f xkm
h h
y y = ( )f x
Q(x+h,f(x+h))
P(x,f(x)) h
x
k
As Q gets closer to P :
h gets smaller and
the gradient of the secant line gets closer to the tangent line at point P.
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So….
0
Gradient of tangent line at P = limh
f x h f x
h
This expression is called the derivative of the function ( )f x and is written as ( )f x
0 = lim
h
f x h f xf x
h
Example 1
If 2( )f x x x , find ( )f x
Solution: 2( )f x x x and
2 2 22( )f x h x h x h x xh h x h
0
2 2 2
0
2
0
= lim
2 lim
lim
h
h
h
f x h f xf x
h
x xh h x h x x
h
x
22xh h x 2h x x
2
0
0
2 lim
lim
h
h
h
xh h h
h
h
2 1x h
h
0
lim 2 1 2 1h
x h x
This process of finding the derivative of a function is called differentiation from first principles.
4.2.3 CARTESIAN NOTATION
On the curve ( )y f x :
P is the point ,x y
Q is the point ,x x y y
means a small change in ;
means a small change in
*
* .
x x
y y
The derivative can be written as dy
dx
y y = ( )f x
P(x,y)) x
x
y
Q ,x x y y
0 0 = lim lim
x x
f x x f xdy yf x
dx x x
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Example 1
Differentiate 25y x from first principles.
Solution:
2 2
22 2
5 5
5 2 5
5
y f x x f x
x x x
x x x x x
2x 2
2 5x x x 2x
2
0
2
0
0
2
lim
2 lim
lim 2 2
x
x
x
x x x
f x x f xdy
dx x
x x x
x
x x x
Exercise 4.2
1. Find from first principles, the derivatives of the following:
(a) 2( )f x x
(b) 2 4 3y x x
(c) 4 3( )f x x x
2. For the function 3 4( )f x x x find
(i) f x h f x
h
(ii)
3 3f h f
h
(iii)
0
3 3limh
f h f
h
The technique used in finding derivatives from first principles is very time-consuming. There are
rules which enable us to write down the derivatives of certain functions. These are called
standard derivatives.
4.3 DERIVATIVE OF ncx
2 2
3 2 3 2
We have observed that the derivative of is 2 this can be written as 2
the derivative of is 3 3 and so on.
:
dx x x x
dx
dx x x x
dx
From this we can make a general rule:
1If then ,n nf x cx f x ncx
where c is any constant, and n is any real number.
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Example 1
Find the derivatives given the following functions:
(a) 61
2( )f x x (b) 4y x
Solution:
6
5 5
1(a)
2
1 6 3
2
( )
( )
f x x
f x x x
12
1 12 2
1
(b) 4 4
1 2 4 2
2
y x x
dyx x
dx x
4.3.1 DERIVATIVE OF A CONSTANT
If then 0,f x c f x
4.3.2 DERIVATIVE OF A POLYNOMIAL
If a function is made up of the sum of a number of terms, the expression is called a polynomial.
e.g. 3 24 5 3 4f x x x x
To find the derivative of a polynomial, differentiate each term in order.
3 2
3 2
2
e.g. If 4 5 3 4
d d d d 4 5 3 4
dx dx dx dx
12 10 3
f x x x x
f x x x x
x x
Example 1
Find the derivatives of the following polynomials:
(a) 4 3 23 2 7 7( )f x x x x x (b) 3 35y x x
x
Solution:
4 3 2
3 2 1 0
3 2
3 2 7 7
3 4 2 3 7 2 1 0
12 6 14 1
( ) ( )a f x x x x x
f x x x x x
x x x
1
31 1 12
2 2 2 2
12
3 1 12 2 2
31 12 2 2
3 3
1 1 1
3 3(b) 5 = 5 5 3
3 1 1 5 3
2 2 2
3 5 3
2 2 2
3 5 3
2 2 2
y x x x x x x xx x
dyx x x
dx
x x x
xx x x
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Exercise 4.3
For questions 1 –10, find the derivatives of the given functions.
1. 4 3( )f x x 2. 24 5 3y x x
3. 4 3 21 1 1
4 3 2( )f x x x x x 4. 24 5 3y x x
5. 1 12 22 3( )f x x x
6.
