Calculus 05-6 ccccccccccccccccccccccc.ppsx

Faculty of Engineering
Calc. and Analytic geom. 1
ENB1101
Lectures 5-6
sinaiuniversity.net
Asst. Prof.
Dr. M. Basseem

Exercise
Consider 𝑦 =
𝑥𝑥 ln3𝑥
𝑒 𝑥sin−1𝑥
7/2
. Find 𝑦′
ln 𝑦 =
7
2
ln
𝑥𝑥 ln3𝑥
𝑒 𝑥sin−1𝑥
ln 𝑦 =
7
2
𝑥 ln 𝑥 + 3 ln ln 𝑥 − 𝑥 − ln sin−1
𝑥
By diff.
Solution
𝑦′
𝑦
=
7
2
1 + ln 𝑥 +
3
𝑥 ln 𝑥
−
1
2 𝑥
−
1
1 − 𝑥2 sin−1𝑥
𝑦′
=
7𝑦
2
1 + ln 𝑥 +
3
𝑥 ln 𝑥
−
1
2 𝑥
−
1
1 − 𝑥2 sin−1𝑥

Exercise
Consider 𝑓 𝑥 = 𝑥2 + 6 𝑥 + 3, 𝑥 ≥ −3.
Find 𝑓−1
′(3). Sketch 𝑓 and 𝑓−1
. Deduce its domain
and range.
∵ 𝑓 0 = 3
Solution
𝑓−1 ′(3) =
1
𝑓′ 𝑓−1
(3)
=
1
𝑓′ 0
=
1
6
.
𝑓′ 𝑥 = 2𝑥 + 6

4
Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
𝑓 𝑥 = 𝑥2
+ 6 𝑥 + 3 = 𝑥 + 3 2
− 6
𝑦 = 𝑥 + 3 2
− 6
𝑦 + 6 = 𝑥 + 3 2
𝑦 + 6 = 𝑥 + 3
𝑦 + 6 − 3 = 𝑥
∴ 𝑓−1
= 𝑥 + 6 − 3
Domain = [−3, ∞[ Range = [−6, ∞[
Domain = [−6, ∞[ Range = [−3, ∞[
(−3, −6)
(−6, −3)

5
1
sinh csch
2 sinh
1
cosh sech
2 cosh
sinh cosh
tanh coth
cosh sinh
x x
x x
e e
x x
x
e e
x x
x
x x
x x
x x



 

 
 
Hyperbolic functions
DEFINITION

6
Note that sinh 𝑥 has domain and range
 cosh 𝑥 has domain and range [1, ∞[.
HYPERBOLIC FUNCTIONS

7
The graph of tanh 𝑥 is shown.
It has the horizontal asymptotes y = ±1.

8
 It can be proved that, if a heavy flexible cable
 is suspended between two points at the same
 height, it takes the shape of a curve with
 equation y = c + a cosh(x/a) called a catenary.
 The Latin word
catena means
‘chain.’
APPLICATIONS

9
 Another application occurs in the
 description of ocean waves.
 The velocity of a water wave with length L moving across a
body of water with depth d is modeled by
the function
where g is the acceleration due to gravity.
2
tanh
2
gL d
v
L


 
  
 
APPLICATIONS

10
 The hyperbolic functions satisfy
 a number of identities that are similar to
 well-known trigonometric identities.
HYPERBOLIC IDENTITIES

11
2 2
2 2
sinh( ) sinh
cosh( ) cosh
cosh sinh 1
1 tanh sech
sinh( ) sinh cosh cosh sinh
cosh( ) cosh cosh sinh sinh
x x
x x
x x
x x
x y x y x y
x y x y x y
  
 
 
 
  
  
HYPERBOLIC IDENTITIES
 We list some identities here.
sinh 2𝑥 = 2 sinh 𝑥 cosh 𝑥
cosh 2𝑥 = cosh2
𝑥 + sinh2
𝑥

12
 Prove:
a. cosh2x – sinh2x = 1
b. 1 – tanh2 x = sech2x
HYPERBOLIC FUNCTIONS Example 1

13
2 2
2 2
2 2 2 2
cosh sinh
2 2
2 2
4 4
4
1
4
x x x x
x x x x
e e e e
x x
e e e e
 
