Chapter Seven
7
Transcendental Functions
TRANSCENDENTAL FUNCTIONS
Inverse Functions and their Derivatives:
DEFINITION: One-to-One Function:
A function f(x) is one-to-one in a domain D if f(x1) ≠ f(x2) whenever x1 ≠ x2 in D.
Example 1:
(a) f ( x)  x is one-to-one on any domain in non-negative numbers because
x1  x2 whenever x1 ≠ x2.
(b) g(x) = sin x is not one-to-one function on the interval [0, ] because for example
sin( /6) = sin( /6). The sine is one-to-one on [0, ], however, it is a strictly
increasing function on [0, 
The Horizontal Line Test for One-to-One Functions
A function y=f(x) is one-to-one if and only if its graph intersects each horizontal line
at most once.
Example 2: Using the horizontal test we see that
(a) y  x 3 and y  x are one-to-one on their
domains (-∞, ∞) and [0, ∞).
(b) y  x 2 and y  sin x are not one-to-one
on their domains (-∞, ∞).
Note: The function defined by reversing a
one-to-one function f is called the inverse
of f. The symbol for the inverse f -1, read "f
inverse". The -1 in f
-1
is not exponent:
f -1(x) does not mean 1/f(x).
University of Kufa\Civil Eng. ………………
(170)
……………………. Mathematics \1st class
Chapter Seven
To express f -1as a function of x
Transcendental Functions
1. Solve the equation y = f(x) for x in terms of y.
2. Switch x and y. The resulting formula will be y = f -1(x).
Features of the inverse functions:
1.
f o f 1 ( x)  x and
f 1o f ( x)  x
2. Domain of f -1 = Range of f
Range of f -1 = Domain of f
and
3. The graphs of the function and its inverse are interchanged by reflection
through the line y = x.
1
2
Example 3: Find the inverse of y  x  1, expressed as a function of x.
y
Sol.: 1. Solve for x in terms of y:
1
x 1
2
2y  x  2
x  2y  2
2. Interchange x and y:
y  2x  2
1
2
The inverse of the function f ( x)  x  1 is the
function f 1 ( x)  2 x  2 . To check, we verify that both
composites give the identity function:
f o f 1 ( x)  f ( f 1 ( x)) 
1
(2 x  2)  1  x  1  1  x o.k.
2
1
2
and f 1o f ( x)  f 1 ( f ( x))  2( x  1)  2  x  2  2  x o.k.
Example 4: Find the inverse of y  x 2 , x  0 , expressed as
a function of x.
Sol.: 1. Solve for x in terms of y:
y  x2
y  x 2  x  x ( x  x because x  0 )
2. Interchange x and y:
y
x
The inverse of the function f ( x)  x 2 , x  0 is the
University of Kufa\Civil Eng. ………………
(171)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
1
function f ( x)  x . To check, we verify that both composites give the identity
function:
f o f 1 ( x)  f ( f 1 ( x))  ( x )2  x o.k.
and f 1o f ( x)  f 1 ( f ( x))  x 2  x o.k.
Note: Unlike the restricted function y  x 2 , x  0 , the unrestricted function y  x 2 is
not one-to-one and therefore has no inverse.
Example 5: Find the inverse of y  3x  2 , expressed as a function of x.
Sol.: 1. Solve for x in terms of y:
y  3x  2
y 2  3x  2
y2 2
 x
3 3
y
2. Interchange x and y:
x2 2

3 3
The inverse of the function f ( x)  3x  2 is the function f 1 ( x) 
x2 2
 . To
3 3
check, we verify that both composites give the identity function:
f o f 1 ( x)  f ( f 1 ( x))  3(
and f 1o f ( x)  f 1 ( f ( x)) 
x2 2
 )  2  x2  2  2  x2  x
3 3
o.k.
( 3x  2 ) 2 2 3x  2 2 3x 2 2
 
 
   x o.k.
3
3
3
3 3 3 3
The range and domain of the original function will be calculated and reflected
for the inverse function:
2
3
The domain of f is: 3x  2  0  x  .
The range of f is:
So,
y  0.
the domain of f -1 is: x  0 .
2
3
The range of f -1 is: y  .
University of Kufa\Civil Eng. ………………
(172)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Derivatives of Inverse Differentiable Functions:
Theorem: The Derivative Rule for Inverses
If f has an interval I as domain and f '(x) exists and is never zero on I, then f
-1
is
df 1
differentiable at every point in its domain. The value of
at a point b in the
dx
domain of f -1 is the reciprocal of the value of f ` at the point a=f -1(b):
df 1
dx

x b  f ( a )
1
df
dx x  a
…(1)
Example 6: Verify Eq.(1) for f(x) = x2, x ≥ 0 and its
inverse f 1 ( x)  x at the point x = 2 in
the domain of f.
Sol.: at x = 2  f(2) = 22 = 4
df 1
dx
df
dx

x f ( a )

x 2
Thus
d
x
dx
d 2
x
dx
df 1
dx

x4
1
2 x

x 4
1
2 4

1
1

2*2 4
2 x x  2  2 * 2  4
x 2

x f ( 2)
1
df
dx
x 2
University of Kufa\Civil Eng. ………………
(173)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
1
df
at x = 6 = f(2) without finding a
dx
Example 7: Let f ( x)  x3  2. Find the value of
formula for f -1(x).
Sol.:
df
d
 ( x3  2) 3x 2 x2  3(2) 2  3 * 4  12
dx x2 dx
x 2
df 1
Thus
dx

x  f ( 2)  6
1
df
dx x  2

1
.
12
Homework
1. Find a formula for the inverse of the function and show that f(f-1(x))=f-1(f(x)).
4x 1
2x  3
a. y  10  3 x
b. y 
2
;x  0
x2
g. y  ( x  1) 2 ; x  1
e. y  x 2  2 x ; x  0
d. y  1 
h. y  x 2 3 ; x  0
c. y  2 x 3  3
1
;x  0
x3
i. y  1 2x  7 2
f. y 
2. Find a formula for the inverse of the function and verify that
evaluating df dx at x=a and df
a. f(x) =2x + 3, a = -1
c. f(x) =(1/5)x + 7, a = -1
1
df 1
dx

x f (a)
1
df
dx
by
xa
dx at x=f(a).
b. f(x) =5-4x , a = 1/2
d. f(x) =2x2, x>0, a = 5
University of Kufa\Civil Eng. ………………
(174)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
1. Logarithmic Functions:
The Natural Logarithm of positive
number x, written as ln x, is the value of integral:
x
1
ln x   dt ; x > 0
t
1
We can see from the figure

ln x  0
if
x>1

ln x  0
if
x=1
 ln x  0
if
0 < x < 1.
Definition
The number e
The number e is that the number in the domain of the natural logarithm satisfy
ln(e) = 1
Geometrically, the number e corresponds to the point on the x-axis for which the area
under graph of (y=1/t) and above the interval [1, e] is the exact area of the unit square.
Rules of arithmetic for logarithms
For any positive numbers a and b and for any exponent n,
a
b
2. ln  ln a  ln b
1. ln ab  ln a  ln b
1
b
3. ln   ln b
4. ln a n  n ln a
(Rule 2 with a=1)
1
n
5. ln n a  ln a
The derivative of y = lnx
By the First Fundamental Theorem of Calculus
x
d
d 1
1
ln x 
dt 

dx
dx 1 t
x
If u is a differentiable function of x whose values are positive,
d
1 du
ln u  .
dx
u dx
University of Kufa\Civil Eng. ………………
(175)
……………………. Mathematics \1st class
Chapter Seven
Examples: Find
Transcendental Functions
dy
for the following functions:
dx
1. y  ln( 3x 2  4)
Sol.:
dy
1
6x
 2
* (2 * 3x)  2
dx 3x  4
3x  4
2. y  ln 3 x  (ln x)3
Sol.:
dy
1
 3(ln x) 2 *
dx
x
3. y  ln x3  3 ln x
Sol.:
dy 3

dx x
Sol.:
dy 3
1
3
 *
( sin x) 
tan x
dx 2 cos x
2
3
4. y  ln cos3 x  ln cos 2 x 
3
ln cos x
2
5. y  x[sin(ln x)  cos(ln x)]
Sol.:
dy
1
1
 x[cos(ln x).  sin(ln x). ]  [sin(ln x)  cos(ln x)]
dx
x
x
 cos(ln x)  sin(ln x)  sin(ln x)  cos(ln x)
 2 cos(ln x)
6. y  ln
x x2
( x  2) 2
Sol.: Simplify the right-hand side
y  ln x  ln x  2  ln( x  2) 2