3
25( )f x x
x
7. 2 1 3y x x x 8. 2y ax bx c
9. 3 24 3 2
0( ) ,x x
f x xx
10.
32( )f x x
11. If 23 7 3y x x , find the value of y for which dy
0dx
12. For 3 2 3 , findf x x x
(i) 0 (ii) the value of for which =0 f x f x
13. Find the gradient of the curve 2 4 3y x x at the y-intercept.
14. Find the gradient of the curve 2 3 10f x x x at the points where the graph cuts the
x-axis.
15. For the function : 2
1 2f x x x find :
(a) 0( )f (b) the value(s) of x for which 0( )f x
(c) ( )f x (d) 0( )f (e) the value(s) of x for which 0( )f x
4.4 THE PRODUCT AND QUOTIENT RULES
4.4.1 DERIVATIVE OF A PRODUCT
When two functions are multiplied together, the derivative of the product can be found using the
product rule. The proof of this rule is in appendix A.
The product rule states:
If where and
then
y uv u u x v v x
dy du dvv u
dx dx dx
In function notation: If then f x u x v x f x v x u x u x v x
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Example 1
Find if 5 3 2 1dy
y x xdx
Solution:
Let 5 3 5 and let 2 1 2
2 1 5 5 3 2
10 5 10 6
20 1
du dvu x v x
dx dx
dy du dvv u x x
dx dx dx
x x
x
Example 2
Find 2 if 2 3 2 5dy
y x x xdx
Solution
2
2
2 2
2
Let 2 3 4 3 and let 2 5 2
2 5 4 3 2 3 2
8 6 20 15 4 6
12 32 15
du dvu x x x v x
dx dx
dy du dvv u x x x x
dx dx dx
x x x x x
x x
Example 3 y = (2x
2 + 6x) (2x
3 + 5x
2)
4.4.2 DERIVATIVE OF A QUOTIENT
When one function is divided by another, the derivative of the quotient can be found using the
quotient rule. The proof of this rule is in appendix C.
The quotient rule states:
2
If where and
then
u
y u u x v v xv
du dvv u
dy dx dx
dx v
In function notation:
2If then
u x v x u x u x v xf x f x
v x v
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Example 1
Find 3
if 4 1
dy xy
dx x
Solution:
2 2
2 2
Let 3 1 and let 4 1 4
4 1 1 3 4
4 1
4 1 4 12 13
4 1 4 1
du dvu x v x
dx dx
du dvv u x xdy dx dx
dx v x
x x
x x
Example 2
Find
34 if
2 6
dy xy
dx x
Solution:
3 2
2 3
22
3 2 3
2
3 2
2
Let 4 12 and let 2 6 2
2 6 12 4 2
2 6
24 72 8
2 6
416 72
2 6
du dvu x x v x
dx dx
du dvv u x x xdy dx dx
dx v x
x x x
x
x x
x
3 24 18
4
x x
2
2 2
2 2 9
3 3
x x
x x
Exercise 4.4
Find the derivative of each of the following functions.
1. 7 4 3y x x 2. 2 42 4f x x x
3. 3 2
2 3
xy
x
4. 1²y x x
5. 3
4
xy
x
6.
2
4
2
4
xy
x
7. 2 3f x x x 8. 2 3
1 2f xx x
9. 2 42 5y x x 10. 2 1x
yx
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4.5 THE CHAIN RULE
Some functions are composed of two functions, one inside the other. For example:
Example 1
2
4 4 is subtraced from ; then the result is squared.f x x x ( ) .
This can be written as 2 where 4( ) ,f x u u x
Example 2
2 4 4 is added to the square of ; then the square root of the result is taken.y x x .
This can be written as 2 where 4,y u u x
Functions like these are called composite functions.
The derivative of a composite function can be found by means of the chain rule. The proof of this
rule is in appendix B.
The chain rule states:
If where theny y u u u x
dy dy du
dx du dx
In function notation: If where then ' f x f u u u x f x f u u x
Note: To solve problems, first decide on an appropriate substitution.
Example 3
Use the chain rule to find the derivative of the function 5
3 2y x
Solution:
5 4
44 4
Let 3 2 =3
So =5
5 3 15 15 3 2. .
duu x
dx
dyy u u
du
dy dy duu u x
dx du dx
Example 4
Find the derivative of the function 2 3( )f x x x
Solution:
1 12 2
12
12
2Let 3 2 3
1So
2
1 2 3 2 32 3
2 22
( )
( ) .
u x x u x x
f u u u f u u
x xf x f u u x u x
uu
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Exercise 4.5A
In each of the following examples, find the derivative of the given function.