 
   
 
  
   
   
   
 
 
HYPERBOLIC FUNCTIONS Example 1 a

14
 We start with the identity proved in (a):
 cosh2x – sinh2x = 1
 If we divide both sides by cosh2x, we get:
2
2 2
2 2
sinh 1
1
cosh cosh
or 1 tanh sech
x
x x
x x
 
 
HYPERBOLIC FUNCTIONS Example 1 b

15
 The identity proved in Example 1 a
 gives a clue to the reason for the name
 ‘hyperbolic’ functions, as follows.

16
 The derivatives of the hyperbolic
 functions are easily computed.
For example,
(sinh ) cosh
2 2
x x x x
d d e e e e
x x
dx dx
 
 
 
  
 
 
DERIVATIVES OF HYPERBOLIC FUNCTIONS

17
 We list the differentiation formulas for
 the hyperbolic functions here.
DERIVATIVES

18
 Any of these differentiation rules can be combined with the
Chain Rule.
sinh
(cosh ) sinh
2
d d x
x x x
dx dx x
  
1.
𝑑
𝑑𝑥
sinh 𝑓 = cosh 𝑓 . 𝑓′
2.
𝑑
𝑑𝑥
cosh 𝑓 = sinh 𝑓 . 𝑓′
3.
𝑑
𝑑𝑥
tanh 𝑓 = sech2 𝑓 . 𝑓′
4.
𝑑
𝑑𝑥
sech 𝑓 = −sech 𝑓 tanh 𝑓 . 𝑓′
5.
𝑑
𝑑𝑥
csch 𝑓 = −csch 𝑓 coth 𝑓 . 𝑓′
6.
𝑑
𝑑𝑥
coth 𝑓 = −csch2 𝑓 . 𝑓′
Consider
𝑓 = 𝑓(𝑥)
Example:

19
 You can see from the figures that sinh
 and tanh are one-to-one functions.
So, they have inverse functions denoted by
sinh−1
𝑥 and tanh−1
𝑥.
INVERSE HYPERBOLIC FUNCTIONS

20
 This figure shows that cosh 𝑥 is not one-to-one.
However, when restricted to the domain
[0, ∞], it becomes one-to-one.
INVERSE FUNCTIONS

21
1
1
1
sinh sinh
cosh cosh and 0
tanh tanh
y x y x
y x y x y
y x y x



  
   
  
Definition 2
INVERSE FUNCTIONS
 The remaining inverse hyperbolic functions
are defined similarly.

22
 The graphs of sinh−1
𝑥,
cosh−1
𝑥 and tanh−1
𝑥 are
displayed.
INVERSE FUNCTIONS
𝑦 = sinh−1
𝑥
𝑦 = tanh−1 𝑥
𝑦 = cosh−1
𝑥

23
 Since the hyperbolic functions are defined
in terms of exponential functions, it’s not
surprising to learn that the inverse hyperbolic
functions can be expressed in terms of logarithms.
INVERSE FUNCTIONS

24
In particular, we have:
 
 
1 2
1 2
1 1
2
sinh ln 1
cosh ln 1 1
1
tanh ln 1 1
1
x x x x
x x x x
x
x x
x



   
   

 
   
 

 
INVERSE FUNCTIONS

25
Show that .
Let y = sinh-1 x. Then,
So, e y – 2x – e -y = 0
Or, multiplying by e y, e 2y – 2xe y – 1 = 0
This is really a quadratic equation in e y:
(e y )2 – 2x (e y ) – 1 = 0
 
1 2
sinh ln 1
x x x

  
sinh
2
y y
e e
x y


 
INVERSE FUNCTIONS

26
 Solving by the quadratic formula,
 we get:

Note that e y > 0, but
So, the minus sign is inadmissible and we have:
Thus,
2
2
2 4 4
1
2
y x x
e x x
 