1
y  ln x  ln( x  2)  2 ln( x  2)
2
3
y  ln x  ln( x  2)
2
Differentiate:
dy 1 3
1
  *
dx x 2 x  2
dy 1
3
 
dx x 2( x  2)
Logarithmic Differentiation:
The derivatives of positive function given by formulas that involve products,
quotients and powers can often be found more quickly if we take natural logarithm of
both sides before differentiating. This enables us to use laws of logarithms to simplify
the formulas before differentiating. The process is called logarithmic differentiation.
University of Kufa\Civil Eng. ………………
(176)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Examples: Find
1. y 
dy
for the following functions:
dx
( x 2  1)( x  3)1 2
x 1
Sol.: We take the natural logarithm of both sides and simplify the result with
properties of logarithms:
y
( x 2  1)( x  3)1 2
x 1
 ( x 2  1)( x  3)1 2 
ln y  ln 
 ln[( x 2  1)( x  3)1 2 ]  ln( x  1)

x 1


 ln( x 2  1)  ln( x  3)1 2  ln( x  1)
1
 ln( x 2  1)  ln( x  3)  ln( x  1)
2
Then we differentiate by implicit differentiation
1 dy
2x
1
1
1
*  2
 *

y dx x  1 2 x  3 x  1
dy
1
1 
 2x
 y 2


dx
 x  1 2 x  6 x  1
dy  ( x 2  1)( x  3)1 2   2 x
1
1 




2


dx 
x 1
  x  1 2 x  6 x  1
2. y 2 3 
( x 2  1)(3x  4)1 2
5
(2 x  3)( x 2  4)
Sol.: We take the natural logarithm of both sides and simplify the result with
properties of logarithms:
ln y
23
 ( x 2  1)(3x  4)1 2 
 ln 

2
5
 (2 x  3)( x  4) 
2
ln y  ln[( x 2  1)(3x  4)1 2 ]  ln 5 (2 x  3)( x 2  4)
3
2
1
ln y  ln( x 2  1)  ln( 3x  4)1 2  ln[( 2 x  3)( x 2  4)]
3
5
2
1
1
1
ln y  ln( x 2  1)  ln( 3x  4)  ln( 2 x  3)  ln( x 2  4)
3
2
5
5
Then we differentiate by implicit differentiation
University of Kufa\Civil Eng. ………………
(177)
……………………. Mathematics \1st class
Chapter Seven
2 1 dy
2x
1
3
1
2
1
2x
* *
 2
 *
 *
 * 2
3 y dx x  1 2 3x  4 5 2 x  3 5 x  4

Transcendental Functions
dy 3  2 x
1
3
1
2
1
2x 
 y 2
 *
 *
 * 2
dx 2  x  1 2 3 x  4 5 2 x  3 5 x  4 
32
dy 3  ( x 2  1)(3 x  4)1 2   2 x
3
2
2x 
 


 2
  2
2
5
dx 2  (2 x  3)( x  4)   x  1 6 x  8 10 x  15 5 x  20 
x( x  1)( x  2)
( x 2  1)( 2 x  3)
3. y  3
Sol.: We take the natural logarithm of both sides and simplify the result with
properties of logarithms:
ln y  ln 3
x( x  1)( x  2)
( x 2  1)( 2 x  3)
1 x( x  1)( x  2)
ln y  ln 2
3 ( x  1)( 2 x  3)
ln y 

1
ln x  ln( x  1)  ln( x  2)  ln( x 2  1)  ln( 2 x  3)
3

Then we differentiate by implicit differentiation
1 dy 1  1
1
1
2x
2 
*
  

 2

y dx 3  x x  1 x  2 x  1 2 x  3 
dy y  1
1
1
2x
2 
  

 2

dx 3  x x  1 x  2 x  1 2 x  3 
dy 1 x( x  1)( x  2)  1
1
1
2x
2 
 3 2


 2


dx 3 ( x  1)( 2 x  3)  x x  1 x  2 x  1 2 x  3 
The Integral
1
 u du
If u is a differentiable function that is never zero,
1
 u du  ln u  C
Examples: Evaluate the following integrals:
1.
 3x
x
dx
4
2
Sol.: Let u  3 x 2  4  du  6x.dx
University of Kufa\Civil Eng. ………………
(178)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
 x.dx 
So
2
2.
x
0
du
6
xdx
1 du 1 du 1
1
 *
 
 ln u  C  ln 3x 2  4  C
2
4
u 6
6 u
6
6
 3x
2x
dx
5
2
Sol.: Let u  x 2  5  du  2x.dx
2x.dx  du
When x  0  u(0)  (0)2  5  5
And
x  2  u(2)  (2)2  5  1
1
2
So
 2
2x
du
0 x 2  5 dx  5 u  ln u
1
5
 ln  1  ln  5  ln 1  ln 5   ln 5
4 cos
3. 
d
 2 3  2 sin 
Sol.: Let u  3 2 sin   du  2 cosd
 4 cosd  2.du
When   
And  


 u ( )  3  2 sin(  )  3  2  1
2
2
2



 u ( )  3  2 sin( )  3  2  5
2
2
2

5
5
4 cos
2.du
d


 3  2 sin 
 u  2 ln u 1  2[ln 5  ln 1 ]  2 ln 5
 2
1
 2
So
4.

5x  2
dx
x 1
Sol.: For rational functions when the degree of denominator is
5
equal or greater the degree of numerator, us the long x  1 5 x  2
division to simplify the integral.
 5x  5
3
5x  2
3 

 x  1 dx    5  x  1 .dx  5x  3 ln x  1  C
5.
x2  2x  2
 x  2 dx
x
x  2 x2  2x  2
x 2  2x  2
2 
x2

Sol.: 
dx    x 
 2 ln x  2  C
.dx 
x2
x2
2

University of Kufa\Civil Eng. ………………
(179)
 x2  2x
2
……………………. Mathematics \1st class
Chapter Seven
dx
6. 
x ln x
Transcendental Functions
Sol.: Let u  ln x  du 

7.

dx
x
dx
du

 ln u  C  ln ln x  C
x ln x
u
dx
x (1  x )
Sol.: Let u  x  u 2  x  2udu  dx

dx
2udu
2du
du


 2
u (1  u )
(1  u )
(1  u )
x (1  x )
Let z  1  u  dz  du  du  dz
 2
du
 dz
 2
 2 ln z  C  2 ln 1  u  C  2 ln 1  x  C
(1  u )
z
The Integral of tan x, cot x, secx and cscx:
Examples: Evaluate the following integrals,
1.  tan x.dx  
sin x
dx
cos x
Let u  cos x  du   sin xdx or sin xdx  du
sin x
 cos x dx  
 du
  ln u  C   ln cos x  C
u
 ln (cos x) 1  C  ln
2.  cot x.dx  
1
 C  ln sec x  C
cos x
cos x
dx
sin x
Let u  sin x  du  cos x
cos x
 sin x dx  
du
 ln u  C  ln sin x  C
u
3.  sec x.dx   sec x.dx *
 sec x.dx *
Let
sec x  tan x
sec x  tan x
(multiply and divide by secx + tanx)
sec x  tan x
sec 2 x  sec x tan x