1. 3
3 2y x x 2. 6
33 1( )f x x
3. 8
42 4y x 4. 4
26 3( )f x x x
5. 5 4y x 6.
3
2
2
3 1y
x x
7. 4 32y x x 8. 15210 3( )f x x x
9.
5
3 2
4
2 3 15y
x x x
10. 3 2 5y x
4.5.1 DERIVATIVES INVOLVING CHAIN RULE AND OTHER RULES
Example 1
Find the derivative of the function
2
2 3
3 1
xy
x
Solution:
2
Let 2 3 and let 3 1
2 2(3 1) (3 1)
6(3 1)
u x v x
du dv dx x
dx dx dx
x
du dv u
dy dx
dx
2
42
3 1 (2) (2 3)6(3 1)
3 1
(3 1)
vx x xdx
v x
x
4
2(3 1) 6(2 3)
3 1
x x
x
3
3
3
6 2 12 18
3 1
2(8 3 )
3 1
x x
x
x
x
Exercise 4.5B
Find the derivatives of the following functions:
1. 3
5 7 5y x x 2. 5
32 7 3 5y x x
3.
32
5
xf x
x
4.
2
2 3
4
xy
x
5. 22 5y x x 6. 22 3 6y x x
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4.6 DERIVATIVE OF TRIGONOMETRIC FUNCTIONS
The derivation of the derivatives of sin, cos and tan can be found in appendix 1 at the end of the
notes.
DERIVATIVE OF sin x
sin cos d
x xdx
DERIVATIVE OF sin ax
sin
Let
sin cos
By chain rule cos cos : .
y ax
duu ax a
dx
dyy u u
du
dy dy duu a a ax
dx du dx
sin cos d
ax a axdx
DERIVATIVE OF cos x
cos sin d
x xdx
DERIVATIVE OF cos ax
The derivation of the derivative of cos ax can be found using the chain rule:
cos sin d
ax a axdx
DERIVATIVE OF tan x
2tan sec d
x xdx
DERIVATIVE OF tan ax
Similarly if we use the chain rule we obtain:
2tan sec d
ax a axdx
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Example 1
Find down the derivatives of the following functions:
2
tan 2(a) sin 3 (b) cos sin 3 (c) .
xy x f x x x y
x
Solution:
(a) sin 3
3 cos 3
(b) cos sin 3
cos sin 3 + sin 3 cos (product rule)
cos .3cos 3 sin 3 sin
3
y x
dyx
dx
f x x x
d df x x x x x
dx dx
x x x x
.
2
2 2
22
2 2
4
2
cos cos 3 sin sin 3
tan 2(c)
tan 2 tan 2
(quotient rule)
2 sec 2 2 tan 2
2 sec 2 ta
x x x x
xy
x
d dx x x x
dy dx dx
dx x
x x x x
x
x x x
.
4
2
3
n 2
2 sec 2 tan 2
x
x
x x x
x
Exercise 4.6A
Find the derivatives of the following functions
12
sin 2 . 4 tan
cos 3
1. 2
3. 4.
y x y x
y x f
2
2 12
2
6 sin3
cos 3 2 1 tan .
sin
3 cos 3. .
tan 2
5. 6
7 8
x x x
xf x x x x y
x
x xy y
x x
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The derivatives of composite functions involving sine, cosine and tangent can be found by using
substitution and the chain rule. Examples of composite functions are 3 2sin cos 2 5,x x x
3and tan x
Example 2
Find the derivatives of the following functions:
3 2 3a sin b cos 2 5 c tan y x y x x y x
Solution:
3
3 2
2 2 3
a sin
Let 3
sin cos
cos 3 3 cos . .
y x
duu x x
dxdy
y u udu
dy dy duu x x x
dx du dx
2
2
2
3
1 2
3 3 3
b cos 2 5
Let 2 5 2 2
cos sin
sin 2 2 2 1 sin 2 5
c tan
1 Let
3
y x x
duu x x x
dxdy
y u udu
dy dy duu x x x x
dx du dx
y x
duu x x x
dx
. .