   
2
1 0
x x
  
2
1
y
e x x
  
 
2
ln( ) ln 1
y
y e x x
   
INVERSE FUNCTIONS

27
, |𝑥| < 1 , |𝑥| > 1

28
Prove that .
 Let y = sinh-1 x. Then, sinh y = x.
 If we differentiate this equation implicitly
with respect to x, we get:
 As cosh2 y - sinh2 y = 1 and cosh y ≥ 0, we have:
 So,
1
2
1
(sinh )
1
d
x
dx x



cosh 1
dy
y
dx

2
cosh 1 sinh
y y
 
2 2
1 1 1
cosh 1 sinh 1
dy
dx y y x
  
 
DERIVATIVES

29
 By another way, we have:
   
 
 
1 2
2
2
2 2
2
2
2 2
sinh ln 1
1
1
1
1
1
1 1
1 1
1
1 1
d d
x x x
dx dx
d
x x
dx
x x
x
x x x
x x
x
x x x

  
  
 
 
 
 
  
 
 
 

  
DERIVATIVES

31
Implicit Differentiation
So far, all the equations and functions we
looked at were all stated explicitly in terms
of one variable:
x
y
1

In this function, y is defined
explicitly in terms of x. If we re-
wrote it as 𝑥𝑦 = 1, y is now
defined implicitly in terms of x.
It is easy to find the derivative of an explicit
function, but what about:
5
3
4 2
3



 y
y
y
x

32
2 2
1
x y
 
This is not a function,
but it would still be nice
to be able to find the
slope.
2 2
1
d d d
x y
dx dx dx
  Do the same thing to both sides.
2 2 0
dy
x y
dx
 
Note use of chain rule.
2 2
dy
y x
dx
 
2
2
dy x
dx y


dy x
dx y
 


33
2
2 sin
y x y
 
2
2 sin
d d d
y x y
dx dx dx
 
This can’t be solved for y.
2 2 cos
dy dy
x y
dx dx
 
2 cos 2
dy dy
y x
dx dx
 
  2
2 cos
dy
x
y
dx


2
2 cos
dy x
dx y


This technique is called
implicit differentiation.
1 Differentiate both sides w.r.t. x.
2 Solve for .
dy
dx 

34
4
5
2
3
3
4
i) x
x
y
y 


Find dy/dx if:
3
4
2
12
5
8
3 x
x
dx
dy
y
dx
dy
y 


  3
4
2
12
5
8
3 x
x
y
y
dx
dy



 
y
y
x
x
dx
dy
8
3
12
5
2
3
4



Examples.

35
Find
dy/dx
if:
2
2
2
2
sin
cos
cos
sin
ii) x
y
x
y 


)
2
(
cos
2
sin
)
2
(
sin
2
cos 2
2
2
2
x
x
dx
dy
y
y
x
x
dx
dy
y
y 















2
2
2
2
sin
2
cos
2
sin
2
cos
2 x
x
x
x
dx
dy
y
y
dx
dy
y
y 


  2
2
2
2
sin
2
cos
2
sin
2
cos
2 x
x
x
x
y
y
y
y
dx
dy



 
2
2
2
2
sin
2
cos
2
sin
2
cos
2
y
y
y
y
x
x
x
x
dx
dy



 
 
2
2
2
2
sin
cos
sin
cos
y
y
y
x
x
x
dx
dy




36
Find
dy/dx
if:
8
4
5
3
iii) 3
2
2


 y
xy
x
0
12
10
5
6 2
2




dx
dy
y
dx
dy
xy
y
x
Product
Rule!
  2
2
5
6
12
10 y
x
y
xy
dx
dy




 
2
2
12
10
5
6
y
xy
y
x
dx
dy





37
We need the slope. Since we can’t solve for y, we use implicit
differentiation to solve for .
dy
dx
2 2
7
x xy y
   ( 1,2)

2 2
7
x xy y
  
2 2 0
dy
dy
x y
x y
dx
dx
 
  

 
 
Note product rule.
2 2 0
dy dy
x x y y
dx dx
   
  2
2
dy
y x
y x
dx
 

2
2
dy y x
dx y x



 
 
2 2 1
2 2 1
m
 

  
2 2
4 1



4
5


Find the equations of the lines tangent and normal to the curve
at .