.dx
sec x  tan x
sec x  tan x
u  sec x  tan x  du  (sec x tan x  sec 2 x).dx
(sec x tan x  sec 2 x).dx
du

 ln u  C  ln sec x  tan x  C

sec x  tan x
u
University of Kufa\Civil Eng. ………………
(180)
……………………. Mathematics \1st class
Chapter Seven
  sec x.dx  ln sec x  tan x  C
Transcendental Functions
And in the same manner we can show that  csc x.dx   ln csc x  cot x  C
So
 tan udu   ln cos x  ln sec u  C
 cot udu  ln sin u  C   ln csc u  C
 sec u.du  ln sec u  tan u  C
 csc u.du   ln csc u  cot u  C
 6
Example: Evaluate
 tan 2 xdx
0
Sol.: Let u  2 x  du  2dx or dx 
du
2
When x  0  u (0)  2 * 0  0
x


 
 u( )  2 * 
6
6 3
6
 6
 3
 tan 2 xdx  
0
0

du 1
tan u.
 ln sec u
2 2
 3

0

1

ln sec  ln sec 0 

2
3

1
ln 2  ln 1  1 ln 2
2
2
The Graph of Natural Logarithmic Function (y = lnx)
1. Domain:
Df = (0, ∞)
2. The first derivative test:
d
1
1
(ln x )  , since x > 0 so  0 too
dx
x
x
So lnx is an increasing function on its domain and has
y
no critical function.
2
3. The second derivative test:
1
d2
d 1
1
(ln x) 
( )   2  0 (is negative for all x),
2
dx
dx x
x
y=lnx
(1,0)
0
-1
0
1
x
2
3
-1
so lnx is concave down on its domain.
4. Asymptotes:
-2
lim ln x   , so there is no horizontal asymptote
x 
lim ln x   , so x = 0 is vertical asymptote.
x 0 
University of Kufa\Civil Eng. ………………
(181)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
5. Symmetry: since x > 0 the functions has no symmetry.
6. additional points on the graph:
at x = 1  lnx = 0,  (1,0) is x- intercept.
Thus the Range: Rf = (-∞, ∞).
Examples: Graph the following functions:
y
2
y=ln2x
1. y =ln2x
y=lnx
1
Sol.: Shrinking the graph of (y = lnx) two units along
x-axis
-1.0
-0.5
Df = (0, ∞), and Rf = (-∞, ∞).
0
0.0
(0.5,0) (1,0)
0.5
1.0
x
1.5
2.0
2.5
3.0
-1
-2
2. y = ln(x+1)
y
x=-1
2
Sol.: Shifting the graph of (y = lnx) one unit left
y=ln(x+1)
Df = (-1, ∞), and Rf = (-∞, ∞)
y=lnx
1
-2.0
0
0.0
-1.0
(1,0)
(0,0)
1.0
x
2.0
3.0
4.0
-1
-2
y
3. y = ln|x|
2
 ln x
ln(  x)
Sol.: y  ln x  
when x  0
y=ln(-x)
when x  0
y=lnx
1
(-1,0)
(1,0)
0
-3
-2
-1
0
1
x
2
3
4
Note: ln(-x) can be obtained by reflecting ln(x)
-1
across y-axis
-2
Df = R\{0}, and Rf = (-∞, ∞)
y
4
4. y =|lnx|
y=-lnx
3
 ln x
 ln x
Sol.: y  ln x  
when ln x  0
when ln x  0
x 1
x 1
2
y=lnx
1
(1,0)
0
Df = (0, ∞), and Rf = [0, ∞).
-1
0
1
2
x
3
4
5
6
7
8
-1
-2
University of Kufa\Civil Eng. ………………
(182)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Homework
I. Find dy/dx of the following:
1. y  ln 2 x
2. y  (ln x) 2
3. y  ln(tan x)
4. y  1 ln 2 x
5. y  sin 2 (ln x)
6. y  ln(ln x)
x 1
x 1
9. y  ln xy  1
7. y  ln
cos x
4  3x 2
10. y  ln( x tan y )
13. y 
8. y  ln
11. y  x.3 1  x 2
12. y 
sin x cos x tan 3 x
14. 3 y  x 2 ln xy  2
x
( x 2  8)1 3 x 3  1
x6  7x  5
15. ln x ln y  xy  1
II. Evaluate the following integrals:
1.
x
2
x
dx
1
2.
1
1
dx
4. 
2x  7
2
7.
(2  ln x) 3
 x dx
1
 8 x  3 dx
4
5.

1
8.
3.
2
dx
x ( x  4)
ln x
 x dx
0
5.
7 x 2  3x  1
 2 x dx
11.
1
 x ln 2 x dx
13.
sin x
 2  cos x dx
14.
sec 2 3x
 1  tan 3x dx
2
x2
dx
 4x  9
1
 4  5 x dx
1
9.
10.
x
2 x 2  5x  7
 x  3 dx
12.
( x 2  4) 2
 2 x dx
III. Graph the following functions:
1. y  ln x  3
2. y  ln( x  1)  5
3. y  ln x  2
4. y  ln( x  3)
5. y  ln( x  1) 2
6. y  2 ln( x  1)
6. y  ln x  1
University of Kufa\Civil Eng. ………………
(183)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
2. The Exponential Function:
Definition
The Natural Exponential Function
For every real number x, ex=ln-1x =exp(x)
(e = 2.718281828459045)
Note1: Because ex is the inverse of lnx, its graph can be
obtained by reflecting the graph of lnx across the line y = x .
So lim e x   and lim e x  0
x

x
The domain of ex is the range of ln x = (-∞, ∞)
And the range of ex is the domain of ln x = (0, ∞)
Inverse Equations for ex and ln x
e ln x  x (for all x > 0)
ln( e x )  x (for all x  R )
Note 2:
 To remove logarithms from an equation, take the exponential of both sides.
 To remove exponential from an equation, take the logarithm of both sides.
Laws of Exponents
For all real numbers a and b, the natural exponential obeys the following laws:
2. e a 
1. e a .eb  e ab
3.
ea
 e a b
eb
1
ea
4. (e a )b  e a.b  (eb ) a
To proof of the first law:
let
y1  e a
and y2  e b
Then ln y1  ln e a  a and ln y2  ln eb  b (take the logarithm of both sides)
Add the two equations: ln y1  ln y2  a  b

ln( y1. y2 )  a  b

e ln( y1 . y2 )  e a b

y1. y2  e ab
So
e a .e b  e a b
University of Kufa\Civil Eng. ………………
(exponentiate for both sides)
(184)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Examples: Solve for x the following:
1. ln (x+2)+ln (x-3)=ln 6
Sol.: ln (x+2)(x-3)=ln 6
Exponentiate both sides:
eln(x+2)(x-3) = eln 6  (x+2)(x-3)=6  x2-3x+2x-6=6  x2-x-12=0
 (x-4)(x+3)=0
So either x=4
x = -3 neglect (don’t satisfy the equation)
or
2. e2x - 3ex-10 = 0
Sol.: (ex-5)( ex+2)=0
So either ex=5
or
 x=ln 5
ex = -2 impossible
The Derivative of ex:
Let y = ex
 ln y = x (take
the logarithm of both sides)
d
dy
1 dy
(ln y )  1 
 y  ex
. 1 
dx
dx
y dx
d x
e  ex
dx

In general:
d u
du
e  eu .
dx
dx
where u is a differential function of x.
Examples 1: Find dy/dx of the following functions:
a. y  e x
Sol.:
dy
 e  x (1)  e  x
dx
b. y  e x
Sol.:
2
2
2
dy
 e x (2 x)  2 xex
dx
c. y  esin x
Sol.:
dy
 esin x cos x  cos xesin x
dx
University of Kufa\Civil Eng. ………………
(185)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
2
2
Examples 2: Find d y/dx of the following functions:
a. y  e  x ln x
Sol.:
dy
1
1
 e  x .  e  x ln x  e  x (  ln x)
dx
x
x
d2y
1 1
1
1 2
 e  x ( 2  )  e  x (  ln x)  e  x ( 2   ln x)
2
dx
x
x
x
x
x

e x
(1  2 x  x 2 ln x)
2
x
b. y  e2 x sin 3x
Sol.:
dy
 e 2 x (3 cos 3x)  2e 2 x sin 3x  e 2 x (3 cos 3x  2 sin 3x)
dx
d2y
 e 2 x (9 sin 3x  6 cos 3x)  2e 2 x (3 cos 3x  2 sin 3x)
2
dx
 e2 x (9 sin 3x  6 cos 3x  6 cos 3x  4 sin 3x)
 e2 x (5 sin 3x  12 cos 3x)  e2 x (5 sin 3x  12 cos 3x)
The Integral of ex:
 e .du  e
u
u
C
Examples: Evaluate the following integrals:
1.  e3 x dx
Sol.: let u  e3 x  du  3e3 x dx
So
du
3

 e3 x dx 

 dx 

cos xdx  du
du u
e3 x


C

C
3 3
3
Another solution:
Let u  3x  du  3dx
So  eu
du
3
du e u
e3x

C 
C
3
3
3
2.  e sin x . cos xdx
Sol.: let u  sin x  du  cos xdx
So  eu du  eu  C  esin x  C
University of Kufa\Civil Eng. ………………
(186)
……………………. Mathematics \1st class
Chapter Seven
3.  e3 x1dx
Transcendental Functions
Sol.: let u  3x 1  du  3dx
So  eu
du
3

 dx 

 xdx 
du
2

 xdx 
du
2
du eu
e3 x 1
 C 
C
3
3
3
4.  xex 3dx
2
Sol.: let u  x 2  3  du  2xdx
du eu
e x 3
 C 
C
2
2
2
2
So  eu
5.
ex
 e x  1 dx
Sol.: let u  e x  1  du  e x dx
So
du
 ln u  C  ln e x  1  C
u