2
2 32 2
2 2 33 3
3 2
tan sec
sec1 1 sec sec
3 3 3
dyy u u
du
xdy dy duu x x x
dx du dx x
. .
Exercise 4.6B
Find the derivatives of the following functions
3
2
sin 3 5 . 3 tan 5
cos si
1. 2
3. 4.
y x y x
xy x f x
2
2
n 4cos
tan . sin
. 3 sin 2 . cos sin 4
5. 6
7 8
xx
f x x x yx
y x y x
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4.7 DERIVATIVE OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS
The derivation of the derivatives of and xln x e can be found in the appendix 2 at the end of the
notes.
DERIVATIVE OF ln x
1
d
ln xdx x
DERIVATIVE OF ex
x xde e
dx
DERIVATIVE OF eax
ax axde a e
dx
Example 1
Find down the derivatives of the following functions:
33 2 1
2(a) (b) 3 (c) (d) sin xx xy e f x ln x y e y ln
Solution:
3 2
3 2
(a) (b) 3
3 Let 3 2
x
x
y e f x ln x
dy due u x x
dx dx
3
2
1
1 2 2
3
(c)
. .
x
dyy ln u
du u
dy dy du xx
dx du dx u x
y e
12
3 2 1 1 12 2 2
(d) sin x
Let 3 Let sin x cos
u u
y ln
du duu x x u x
dx dx
dyy e e
du
2 2 1 1
2 2
1
1 3 3 cos
. . . .u u
dyy ln u
du u
dy dy du dy dy due x x e x
dx du dx dx du dx u
1 1
2 2 1 1
2 21
2
cos cot x
sin x
x
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Exercise 4.7
Find the derivatives of the following functions
2 1
4
5 2 .
1. 2
3. 4.
x
x x
y ln x y e
y e f x e
3 2 2
2
3
2 .
6. .
5. 6
7 8
x
x lnx
e
f x ln x y x ln x xx
ln xy y x e
x
4.8 DERIVATIVE OF INVERSE TRIGONOMETRIC FUNCTIONS
The derivation of the derivatives of 1 1 1 Sin , Cos and Tanx x x can be found in the appendix 3
at the end of the notes.
DERIVATIVE OF -1
sin x
1
2
1sin for 1 < < 1
1
dx x
dx x
DERIVATIVE OF -1
cos x
1
2
1cos for 1 < < 1
1
dx x
dx x
DERIVATIVE OF -1
tan x
1
2
1tan
1
dx
dx x
Example 1
Find down the derivatives of the following functions:
1 1 2 1(a) sin 5 (b) cos (c) tanxy x y x y e x
Solution:
1 1 2
2
(a) sin 5 (b) cos
Let 5 5 Let 2
y x y x
du duu x u x x
dx dx
1 1
2 2
2 2 2 4
1
1 1 sin cos
1 15 5 2 2
1 1 25 1 1
(c) tan x
dy dyy u y u
du duu udy dy du dy dy du x x
dx du dx dx du dxu x u x
y e x
1 1
1 1
2 2
(Product rule)
tan +tan
1 1 tan tan
1 1
x x
x x x
dy d de x x e
dx dx dx
e e x e xx x
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Exercises 4.8
Find the derivatives of the following functions
1 1 12
1 3
sin 3 2 . tan
cos sin
1. 2
3. 4.
y x y x
y x f x x
1
11tan
. sin 2 1
5. 6
x
xf x y ln x
x
4.9 IMPLICIT FUNCTIONS
We have found derivatives for functions where y f x ; these functions are called explicit
functions ( is in terms of y x ).
However many functions do not have the x and y terms separated. These are called implicit
functions. Examples of implicit functions are : 2 23 5 8 ; 12 ; 6x y xy x y xy x
In some cases the (eg 3x+5y=12) the expressions can be easily separated.
In other equations ( eg. 2 2 6x y xy x ) this cannot be done.
All the above equations can be solved by a method called implicit differentiation.
Example 1
Find the derivative of 3 5 8x y by means of implicit differentiation
Solution:
Differentiate both sides of the equation with respect to x.
3 5 8
3 5 8
3 5 8
3 33 5 0
5 5
d dx y
dx dx
d d dx y
dx dx dx
d d dx y
dx dx dx
dy dy
dx dx
Example 2
Find the derivative of xy =12 by means of implicit differentiation
Solution:
Differentiate both sides of the equation with respect to x.