38
2 2
7
x xy y
   ( 1,2)

4
5
m 
tangent:
 
4
2 1
5
y x
  
4 4
2
5 5
y x
  
4 14
5 5
y x
 

Find the equations of the lines tangent and normal to the curve
at .
normal:
 
5
2 1
4
y x
   
5 5
2
4 4
y x
   
5 3
4 4
y x
  
Normal line is
perpendicular
to tangent
)
( 1
1 x
x
m
y
y 


Eq. of straight line

39
Example. Find derivative at (1, 1)
x
x
y
x
y



2
3
3
2
Product Rule is
easier than quotient
rule, so let’s cross
multiply!
3
3
3
2
x
xy
x
y 


2
3
2
2
3
3
3
2 x
y
dx
dy
xy
x
dx
dy
y 



  2
3
2
6
3
2 x
y
xy
y
dx
dy



 
2
2
3
3
2
6
xy
y
x
y
dx
dy



 
2
2
3
)
1
)(
1
(
3
)
1
(
2
)
1
(
6
)
1
(



dx
dy
5
1
5




dx
dy

40
Higher Order Derivatives
Find if .
2
2
d y
dx
3 2
2 3 7
x y
 
3 2
2 3 7
x y
 
2
6 6 0
x y y
 
2
6 6
y y x

  
2
6
6
x
y
y

 

2
x
y
y
 
2
2
2
y x x y
y
y

 
 
2
2
2x x
y y
y y
 
 
2 2
2
2x x
y
y
x
y
y
   
4
3
2x x
y
y y
  
Substitute
back into the
equation.
y


Derivatives of
Parametric Equations
Lesson 10.2

Derivative of Parametric Equations
Consider the graph of
x = 2 sin t, y = cos t
We seek the slope, that
is
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑡
.
𝑑𝑡
𝑑𝑥
For parametric equations:
𝑑𝑦
𝑑𝑥
=
𝑑𝑦/𝑑𝑡
𝑑𝑥/𝑑𝑡
 For our example :
𝑑𝑦
𝑑𝑥
=
− sin 𝑡
2 cos 𝑡
=
− tan 𝑡
2

Second Derivatives
The second derivative is the derivative of the
first derivative
But the first derivative is a function of t
 We seek the derivative with respect to x
 We must use the chain rule
𝑑2𝑦
𝑑𝑥2 =
𝑑
𝑑𝑥
𝑦′
=
𝑑
𝑑𝑡
𝑦′
.
𝑑𝑡
𝑑𝑥
=
𝑑
𝑑𝑡
𝑦′
𝑑𝑥
𝑑𝑡

Second Derivatives
Find the second derivative of the parametric
equations
 x = 3 + 4cos t
y = 1 – sin t
 First derivative:
𝑑𝑦
𝑑𝑥
=
𝑦.
𝑥. =
− 𝑐𝑜𝑠 𝑡
−4 sin 𝑡
=
cot 𝑡
4
 Second derivative:
𝑑2𝑦
𝑑𝑥2 =
𝑑
𝑑𝑡
𝑦′
.
𝑑𝑡
𝑑𝑥
=
−1
4
csc2𝑡
−4 sin 𝑡
∴ 𝑦′′
=
1
16 sin3𝑡

Try This!
Where does the curve described by the
parametric equations have a horizontal tangent?
 x = t – 4
y = (t 2 + t)2
 Find the derivative
 For what value of t does dy/dx = 0?

Exercise
Consider 𝑥 = 2 sin
𝑡+1
𝑡−1
, 𝑦 = cos
𝑡+1
𝑡−1
. Prove 𝑦′′ =
−1
4 𝑦3
∵ cos2
𝜃 + sin2
𝜃 = 1
∴
𝑥2
4
+ 𝑦2
= 1
𝑥
2
+ 2𝑦𝑦′ = 0
∴ 𝑦′
=
−𝑥
4𝑦
By diff.
4𝑦′′
=
−𝑦+𝑥𝑦′
𝑦2
4𝑦′′
=
−𝑦+𝑥
−𝑥
4𝑦
𝑦2
By diff.
4𝑦′′
=
−𝑦2−
𝑥2
4
𝑦3
𝑦′′
=
−1
4𝑦3
Solution