6.  e xe dx   e x * ee dx
x
x
Sol.: let u  e x  du  e x dx
So  eu du  eu  C  ee  C
x
Homework:
I. Find dy/dx of the following:
1. y  e5x
2
2. y  e
15 x3
5. y  ln cos e x
6. y  ln( 1  xe x )
9. y  sin( e x )
II. Evaluate the following integrals:
ex
dx
2. 
1  ex
1.  e 2 x dx
y .e
y
6.
dx
0 x  e
ln 3
9.
ex
 e x  4 dx
ln 3
e6
13.

e6
36  ln x
dx
x
2
10.
 xe
 x2
3.  x e dx
7.
dx
ex
ln x
3x
8. y  e xe
4. y 
3 x4
e
dy
5. 
e x  e x
e x  ex
7. y  e x tan x
3. y 
e x  e x
 e x  ex dx
11.  e 2 ln x dx
y 1
e
4.

8.
e
y 1
dy
dx
x
12.  [ln( e x )  ln( e  x )]dx
1
14 . 
sec xdx
ln(sec x  tan x)
University of Kufa\Civil Eng. ………………
(187)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Example 1: Generate or sketch the graph of y  e x 2 .
2
Sol.:
1. Domain: Df = (-∞, ∞).
2. Symmetry: f ( x)  e(  x )
2
 f ( x )  e  x
2
2
 ex
2
 f ( x) not o.k .
2
2
 f ( x) o.k .
So it is an even function (It has symmetry about y-axis).
3. x and y-intercept:
When x = 0  y  e0 2  e0  1
2
Assume y = 0  e x
2
2
0
 (0,1) is y-intercept.
So there is no x-intercept.
4. Asymptotes:
- Horizontal asymptotes:
lim f ( x)  lim e  x
x
2
2
x
lim f ( x)  lim e  x
x
 lim e 
2
2
 e   0
 e  (  )
2
2
 e   0
x
2
x
2
 y  0 (x-axis) is horizontal asymptote
- Vertical asymptotes:
There is no vertical asymptote because that f (x)   at x=a.
5. First derivative test:
2
2
dy
 2x
 ex 2 *
  xe x 2
dx
2
Put
2
2
dy
 0   xe x 2  0 since e  x 2  0
dx
 at x = 0 there is a critical point.
Note: Sign of  xe x 2 depend only on
2
(-x) because that e  x
2
2
 0 for all x.
So (0,1) is max. point.
Sign of(-x)
2
Sign of ( e x 2 ) -∞
Sign of y`
0-----------------+∞
0
++++++++++++++++++++++ +∞
0
+∞
-------------++++++
-∞ ++++++++
-∞
increase
The function increases on (-∞,0].
max. decrease
at x=0
The function decreases on [0,∞).
University of Kufa\Civil Eng. ………………
(188)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
6. Second derivative test:


2
2
2
d2y
  x  xe x 2  e  x 2  ( x 2  1)e  x 2
2
dx
2
d2y
Put 2  0  ( x 2  1)e  x 2  0
dx
since e  x
2
2
0 
( x 2  1)  0
 ( x  1)( x  1)  0
When x = -1  y  e ( 1)
and x = 1  y  e (1)
2
2
2
2
 e 1 2 
 e 1 2 
+1
++++++ +∞
-----------------Sign of (x-1) -∞
-1
Sign of (x+1) -∞ ------+++++++++++++++ +∞
+1
2
-1
Sign of e x 2 -∞ ++++++++++++++++++++++ +∞
-1
+1
++++++
---------- +++++++ +∞
∞
Sign of y``
1
e1 2
Concave up
I.P.
at x=-1
Concave down
1
e1 2
1
So the curve has inflection points at (-1, 1 2 )
e
1
and (1, 1 2 )
e
It concaves up on (-∞,-1) and (1, ∞),
(-1,
1
12
(0,1)
)
I.P.
at x=1
Concave up
y  e x
(1,
e
1
12
2
2
)
e
And concaves down on (-1, 1)
So R f  (0,1]
Example 2: Generate or sketch the graph of y 
ln x
.
x
Sol.:
1. Domain: Df = (0, ∞).
2. Symmetry: Because x > 0 for all x, the function has no symmetry.
3. x and y-intercept:
When x ≠ 0  So there is no y-intercept
Assume y = 0 
ln x
 0  ln x  0  x  e0  1  (1,0) is x-intercept.
x
4. Asymptotes:
a. Horizontal asymptotes:
ln x
1x
1
 lim
 lim  0
x x
x 1
x x
lim f ( x)  lim
x
lim f ( x) impossible because x > 0.
x
 y  0 (x-axis) is horizontal asymptote.
University of Kufa\Civil Eng. ………………
(189)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
b. Vertical asymptotes:
lim f ( x)  lim
xo
xo
ln x  

 
x
0
 x  0 (y-axis) is vertical asymptote.
5. First derivative test:
dy  ln x 1 1 1

 *  2 (1  ln x)
dx
x2
x x x
dy
0 
Put
dx
1
(1  ln x)  0
x2
 (1  ln x)  0

ln x  1
0
e
++++ --------------- +∞
--0
+++++++++++ +∞
0
e
++++ --------------+∞
Sign of(1-lnx) -∞
Sign of ( 1

Sign of y`
x2
)-∞
-∞
x  e1  e

increase
decrease
max.
at x=e
at x = e there is a critical point.
y 
ln e 1
  e 1
e
e
So (e, e-1) is max. point.
The function increases on (0, e]
The function decreases on [e, ∞)
6. Second derivative test:
d2y 2
1 1
 3 (1  ln x)  3  3 (2  2 ln x  1)
2
dx
x
x
x
2 ln x  3
x3

d y
2 ln x  3
0
Put 2  0 
x3
dx
2
Concave down
at
(e,1/e) (e
(1,0)
 y
3
2
32
,
2e
x  e3 2
Concave up
3
)
32
 x  e3 2
ln e3 2 3 2
 32
e3 2
e
Sign of y`` depends on (2lnx-3)
So the curve has inflection
University of Kufa\Civil Eng. ………………
+∞
I.P.
since x 3  0  2 ln x  3  0
 ln x 
0 ------------- e3 2 ++++++
Sign of (2lnx-3)-∞
(190)
……………………. Mathematics \1st class
Chapter Seven
points at ( e3 2 ,
Transcendental Functions
3
)
2e 3 2
1
e
It concaves down on (0, e3 2 ) and concaves up on ( e3 2 , ∞).  R f  (, ] .
Homework: Sketch the graph of the following functions:
2. y 
ex
x
3. y  x 2e2 x
4. y  x ln x
5. y 
ln x
x2
6. y  x 2 ln( 2 x)
7. y  x ln 2 x
8. y  x 2e x
1. y  xex
2
3. Other Exponential and Logarithmic Function
a. The Function ax
If a is a positive number and x is any number, we define the function ax "a to
the x" by the equation:
ax = exlna
Assume y = ax
 lny = lnax
"take the logarithm of both sides"
lny = xlna
elny= exlna
"exponentiate of both sides"
 y = exlna
Laws of Exponents
If a > 0 any x and y then:
1. a x .a y  a x  y
3.
ax
 a x y
y
a
2. a  x 
1
ax
4. (a x ) y  (a y ) x  a xy
The Derivative of ax:
Let y = ax = exlna