12 Use the product rule for
0
0
:d d d
xy xydx dx dx
d dx y y x
dx dx
dyx y
dx
dy y
dx x
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Example 3
Find the derivative of 2 3 2 3x y y by means of implicit differentiation
Solution:
2 3
2 3 3
3
2
2
2 2
2 (3)
2 0 use chain rule for
2 (2 ) 0
2 3 2 0
3 2 2
2 2 or
3 2 2 3
d dx y y
dx dx
d d d dx y y y
dx dx dx dx
d dy d dyx y y
dy dx dy dx
dy dyx y
dx dx
dyy x
dx
dy x x
dx y y
Example 4
Find the derivative of 2 2 6x y xy x by means of implicit differentiation
Solution:
Differentiate both sides of the equation with respect to x.
2 2
2 2 2 2
2 2 2 2 2
2 2 2
2
6
0 Use the product rule for and
0 Use chain rule for
2 1 1 0
d dx y xy x
dx dx
d d d d dx y xy x x y xy
dx dx dx dx dx
d d d d d dx y y x x y y x x y
dx dx dx dx dx dx
dy d dyx y x x y y
dx dy dx
dyx
d
2
2 2
2 2
2 2
2
2
2 2 1 0
2 2 1 0 Rearrange terms
+2 2 1
2 2 1
2 1
2
dyxy x y y
x dx
dy dyx xy xy y
dx dx
dy dyx xy xy y
dx dx
dyx xy xy y
dx
dy xy y
dx x xy
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Example 5
Differentiate xy a where a is a constant.
Solution:
ln ln
ln
Differentiate with respect to :
ln ln
1 ln ln
0 ln 1
ln
x
x
y a
y a
x a
x
d dy x a
dx dx
dy d dx a a x
y dx dx dx
x a
dyy
dx
lnx
a
a a
Exercise 4.9
Find the derivatives of the following using implicit differentiation
2
2 3 2
4 +3 7 2 5
6
x y x y
x y x
1. 2.
3. 4.2
2 2 2 2
2
16
1 3 4 12
3 4 2 3 2 4
y
y x x y
x y y xy x y
5. 6.
7. 8.
2 2 2
2
3
2 4 3 7 2
1
5 2
3
x x
x
xx y xy y
x
y y
y
9. 10.
11. 12.
13.
2 2 3 2 2
2
sin 10
2 = 3 4 2
2
sin
xy y
x yx y x y x y x xy y
x
xy
14.
15. 16.
17. cos x y x ln y y ln x x 18.
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4.10 PARAMETRIC EQUATIONS
A function y f x has variables and x y .Sometimes x and y can both be expressed in terms of
another variable. This variable is called a parameter.
A curve given by , x x t y y t is called a parametrically defined curve and the functions
and x x t y y t are called the parametric equations for the curve.
Example 1 2
2 2
2 2
and
For square both sides :
So the parametric equations and represent the parabola
x t y t
x t
x t y
x t y t y x
:
Example 2
2 2 2 2
2 2 2 2
3cos and 3sin
Square both equations :
9cos ; 9sin Add:
9cos 9sin 9
The parametric equations 3cos and 3sin represent a circle of ra
x t y t
x t y t
x y t t
x t y t
dius 3 units.
4.10.1 DERIVATIVES OF PARAMETRIC EQUATIONS
The derivative of a function which is in parametric form can be found by applying the chain rule.
If where , :y f x x x t y y t
.dy dy dt dy dt
dx dt dx dx dt
Example 1
2Find the derivative for the parametric equations: dy
x t and y tdx
Solution:
2
1
2
22
1
Note : this derivative is in terms of the parameter . We can also express in terms of .
2
dxx t
dt
dyy t t
dt
dy dy dt tt
dx dx dt
dyt x
dx
dyx
dx
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Example 2
Find the derivative for the parametric equations: 4cos 4sin dy
x t and y tdx
Solution:
4 cos 4 sin
4 sin 4 cos
4 cos cot
4 sin
Note : this derivative is in terms of the parameter . We can also express in terms of .