Exercise
i) Consider
∵ cos 2𝛽 = 1 − 2 sin2
𝛽
∴ 𝑦 = 1 − 2𝑥2
Solution
𝑦′
= −4 𝑥 → 𝑦′ 2
= 16𝑥2
𝑦′′
= −4
𝑦′ 2
𝑦′′
= −64𝑥2
= 32 −2𝑥2
= 32(𝑦 − 1)
𝑦 = cos 2
𝑡
1−𝑡
, 𝑥 = sin
𝑡
1−𝑡
. Prove
𝑦′ 2𝑦′′ = 32(𝑦 − 1)

48
Find
dy/dx
if:    .
tanh
sin
5 2
1
y
xy
xy

 
  y
h
y
y
x
x
yy
y
x
y
x
y
e x
y 2
4
2
2
ln
sec
'
1
1
'
2
5
ln
' 


















.
1
10
ln
sec
1
5
'
4
2
ln
2
4
2
2
ln
y
x
yx
x
e
y
h
y
x
y
e
x
y
y
x
y
x
y






ii) Find 𝑦’ for:

49
Introduction to integrtion
 In differentiation, the differential coefficient
𝑑𝑦
𝑑𝑥
indicates
that a function of x is being differentiated with respect
to x, the dx indicating that it is “with respect to x”.
 In integration, the variable of integration is shown by
adding d(the variable) after the function to be
integrated. When we want to integrate a function, we
use a special notation: 𝒇 𝒙 𝒅𝒙.
 Thus, to integrate 4x, we will write it as follows:
49

50
Introduction
4𝑥 𝒅𝒙 = 2x2
+ c , c ∈ ℝ.
Integral
sign
This term is called
the integrand
There must always be
a term of the form dx
Constant of
integration
50

51
Indefinite integral
 Note that along with the integral sign ( 𝑑𝑥), there is a term of the form dx,
which must always be written, and which indicates the variable involved, in
our example x.
 We say that 4x is integrated with respect x, i.e: 𝟒𝒙 𝒅𝒙
 The function being integrated is called the integrand.
 Technically, integrals of this type are called indefinite integrals, to
distinguish them from definite integrals, which we will deal with later.
 When you are required to evaluate an indefinite integral, your answer must
always include a constant of integration.; i.e:
𝟒𝒙 𝒅𝒙 = 2𝑥2 + c; where c ∈ ℝ
51

52
Definition of anti-derivative
 Formally, we define the anti-derivative
as: If f(x) is a continuous function and F(x)
is the function whose derivative is f(x), i.e.:
𝑭′
(x) = f(x) , then:
𝒇 𝒙 𝒅𝒙 = F(x) + c;
where c is any arbitrary constant.
52

53
Table of
INTEGRATION

54
i)
𝑑𝑥
𝑥 1+ln 𝑥 3 •
𝑓′𝑑𝑥
𝑓 𝑛 =
𝑓−𝑛+1
−𝑛+1
+ 𝑐
=
1+ln 𝑥 −2
−2
+c
ii)
1+tan 𝑥 3𝑑𝑥
1−𝑠𝑖𝑛2 𝑥
• 𝑓𝑛
𝑓′
𝑑𝑥 =
𝑓𝑛+1
𝑛+1
+ 𝑐
=
1+tan 𝑥 4
4
+c
=
1 + tan 𝑥 3𝑑𝑥
𝑐𝑜𝑠2 𝑥
= sec2 𝑥 1 + tan 𝑥 3𝑑𝑥
Examples. Find the following integrals:

55
iii)
1−2𝑥3 𝑑𝑥
2𝑥−𝑥4+1
•
𝑓′𝑑𝑥
𝑓
= 2 𝑓 + 𝑐
=
1
2
2 1−2𝑥3 𝑑𝑥
2𝑥−𝑥4+1
=
1
2
. 2 2𝑥 − 𝑥4 + 1 + 𝐶
•
𝑓′𝑑𝑥
𝑎2−𝑓2
= sin−1 𝑓
𝑎
+ 𝑐
iv)
𝑑𝑥
𝑥 4−9 ln2 𝑥
=
1
3
. sin−1 3 ln 𝑥
2
+ 𝐶
=
1
3
3𝑑𝑥
𝑥 4− 3 ln 𝑥 2