dy d x
d
 (a )  (e x ln a )  e x ln a * ln a  a x * ln a
dx dx
dx

d x
a  ln a * a x
dx
University of Kufa\Civil Eng. ………………
(191)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
In general:
d u
du
a  ln a * a u *
dx
dx
where u is a differential function of x.
Examples: Find dy/dx of the following:
1. y  3 x
Sol.:
dy
 3 x * ln 3(1)   ln 3 * 3 x
dx
2. y  3sin x
Sol.:
dy
 3sin x * ln 3 * (cos x)
dx
Another Solution:
y  3sin x
ln y  ln 3sin x "take the logarithm of both sides"
ln y  sin x * ln 3
1 dy
.  cos x * ln 3
y dx

dy
 y * cos x * ln 3
dx
 3sin x * cos x * ln 3
Other Power Functions:
Examples: Find dy/dx of the following:
1. y = xx ,
x>0
Sol.: Take the logarithm of both sides
lny = lnxx
lny = xlnx
Differentiate using implicit differentiation
1 dy
1
.  x *  ln x
y dx
x

dy
 1

 y  x *  ln x   x x 1  ln x
dx
 x

2. y = xlnx , x > 0
Sol.: ln y = (ln x) ln x  ln y = ln x* ln x  ln y = ln2 x
University of Kufa\Civil Eng. ………………
(192)
……………………. Mathematics \1st class
Chapter Seven
1 dy
1
.  2 ln x *
y dx
x

Transcendental Functions
dy
 2 ln x 
 ln x 
 y
 2 x ln x 

dx
 x 
 x 
3. y  (sin x) tan x
Sol.: ln y = tan x ln (sin x)
1 dy
cos x
.  tan x *
 sec 2 x ln sin x
y dx
sin x




dy
 y 1  sec 2 x ln sin x  sin x tan x 1  sec 2 x ln sin x
dx

4. y  ( x ) x
1
2
Sol.: ln y  x ln x  ln y  x ln x1 2  ln y  x ln x
1 dy 1  1
.

x.  ln
y dx 2  x


x

( x)x
dy y
1  ln x
 1  ln x  
dx 2
2
5. y  x ( x1)
Sol.: ln y  ( x  1) ln x
1 dy
1
.
 ( x  1).  ln x
y dx
x
dy
x 1
 y
 ln
dx
 x

x 1
x   x ( x 1) 
 ln

 x

x

The Integral of ax:
If a ≠ 1 so that lna ≠ 0, you know that:
d u
du
a  a u * ln a *
dx
dx

1  d u
du
a   au

ln a  dx 
dx
"divide both sides by lna"
Integrate of both sides:
 a
u
du 
1  d u
1  d u
dx  
a  dx 
a dx


dx 
ln a  dx 
ln a   dx 
University of Kufa\Civil Eng. ………………
(193)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
1
 a du  ln a a
u
u
C
Examples: Evaluate the following integrals:
1.
1
 2 .dx  ln 2 .2
x
x
C
Another solution:
 2 .dx   e
x
x ln 2
.dx 
1
1 x ln 2
1 x
e x ln 2 . ln 2.dx 
e
C 
2 C

ln 2
ln 2
ln 2
2.  2sin x cos x.dx
Sol.: let u= sinx  du= cosx.dx
1
 2 .du  ln 2 .2
u
u
C 
1 sin x
.2  C
ln 2
3.  3tan 7 x sec 2 7 x.dx
Sol.: let u= tan7x  du= sec2 7x* 7dx
 sec 2 7 x.dx 
du
1
1
1
du
7
1
1
u
u
tan 7 x
C 
* 3tan 7 x  C
 3 . 7  7 * ln 3 * 3  C  7 * ln 3 * 3
7 ln 3
4.  2 x cos 2 x.dx
Sol.: let u= 2x  du= ln2*2x dx
 2 x.dx 
 cos u.
du
ln 2
du sin u
sin( 2 x )

C 
C
ln 2 ln 2
ln 2
 sec3 x esin x 
sec3 x  esin x
5. 
.dx   

.dx   sec 2 x  esin x . cos x .dx  tan x  esin x  C

sec x
 sec x sec x 

6.

e x  e x
 e x  ex .dx
Sol.: let u = e x + e-x  du = (e x - e-x). dx

du
 ln | u | C  ln | e x  e  x | C
u
University of Kufa\Civil Eng. ………………
(194)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
7.  e2t 1  e2t .dt   e2t (1  e2t )1 2 .dt
Sol.: let u = 1 + e2t  du = 2e 2t. dt
 e 2t .dt 
12
u .
du
2
du 1 u 3 2
1 2u 3 2
1
 .
C  .
 C  (1  e 2t )3 2  C
2 2 32
2 3
3
8.  x4  x .dx
2
Sol.: let u  4 x  du  ln 4 * 4 x * (2x).dx
2
2
 x 4  x .dx  
2
du
2 ln 4
du
u
 4 x



C

C
 2 ln 4 2 ln 4
2 ln 4
2
b. Base a Logarithms:loga x = inverse of ax
 log a a x  x
and
where a > 0 and a ≠ 1
for all x
a loga x  x
for all x > 0
Evaluation of loga x:
Use a log x  x
a
So a log x  x "take the logarithm of both sides"
a
ln a loga x  ln x
log a x  ln a  ln x
 log a x 
ln x
ln a
Rules of arithmetic for base a logarithms:
For any positive numbers u and v and for any exponent n,
u
 log a u  log a v
v
1. log a u.v  log a u  log a v
2. log a
3. log a u n  n log a u
4. log a n u  log a u
University of Kufa\Civil Eng. ………………
1
n
(195)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Derivatives and integrals involving loga x:
If u is a positive differentiable function of x, then
d
d ln u
1 d
1 1 du
(log a u ) 
(
)
. (ln u ) 
.
dx
dx ln a
ln a dx
ln a u dx
d
1 1 du
(log a u ) 
.
dx
ln a u dx
Examples 1: Find dy/dx of the following functions:
a. y  log 10 (3x  1)
Sol.: y  log 10 (3x  1) 

ln( 3 x  1)
ln 10
dy
1
3

.
dx ln 10 3 x  1
b. y  log 5 x 2
Sol.:
ln x 2 2 ln x
y  log 5 x 

ln 5
ln 5
2

dy
2 1
2

.  
dx ln 5  x  x ln 5
c. y  log 10 e x
Sol.: y  log 10 e x 

ln e x
x

ln 10 ln 10
dy
1

dx ln 10
d. y  log 5 ( x 2  1) 2
Sol.: y  log 5 ( x 2  1) 2  2 log 5 ( x 2  1) 

2 ln( x 2  1)
ln 5
dy
2
2x
4x

. 2

dx ln 5 x  1 ln 5( x 2  1)
Examples 2: Evaluate the following integrals:
a.

log 2 x
.dx
x
Sol.:
log 2 x
ln x dx
1
dx
1 (ln x) 2
ln 2 x
.
dx

.

ln
x
.


C

C
 x
 ln 2 x ln 2  x ln 2 2
2 ln 2
University of Kufa\Civil Eng. ………………
(196)
……………………. Mathematics \1st class
Chapter Seven
log ( x  2)
.dx
b.  10
x2
Sol.:

Transcendental Functions
log 10 ( x  2)
ln( x  2) dx
.dx  
.
x2
ln 10 ( x  2)
Let u = ln(x+2)  du 
dx
( x  2)
1
1 u2
ln 2 ( x  2)
u.du 
. C 
C
ln 10 
ln 10 2
2 ln 10
The Graph of ax
If y = ax and both x and a are positive and a ≠ 1, then
 The domain and range of y = ax is the same as the domain and range of the
function y = ex.
 Df=(-∞, ∞) and Rf=(0, ∞)
 First derivative test
d x
a  ln a * a x
dx
So
if a > 1  lna*ax > 0  ax is an increasing function
if a < 1  lna*ax < 0  ax is a decreasing function
 Second derivative test
d2 x d
a  (ln a * a x )  ln 2 a * a x which is positive for all a.
2
dx
dx
So ax is a concave up function on its domain.
Examples 3: Sketch the graph of the following functions
a. y  2 x
b. y  2 x
Sol.: a. y  2 x
 Symmetry: f ( x)  2(  x )  2 x  f ( x)
 f ( x)  2 x  f ( x)
So the function has no symmetry.
1
1
 Asymptote: lim 2 x  2   ; lim 2 x  2     0,
2

x
x
so, y = 0 is horizontal asymptote
 a = 2 > 1  it is an increasing and concave up function.
University of Kufa\Civil Eng. ………………
(197)
……………………. Mathematics \1st class
Chapter Seven
b. y  2 x
Transcendental Functions
 Symmetry: f ( x)  2(  x )  2 x  f ( x)
 f ( x)  2 x  f ( x)
So the function has no symmetry.
y=2-x
 Asymptote:
y=2x
2 x  2  0 ; lim 2 x  2 (  )  2  ,
lim
x
x
so, y = 0 is horizontal asymptote
(0,1)
x
y2
x
 a=
1 1
 x    "x must be positive"
2
2
1
< 1  it is a decreasing and concave up function.
2
The Graph of logax
y=2x
The graph of y = logax can be obtained by
reflecting the graph of
y=x
y = ax across the line y =x.
Examples 4: Sketch the graph of the function y  log 2 x
y=log2x
Sol.:
log 2 x   , lim log 2 x  
So lim
x 
x 0 
Homework
I. Find dy/dx of the following functions:
1. y  7 x
2. y  8x 1
3. y  9
4. y  log 10 ( x 4  3x 2  1)
5. y  log 5 x 2  1
6. y  log 10
7. y  ( x  1) x
8. y  x . x
9. y  ln log 2 x
2
10. y  (3  sec x)
2
tan2 x
(1sin x )3
11. y  (1  cos x)
2
sin2 x
(1cos x )3
x
1  x2
2  5x3
(Hint: use logarithmic differentiation)
II. Evaluate the following integrals:
1.  103 x.dx
4.
(2 x  1)
 2 x .dx
3.  x(3 x ).dx
2.  55 x.dx
5.

1
3x
2x  4
University of Kufa\Civil Eng. ………………
2
.dx
(198)
6.  33 x1.dx
1
……………………. Mathematics \1st class
Chapter Seven
10 x
7. 
.dx
x
Transcendental Functions
8.  (3x  3 x ) 2 .dx
9.  4 x (4 x  1).dx
10.
log (10 x)
1 10 10x .dx
11.
2 log ( x  1)
0 x10 1 .dx
12.

13.
x 1
 x  x .dx
14.
e3 x  e 2 x
 e x  1 .dx
15.
ln 3 x  1
 x ln x  2 x .dx
9
10
2 log 2 ( x  1)
.dx
x

1
2
3
III. Find the limits of the following:
log 2 x
1. lim
x
2. lim 3x
3. lim 3 x
4. lim log 10 x
1
5. lim log 10  
x0
1
log 2  
6. lim
x
x0
x


x
 x
 x
IV. Solve for x the following:
1. 3log 7  2log 5  5log x
3
2
5
University of Kufa\Civil Eng. ………………
2. 8log 3  eln 5  x 2  7log
8
(199)
7
3x
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
L`Hopital`s Rule:
The forms 1∞, ∞0, and 00 (which are also indeterminate forms):
ln f ( x)  L ,
If: lim
x a
DEFINTION:
Then: lim f ( x)  e L
xa
Example: Show that lim (1  x)1 x  e .
x0
1
x
Sol.: Let f ( x)  (1  x)1 x  ln f ( x)  ln( 1  x)1 x  ln( 1  x) 
 lim ln f ( x)  lim
x 0
x 0
ln( 1  x)
x
ln( 1  x) 0
 (Indeterminate form)
x
0
1
1
ln( 1  x)
(1  x) (1  0) 1
 lim

 1
By L`Hopital`s Rule  lim
x0
x0
x
1
1
1
Therefore lim f ( x)  lim (1  x)1 x  e1  e
x0
o.k.
x0
Examples: Find the limits of the following:
(sec 3 2 x) cot 3 x  (sec 3 (2 * 0)) cot 3*0  1 (indeterminate form)
1. lim
x 0
2
2
Sol.: Let y  (sec 3 2x)cot 3x
2
 ln y  ln(sec 3 2 x) cot
2
3x
 ln(sec 2 x) 3 cot
2
3x
 3 cot 2 3 x ln sec 2 x 
3 ln sec 2 x
tan 2 3x
3 ln sec 2 x 3 ln sec 2 * 0 3 ln 1 0



(also indeterminate form)
x 0 tan 2 3 x
tan 2 3 * 0
0
0
 lim ln y  lim
x 0
sec 2 x tan 2 x * 2
tan 2 x
0
sec 2 x
 lim
 lim

x0 2 tan 3 x sec 2 3 x * 3
x0 tan 3 x sec 2 3 x
0
3
 lim
x0
2 sec 2 2 x
tan 3x * [2 sec 3x * sec 3x tan 3x * 3]  sec 2 3x[sec 2 3x * 3]
2 sec 2 2 x
2 sec 2 0
2 *1
2



2
2
4
2
2
4
x0 6 sec 3 x tan 3 x  3 sec 3 x
6 sec 0 tan 0  3 sec 0 6 * 0  3 * 1 3
 lim
 lim ln y 
x 0
2
3
 lim y  lim (sec 3 2 x) cot
x 0
x 0
University of Kufa\Civil Eng. ………………
2
3x
 e2 3
(200)
……………………. Mathematics \1st class
Chapter Seven
x
2. lim
x 1
 1 


 x 1 
Transcendental Functions
 1 


 11 
1
1
 
0
1
 1
(indeterminate form)
 1 


 1 


Sol.: Let y  x  x1   ln y  ln x  x1  
ln x ln 1 0


(also indeterminate form)
x 1 1 1 0
 lim ln y  lim
x 1
1
ln x
ln x 
x 1
x 1
x 1
1 x 11

1
x 1 1
1
 lim
 lim y  lim x
x 1
 1 


 x 1 
x 1
3. lim (tan x) cos x   0
x
 e1  e
(indeterminate form)

2
Sol.: Let y  (tan x) cos x  ln y  ln(tan x) cos x  cos x ln(tan x) 
 lim  ln y  lim 
x

x
2

ln(tan x)
sec x
ln(tan x) ln  
(also indeterminate form)


sec x


2
sec 2 x
1
sec
x
 lim  tan x  lim 
 lim  cos x 2
2
 sec x tan x
 tan x

 sin x 
x
x
x
2
2
2 

 cos x 
1 cos 2 x
cos x 0
 lim 
. 2  lim 
 0
2
 cos x sin x
 sin x
1
x
x
2
2
 lim  y  lim  (tan x) cos x  e 0  1
x

x
2
4. lim x sin x  00

2
(indeterminate form)
x0
Sol.: Let y  x sin x  ln y  x sin x  sin x ln x 
ln x
csc x
ln x ln 0  


(also indeterminate form)
x0 csc x


 lim ln y  lim
x0
1
1x
 sin 2 x 0
x
 lim
 lim
 lim

x0  csc x cot x
x0
1 cos x x0 x cos x 0

.
sin x sin x
 lim
x 0
 2 sin x cos x
 sin 2 x
0
 lim

0
x

0
 x sin x  cos x
 x sin x  cos x 0  1
 lim y  lim x sin x  e0  1
x0
x0
University of Kufa\Civil Eng. ………………
(201)
……………………. Mathematics \1st class
Chapter Seven
ex 1 1 1 0
5. lim 2 
 (indeterminate form)
x0
x
0
0
Sol.: lim
x0
Transcendental Functions
ex 1
   so the limit does not exist.
2x 0
ln( x)  ln( x  1)  ln( )  ln(   1)     (indeterminate form)
6. lim
x

ln( x)  ln( x  1)  lim ln 
Sol.: lim
x
x
x 

 x 1
x 
Let y  ln 

 x 1
lim e y  lim
x
x
 ey  e
(from ln
 x 
ln 

 x 1 
a
 ln a  ln b )
b
 x 

 (exponentiate of both sides)
 x 1
x
x x
1
1
 lim
 lim

1
x


x


x 1
x x 1 x
11 x 1 0
 x 
lim y  lim ln 
  ln 1  0
x
x
 x 1
Or by another solution

x 
x x
 x 

lim ln 
  ln  lim
  ln  lim
x
x


x


x 1
x x 1
 x 1


7. lim
x0


1
  ln  lim
x


x
11


 1 
  ln 
  ln 1  0
x
1 0 
e x  ex  x2  2 11 0  2 0


00
0
sin 2 x  x 2
Sol.: lim
x0
e x  e x  2 x
e x  e x  2 x 1  1  0 0
 lim


2 sin x cos x  2 x x0 sin 2 x  2 x
00
0
e x  e x  2 1  1  2 0


x0 2 cos 2 x  2
22
0
 lim
e x  e x
11 0
 lim


x0  4 sin 2 x
0
0
 lim
x0
8. lim
x 
ln x 

x

Sol.: lim
x 
9. lim
x 0

e x  e x
11 2
1



 8 cos 2 x  8  8
4
1x 1 0

 0
1
1
1
ln sin x ln 0  


ln tan x ln 0  
cos x
1
1
1
Sol.:  lim sin2 x  lim tan2 x  lim 2   1
x0 sec x
x0 sec x
x0 sec x
1
tan x
tan x
University of Kufa\Civil Eng. ………………
(202)
……………………. Mathematics \1st class
Chapter Seven
10. lim x 2 ln x  0 * ()
x0
Sol.: lim
x 0 
Transcendental Functions
(indeterminate form)
ln x  