4 cos 4 c
4 sin
dxx t t
dt
dyy t t
dt
dy dy dt tt
dx dx dt t
dyt x
dx
dy t
dx t
2 2 2 2
os 4 cos 4 cos
4 1 cos 16 16 cos 16 16 4 cos
t t t x
t t xt
Exercise 4.10
2 2
Find the derivative for the following pairs of parametric equations.
. 2 1 . sin
. sin 2 cos
1 ; 2 ;
3 ;
dy
dx
x t y t x t y t
x t y t
3 5
2 2
2
.
. 3 . 2 sin
. sin cos 1
4 ;
5 ; 6 ;
7 ;
t t
x t y t
x e y e x t y t
x t y t
2 3
3 2
. 1 1
. ; . 2 1 ; cos
8 ;
9 10 ,
x t y t
x t t y t t x t y t t
4.11 HIGHER ORDER DERIVATIVES
3 2 2
2
is the derivative of the function
In fact is itself a function which may also have its own derivative.
For example, if 2 3 then 6 6
This function 6 6 can also be diffe
.
, .
dyy f x
dx
dy
dx
dyy x x x x
dx
x x
2
rentiated.
6 6 12 6 This is called the derivative. . second d dy d
x x xdx dx dx
We could also find third derivatives, fourth order derivatives and so on. Such derivatives are
called higher order derivatives.
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There are a number of different notations which are used for higher order derivatives.
For the function :y f x
2 2
2 2
3 2
3 2
First derivative is : or or
Second derivative is : or or or
Third derivative is : or
dy df x f x
dx dx
d y d dy df x f x
dx dx dx dx
d y d d y
dx dx dx
3
3 or or
And so on.
df x f x
dx
We can find derivatives of higher. In fact the first and second derivatives will be most important
to us, as they are used in sketching curves.
Example 1
If 2
3
26 sin2 , find and
dy d yy x x
dx dx
Solution: 2
2
218 2 cos 2 36 4 sin 2
dy d yx x x x
dx dx
Example 2
If , find and f x x ln x f x f x
Solution:
12
121
2
2 2
2 2
1 1 1 1 2
2 22
For the second derivative, we use the quotient rule:
2 2 2 2 2 2 2
42
2 4 4
4 4
. .
xf x x
x x xx
d dx x x x x x x
dx dxf xxx
x x x
x x
Exercise 4.11
Find the second derivatives of each of the following:
2 2 5 4 2
32 2
4 7 . 7 7 2 11 3
5. 2 . 3 + .
1. 2 3.
4 5 6
7.
y x x y x x f x x x x
f x x y x x y xx
2 2
. tan 2 3
1. sin sin 3 + cos .
2
8 9.
10 11. 12
x x
x
y e e f x x f x ln x
y x y x x y x e
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4.12 STANDARD DERIVATIVES
We have covered a number of derivatives which you are required to recognise. These are called
standard derivatives.
You do not have to memorise these derivatives; a table of standard derivatives (as shown below)
will be provided.
You should try to become very familiar with these derivatives, as you must be able to recofise the
standard derivative, and the rule which is to be used.
Standard Derivatives
or ( )y f x or dy
f xdx
c 0
nx n 1n x
ex ex axe axa e
ln x 1
x
sin x cos x
sin ax a cos ax
cos x – sin x
cos ax –a sin ax
tan x 2sec x
tan ax 2 seca ax
1sin x 2
1
1 x
1cos x 2
1
1 x
1tan x 2
1
1 x
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APPENDICES
APPENDIX A : PRODUCT RULE
For the product function where and
Let be a small increment in ; this produces small increments in and of
and respectively
,
, .
. . .
y uv u u x v v x
x x u v y
u v y
y y u u v v uv u v v u u v
So as 0 and 0
So
. . .
. . .
, , , ,
y u v v u u v
y v u vu v u
x x x x
y dy v dv u dux u
x dx x dx x dx
dy dv duu v
dx dx dx
APPENDIX B : QUOTIENT RULE
1 1 1
For the product function where and
Let be a small increment in ; this produces small increments in and of
and respectively
,
, .
uy u u x v v x
v
x x u v y
u v y
uy y u u v v u v v u v v
u
v v
u u u uvy
v v v
.v u uv
2
So as 0 and 0
So
. .
.. .
.