56
iv)
3+𝑥 𝑑𝑥
𝑥−5
=
𝑥−5 +3+5
𝑥−5
𝑑𝑥
= 1 +
8
𝑥−5
𝑑𝑥
= 𝑥 + 8 ln |𝑥 − 5| + 𝑐
v)
𝑑𝑥
sin−1 𝑥 1−𝑥2
•
𝑓′𝑑𝑥
𝑓
= ln |𝑓| + 𝑐
= ln | sin−1
𝑥 | + 𝑐

57
vi)
1+sin 𝑥 𝑑𝑥
cos2𝑥
= sec2
𝑥 + tan 𝑥 sec 𝑥 𝑑𝑥
= tan 𝑥 + sec 𝑥 + 𝑐
vii)
1−cos 2𝑥 𝑑𝑥
sin2𝑥
=
1−cos 2𝑥 𝑑𝑥
1
2 1−cos 2𝑥
= 2 𝑑𝑥 = 2𝑥 + 𝑐.
• sin2 𝑥 = 1
2 1 − cos 2𝑥
• cos2
𝑥 = 1
2 1 + cos 2𝑥
Note:

58
viii) 1 + cos 𝑥 2
𝑑𝑥
= 1 + 2 cos 𝑥 + cos2
𝑥 𝑑𝑥
= 1 + 2 cos 𝑥 + 1
2 1 + cos 2𝑥 𝑑𝑥
= 𝑥 + 2 sin 𝑥 + 1
2 𝑥 +
sin 2𝑥
2
+ 𝑐
ix) cos3
𝑥 𝑑𝑥
= cos2
𝑥 cos 𝑥 𝑑𝑥
= 1 − sin2
𝑥 cos 𝑥 𝑑𝑥
= cos 𝑥 − sin2
𝑥 cos 𝑥 𝑑𝑥 = sin 𝑥 −
sin3𝑥
3
+ 𝑐

59
x) cos 𝑥 + sin 𝑥 2
𝑑𝑥
= cos2
𝑥 + sin2
𝑥 + 2 sin 𝑥 cos 𝑥 𝑑𝑥
= 1 + sin 2𝑥 𝑑𝑥
= 𝑥 −
cos 2𝑥
2
+ 𝑐

61
xiii)
sin 𝑥 cos 𝑥
sin4 𝑥+ cos4 𝑥
𝑑𝑥
÷ cos4
𝑥
=
sin 𝑥
cos3𝑥
tan4 𝑥+1
𝑑𝑥
=
1
2
2 tan 𝑥sec2𝑥
tan2𝑥 2+1
𝑑𝑥
=
1
2
tan−1
tan2
𝑥 + 𝑐
•
𝑓′𝑑𝑥
𝑎2+𝑓2 =
1
𝑎
tan−1 𝑓
𝑎
+ 𝑐

62
xiv)
sec 𝑥 cos 2𝑥
sin 𝑥+sec 𝑥
𝑑𝑥
÷ sec 𝑥
=
cos 2𝑥
sin 𝑥 cos 𝑥+1
𝑑𝑥
=
cos 2𝑥
sin 2𝑥
2+1
𝑑𝑥
= ln |
sin 2𝑥
2
+ 1| + 𝑐

63
xv)
sec 𝑥+ tan 𝑥
sec 𝑥− tan 𝑥
𝑑𝑥
=
sec 𝑥+ tan 𝑥
sec 𝑥− tan 𝑥
.
sec 𝑥+ tan 𝑥
sec 𝑥+ tan 𝑥
𝑑𝑥
=
sec 𝑥+ tan 𝑥 2
sec2 𝑥−tan2 𝑥
𝑑𝑥
= sec2
𝑥 + sec2
𝑥 − 1 + 2 sec 𝑥 tan 𝑥 𝑑𝑥
tan2
𝑥
= 2 tan 𝑥 − 𝑥 + 2 sec 𝑥 + 𝑐.
 𝑡𝑎𝑛2𝑥 + 1 = 𝑠𝑒𝑐2𝑥

64

65

66

THANK YOU
@Sinaiunieg www.su.edu.eg
info@su.edu.eg

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