1 x2
1x
1
x2
0
 lim
 lim
 lim

0
3
2
x0  2 x
x0  2 x
x0  2
2

11. lim
 
x 0
1
x
1  1
1 
 
  (  ) (indeterminate form)
e 1  0 1 1
x
e x 1  x
e x 1  x 1 1  0 0

lim


x0 x(e x  1)
x0 xex  x
0 *1  0 0
Sol.:  lim
lim
x0
e x 1
0
ex
1
1


lim


x
x
x
x
x
0 *1  1  1 2
xe  e  1 0 x0 xe  e  e
Homework: Find the limits of the following:
e x  e2
 e2
x2
ln( 2  x)
1
4. xlim
 1
x 1
8 x  2 x ln 2

7. lim
x0
4x
2
1. lim
x 2
2. lim
x 0
xex
 1
1  ex
 csc x ln x 
5. lim
x 1
1

8. lim e  tan x * sec 2 x  0
x

2
1
x  1
ln x

11. lim
0

x
x
 ln x x  1  2

10. lim

x 1
13. lim x x  1
x0
16. lim e x  3x   e 4
1x
x0
19. lim sin x  cos x tan x
x
2
22. lim (tan x) cos x  1
x
5 x  2 ln x
5
x  3 ln x
1
ln cot x
 20. lim csc2 x  0
x0
e
e
17. lim
x 
23. lim x
tan
x
2
x1
1 
 ln x

0
x
 x
x4  x2
0
15. lim
x  e x  1
ex
1

1  ex  
18. lim
x 
e
x
2
e  3x
1

21. lim
x
2
x 4e  2 x
4
12. lim

x 
14. lim (cos x)1 x  1
x0
ex 1 1

tan 2 x 2
e 2 x  e 2 x
6. lim
4
x0
sin x
ln sec 2 x
 e2
9. lim
x 0
ln sec
3. lim
x 0
 e 2 
24. lim (e x  1) cot x
x0
2
x
 1
25. lim
1    e
x 
x

3 x
e
0
28. lim
x0 x 2
26. lim x e  0
2 x
x
29. lim
x
2x
3x
2
0
University of Kufa\Civil Eng. ………………
(203)
e x (1  e x )
27. lim
1
x0 (1  x) ln( 1  x)
ln 5 x
0
30. lim
x  x 2
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
4. Inverse of Trigonometric functions:
1. The arcsine of x (sin-1x) is the angle in   2 ,  2 whose sine is x.
The function y=sinx is one-to-one, if we restrict its domain to the interval


2
x

2
. It has an inverse which is denoted by:
y  sin 1 x
and is sometimes written as y=arcsinx
and for the function y=sin-1x
D f  [1,1]
and
  
R f   , 
 2 2
Note: The graph of sin-1x is symmetric about the origin because that the graph of
sin x is symmetric about the origin this means that
sin-1(-x)=- sin-1x
2. The arccosine of x (cos-1x) is the angle in 0,   whose cosine is x.
The function y=cosx is one-to-one, if we restrict its domain to the interval 0  x  
. It has an inverse which is denoted by:
y  cos 1 x
and is sometimes written as y=arc cos x
and for the function y=cos-1x
D f  [1,1]
and
R f  0,  
Note: The graph of y = cos-1x has no such symmetry
University of Kufa\Civil Eng. ………………
(204)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Note: We can see from the figures
below the following identities
cos 1 x  cos 1 ( x)  
1.
 cos 1 ( x)    cos 1 x
and form the triangle
sin 1 x  cos 1 x 
2.

2
x0
for
3. The arctangent of x (tan-1x) is the angle in   2 ,  2 whose tangent is x.
The function y=tan x is one-to-one, if we restrict its domain to the interval


2
x

2
. It has an inverse which is denoted by:
y  tan 1 x
and is sometimes written as y=arctanx
and for the function y=tan-1x
D f  (, )
and
  
Rf   ,  .
 2 2
Note: The graph of tan-1x is symmetric about the origin because that the graph of
tan x is symmetric about the origin, this means that
tan-1(-x)=- tan-1x
University of Kufa\Civil Eng. ………………
(205)
……………………. Mathematics \1st class
Chapter Seven
4. The arctcoangent of x
(cot-1x)
Transcendental Functions
is the angle in 0,   whose cotangent is x.
The function y=cot x is one-to-one, if we restrict its domain to the interval
0  x   . It has an inverse which is denoted by:
y  cot 1 x
and is sometimes written as y=arccot x
and for the function y=cot-1x
D f  (, )
and
R f  0,   .
5. The function y=sec x is one-to-one, if we restrict its domain to the interval

{x : 0  x   } \ { } . It has an inverse which is denoted by:
2
y  sec 1 x
and is sometimes written as y = arcsec x
and for the function y=sec-1x
D f  R \ (1,1)
and

R f  [0,  ] \ { } .
2
6. The function y=cscx is one-to-one, if we restrict its domain to the interval
{x : 

2
x

2
} \ {0} . It has an inverse which is denoted by:
y  csc 1 x
University of Kufa\Civil Eng. ………………
(206)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
and is sometimes written as y=arccscx
and for the function y=csc-1x
D f  R \ (1,1)
R f  [
and
 
, ] \ {0} .
2 2
Note: To find sec-1x, csc-1x and cot-1x, use the following identities:
1
1. sec 1 x  cos 1  
 x
1
2. csc 1 x  sin 1  
 x
3. cot 1 x 

2
 tan 1 x
1
Example 1: Show that sec 1 x  cos 1   .
 x
1
1
1
Sol.: Let z  right side  cos 1    cos z  cos cos 1   cos z 

 x
 x
x
x
1
 sec z
cos z
sec 1 x  sec 1 (sec z )  sec 1 x  z  left side
Example 2: Show that: cot 1 x 
Sol.: Let z  right side 

2

2
o.k.
 tan 1 x .
 tan 1 x  tan 1 x 

2
z


sin   z 


2
  cos z  cot z
 tan(tan 1 x)  tan   z   x 

 sin z
2

cos  z 
2

 z  cot 1 x  left side o.k.
University of Kufa\Civil Eng. ………………
(207)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Examples: Find the limits of the following:
1. lim sin 1 x  sin 1 1 

x 1

2
2. lim cos 1 x  cos 1 (1 )  
x1
tan 1 x  tan 1  
3. lim
x 

2
tan 1 x  tan 1 ()  
4. xlim


2
1
x
sec 1 x  lim cos 1 ( )  cos 1 0 
5. lim
x 
x 

2
1
x
sec 1 x  lim cos 1 ( )  cos 1 0 
6. xlim

x 

2
1
x
csc 1 x  lim sin 1 ( )  sin 1 0  0
7. lim
x
x 
1
x
csc 1 x  lim sin 1 ( )  sin 1 0  0
8. xlim

x
The Derivative of Inverse Trigonometric Functions:
Example 1: If y=sin-1x, then find dy/dx.
Sol.:
y=sin-1x  x=siny
1  cos y.
dy
dx
(using implicit differentiation)


dy
1
1
1



(remember that cosy>0 for   x  )
2
2
2
2
dx cos y
1  sin y
1 x

d
1
sin 1 x 
dx
1  x2
Example 2: If y=sec-1x, then find dy/dx.
Sol.:
y=sec-1x  x=secy
1  sec y tan y.
dy
dx
(using implicit differentiation)
dy
1
1
1



dx sec y tan y sec y  sec 2 y  1 x  x 2  1


 
d
1
sec 1 x 
dx
x x2  1
University of Kufa\Civil Eng. ………………
(208)