, , , ,
.
u v v u u v
v v v v v v
u vv u
y v u u v x x
x x v v v v v v
y dy u du v dvx v
x dx x dx x dx
du dvv u
dy dx dx
dx v
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APPENDIX C : CHAIN RULE
For the composite function where
Let be a small increment in ; this produces small increments in and of
and respectively
If 0 we can write
So as 0
.
,
. ,
y f u u u x
x x u y
u y
x
y y ux
x u x
0
and
So
,
, ,
.
u
y dy y dy u du
x dx u du x dx
dy dy du
dx dx dx
APPENDIX D : DERIVATIVES OF TRIGONOMETRIC FUNCTIONS
Derivatives for the trigonometric functions can be derived using the using the fundamental limits:
θ 0 θ 0
sin θ cos θ 1lim 1 lim 0
θ θ
Derivative of sin x
0 0
0
0
sin
+ sin sin cos +cos sin
sin sinlim lim
sin cos +cos sin sin lim
sin cos sin +c lim
x x
x
x
y x
y y x x x x x x
x x xdy y
dx x x
x x x x x
x
x x x
0
0 0
0 0
os sin
sin cos 1 +cos sin lim
sin cos 1 cos sin lim lim
cos 1 sin lim cos lim
x
x x
x x
x x
x
x x x x
x
x x x x
x x
xx x
x
sin
sin 0 cos 1
cos
x
x
x x
x
Derivative of cos x
cos
cos sin cos2 2 2
cos 1 sin 12
sin
.
. .
y x
d d dx x x x
dx dx dx
x x
x
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Derivative of tan x
The derivation of the derivative of tan x can be found using the quotient rule.
2
2 2
2 2
sin tan =
cos
sin tan =
cos
cos sin sin cos
= (Quotient rule)cos
cos cos sin sin cos sin = =
coscos
. .
. .
xx
x
d d xx
dx dx x
d dx x x x
dx dx
x
x x x x x x
xx
2
2
1 = sec
cosx
x
APPENDIX E : DERIVATIVE OF EXPONENTIAL AND LOGARITHMIC
FUNCTIONS
DERIVATIVE OF ln x
1
0
+ .
1
1
1 1
1 1
Let
1 1 1. 1 1
As 0 p 0.
lim
,
p
x
y ln x
y y ln x x ln x ln x
ln x x ln xy
x x
ln x x ln xx
x xln
x x
xln
x x
x xln
x x x
xp
x
yln p ln p
x x p x
x
dy y
dx
1
0
1
0
1 lim 1
1 lim 1
1 ***refer back to exponential functions
1
p
p
p
p
x
ln px
ln px
ln ex
x
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DERIVATIVE OF xe
Take natural logs of both sides
Differentiate both sides with respect to :
1
By the chain rule :
1
11
.
.
x
x
x
y e
ln y ln e x
x
d dln y x
dx dx
d dyln y
dy dx
dy
y dx
dyy e
dx
APPENDIX F : DERIVATIVE OF INVERSE TRIGONOMETRIC
FUNCTIONS
DERIVATIVE OF -1
sin x
2 2 2 2
2
sin
cos
1 cos 1 sin 1 cos 1 for 1 <x <1
cos
1 for 1 <x <1
1
x y
dxy
dy
dyy y x y x
dx y
dy
dx x
DERIVATIVE OF -1
cos x
2 2 2 2
2
cos
sin
1 sin 1 cos 1 sin 1 for 1 <x <1
sin
1 for 1 <x <1
1
x y
dxy
dy
dyy y x y x
dx y
dy
dx x
DERIVATIVE OF -1
tan x
2
2 2
2
2
2
tan
sec
1 And 1 + tan sec
sec
1
1 tan
1
1
x y
dxy
dy
dyy y
dx y
dy
dx y
x
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ANSWERS
Exercise 4.1
1. 24 2. –2 3. 5 4. 5 5. –3
6. –2 7. 10 8. 5x2 – 4x 9. 2x
Exercise 4.2
1. (a) 2x (b) 2x –4 (c) 2x –1
2. (i) 2 23 3 4x xh h (ii) 223 9h h (iii) 23
Exercise 4.3
1. –3 2. 8 5x 3. 3 2 1x x x 4. 212 5 1x
5. 2 3
2
x
x x
or
3
1 3
x 2 x 6.
2
2
5 6
2
x
x x
or
5
5 3
2 x x 7. 26 10 3x x
8. 2ax b 9. 3 2
2
8 3 2x x
x
10.
23 2x
11. 7 1
76 12
,x y 12. (i) 9 (ii) 3
4 13. 4
14. –7 , 7 15. (a) –2 (b) –1 , 2 (c) 23 3x (d) –3 (f) 1
Exercise 4.4
1. 25 6x 2. 4 22 3 4 4 x x x 3.
2
13
2 3
x 4. 23 1x
5.
2
2
2 6
4
x x
x 6.
2 4
24
2 4 4
4
x x x
x
7. 3 3
22 2
x
x
8. 3 2
12 1
x x 9. 2 42 5 4 3x x x 10.
23 1
2
x
x x
Exercise 4.5A
1. 2
3 23 2 3 2x x x 2. 5
2 354 1x x 3. 7
3 464 4x x
4. 5
224 6 1 3x x x
5. 5
2 5 4x 6.
4
2
6 2 3
3 1
x
x x
7. 2
4 3
2 3
2
x x
x x
8.
652
2 2 3
3
x
x x
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9.
2
63 2
20 3 4 3
2 3 15
x x
x x x
10.
2
3
2
3 2 5x
Exercise 4.5B
1. 2
2 10 23 5x x 2. 4
3 2 33 32 50 7 2 7x x x
3.
2
2
2 17 2
5
x x
x
4.
3
2 7
4
x
x
5.
2
2
4 5
2 5
x
x
6.
248 10 3
2 6
x x
x
Exercise 4.6A
2 12
2 2 21 1 12 2 2
2 cos2 . 2 sec 3 sin 3 2 9 cos3
3 2 1 sec 2 3 1 tan . cosec
6 tan 2 sec.
x x x x x
x x x x x x
x x x
1. 2 3. 4.
5. 6
7 2
2
2 6 sin3 cos 3 .
2tan 2
x x x x
x xx
8
Exercise 4.6B
2 2 3
2
2
1 3 cos 3 5 . 45 sec 5 sin
2
2 sin 4 2 cos 4 2 sec tan . 2
sin 4 2
x x x xx
x x x x x x xx
x x
1. 2 3.
4. 5. 6
2 2 cosec
. 6 cos 2 . cos sin sin 4
x
x y x x
7 8 .
Exercise 4.7
52 1 4
5
2 4
5 3 4 . 2 4
5 2 2
1 3 62 1. 2 . . 1
1. 2 3. 4. 5.
6 7 8
x x x x
x lnx
xe e e e
x x x
ln xxx e x x l
x x x n x
Exercise 4.8
1. 2
3
12 9 3 x x
2. 2
2
4 x 3.
2
6
3
1
x
x
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4. 1
2sin
1
xx
x
5. 2 1
2 2
2 1 2(1 ) tan
(2 1) (1 )
x x x
x x 6.
2
1
1 ln
x x
Exercise 4.9
2
2
4 2 3 . .
3 43
2 3 2 .
2 13
x x x xx
y y yy
xy y
xx
1. 2 3. 4. 5. 6
7. 8
2
2
2 23
22 2
3 8
2 2 3
1 2 1. . 5 ln 5 . 3 ln 2 2
2 1 2 1
3 ln 3
x x
x
y xy
x xy
xy x
y x y x
9.
10 / 11 12
13. 14.
3 2 2
4 2
2 2
3 2
1 4 4 .
cos 2 2 1
3 3 2 4 cos cos .
sin sin 2 4 2
.
y x y xy
x y x y x y
x y xy y x y y
x x yx x y x
y x y x ln y
x x y ln x
15
16 17.
18
Exercise 4.10
2cos 5. . . 2 tan . . 2
2 3
cos 3 2 1. . 2 sin . .
2 2
1 2 3 4 5
6 7 8 9
tt tt t e
t
t t tt
t t
2 cos sin .
61
10
t t t
t
Exercise 4.11
3 2
3
33
2
3
10 2 . 14 40 132 6 .
8 3 3 2. 2 .
4 94
2 sin . or 2sec t
cos
x x
x xx
x xor e e
x x x xx
xx
x
1. 2 3. 4
5 6 7.
8
2
2
4an . 2 cos 2
2 3
1 1 9 sin 3 cos . 4 2
4 2
x
x xx
x x x x e
9. 10
11. 12
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