(remember that    tan y   for

2
{x: 0  x   }\{ }
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Example 3: If y=tan-1x, then find dy/dx.
y=tan-1x  x=tany
Sol.:
1  sec 2 y.
dy
(using implicit differentiation)
dx
dy
1
1
1


 2
2
2
dx sec y tan y  1 x  1

d
1
tan 1 x  2
dx
x 1
In general: If u is a function of x:
1.
d
du dx
sin 1 u 
dx
1 u2
u 1
2.
d
 du dx
cos 1 u 
dx
1 u2
u 1
3.
d
du dx
tan 1 u 
dx
1 u2
4.
d
 du dx
cot 1 u 
dx
1 u2
5.
d
du dx
sec 1 u 
dx
u u2 1
u 1
6.
d
 du dx
csc 1 u 
dx
u u2 1
u 1
Examples: Find dy/dx of the following functions:
1. y  sin 1 x 2
Sol.:
dy
2x
2x


dx
1  (x2 )2
1  x4
2. y  tan 1 x  1
Sol.:
1
2 x 1
dy

dx 1 

x 1

2

1
1
1
1
*

*
2 x 1 1 x 1 2 x 1 2  x
3. y  sec 1 3x
Sol.:
dy
3
1


dx 3 x (3 x) 2  1 x 9 x 2  1
University of Kufa\Civil Eng. ………………
(209)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
4. y  x sin 1 3x
Sol.:
dy
3
3x
 x*
 sin 1 3 x * 1 
 sin 1 3 x
2
2
dx
1  (3 x)
1  9x
5. y 2 sin x  y  arctan y
Sol.: 2 y. y`sin x  y 2 cos x  y`
y`( 2 y sin x  1 
y`
1  y2
1
)   y 2 cos x
2
1 y
y`(
2 y sin x(1  y 2 )  (1  y 2 )  1
)   y 2 cos x
1  y2
y`(
2 y sin x  2 y 3 sin x  1  y 2  1
)   y 2 cos x
1  y2
 y 2 cos x(1  y 2 )
 y cos x(1  y 2 )
y`

2 y sin x  2 y 3 sin x  y 2 2 sin x  2 y 2 sin x  y
Example: If y  sin 1 (cos x) , show that y``+y`+1=0.
Sol.:
 sin x
y`
1  cos x
2

 sin x

2
sin x
 sin x
 1
sin x
 y`` 0
So y``+y`+1=0+(-1)+1=0
o.k.
1
x
Example: If y  tan 1 x  tan 1 ( ) , show that y``+ y`=0.
Sol.:
y`
1
1 x2
1
1 x2
1
1 x2
1
1






 2
0
2
2
2
2
2
2
2
2
1  x 1  1 x  1  x 1  1 x
1  x ( x  1) x
1 x
x 1
 y`` 0
So y``+y`=0+0=0
o.k.
Homework
I. Verify the following identities.
1. sin 1 x  cos 1 x 


2


2 
1

x


2. sin 1 x  tan 1 
x
University of Kufa\Civil Eng. ………………
(210)
……………………. Mathematics \1st class
Chapter Seven
3. 2 cos 1 x  cos 1 (2 x  1) 0  x  1
Transcendental Functions
II. Find dy/dx of the following.
1. y  tan 1 (3x  1)
2. y  e x sec 1 e x 
3. y 
4. y  x sec 1 ( x )
5. y  x cos 1 4 x  1
6. y  (tan x) tan
7. y  sin 1 (ln x)
8. y  ln tan 1 x 2 
9. y  1  cos 1 3x 
11. x 3  x sin 1 y  ye x
12. ln( x  y)  tan 1 ( xy)
1
10. y    sin 1
x
1

x
tan 2 x
x2  1
1
x
3
4
III. Use L'Hopital's rule to find the limits of the following.
1. lim
x 0
sin 1 x
x
2. lim
x0
sin 1 x
x3
3. lim
x 0
tan 1 x
x
4. lim
x 0
tan 1 x
x3
5. lim
x0
sin 1 2 x
2
x
6. lim
x0
sin 1 x  x 1

x3
6
8. lim
x2  1
sec 1 x
9. lim
x 0
2 tan 1 3x 2
7x2
x tan 1
7. lim
x 
2
x
x1
Integration Formulas
The following formulas can be used.
du
1.

2.
 1 u
3.
u
 sin 1 u  C
1 u
2
du
 tan 1 u  C
2
du
u 1
2
valid for u2 < 1
valid for all u
 sec 1 u  C
valid for u2 > 1
Examples: Evaluate the following integrals:
3 2
dx

1.
2 2
1
2.
1 x
dx
 1 x
2
 sin 1 x

3 2
2 2
 3
 2   
  sin 1 

 sin 1 

 2   3  4  12
2





1
2
 tan 1 x 0  tan 1 1  tan 1 0 
0
2
2
4.
dx

3.

x x 1
2
3
x 2 dx
1  x6


 sec 1 x 2
2
3

4
0 

4
 2    
 sec 1 2  sec 1 
  
 3  4 6 12
x 2 dx
 
1  x3
2
University of Kufa\Civil Eng. ………………
(211)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Let u  x 3  du  3x 2 .dx or x 2 .dx 

5.
du 3
1 u
dx

9  x2
2


Let u 
du
3
1
du
1
1
 sin 1 u  C  sin 1 x 3  C

2
3 1 u
3
3
dx
 x2 
91  
9


1
3
dx
 x
1  
3
2
x
dx
or dx  3du
 du 
3
3
1 3du
x
 sin 1 u  C  sin 1  C

2
3 1 u
3
6.
e
dx
x
1  e 2 x

dx
e x 1  (e  x ) 2

e  x .dx
1  (e  x ) 2
Let u  e  x  du  e  x .dx or e  x .dx  du

7.
 du
1 u
2
 
du
1 u
2
  sin 1 u  C   sin 1 e  x  C
dx
x 1
x
Let z  x  1  z 2  x  1 or 2z.dz  dx

8.

2 z.dz
dz
 2 2
 2 tan 1 z  C  2 tan 1 x  1  C
2
( z  1) z
z 1
cot x.dx
sin x  9
2
Let u 

9.

cos x.dx
1

sin x 9 sin 2 x  1
9


cos x.dx
2
 sin x 
3 sin x 
 1
 3 
sin x
cos x
.dx or cos x.dx  3du
 du 
3
3
1
3du
1
du
1
1
 sin x 
 
 sec 1 u  C  sec 1 
C

2
2
3 3u u  1 3 u u  1 3
3
 3 

sin 1 3x.dx
1  9x2
Let u  sin 1 3x  du 
3dx
or
1  9x2

dx
1  9x2

du
3

2
du
u2
u2
sin 1 3x
  u.  C 
 C 
 C 
C
3
2*3
6
6
University of Kufa\Civil Eng. ………………
(212)
……………………. Mathematics \1st class
Chapter Seven
dx

10.
Transcendental Functions
Completing the square
4x  x2
Sol.: rewrite 4x-x2 by completing the square
2
2
 2
 4   4  

4 x  x  ( x  4 x)   x  4 x        x 2  4 x  4  4  4  ( x  2) 2

 2   2  

2

2
dx
4  ( x  2)
2

dx
 1

41  ( x  2) 2 
 4


dx
 x  2
2 1 

 2 
2
x2
dx
or dx  2.du
 du 
2
2
Let u 
1
2.du
du
 x  2

 sin 1 u  C  sin 1 
C

2
2
2 1 u
 2 
1 u
 4x
11.
2
dx
 4x  2
Sol.: rewrite 4 x 2  4 x  2 by completing the square
2
2
 2

1
1
1 1





4 x  4 x  2  4 x  x         2  4 x 2  x     2

4 4
 2   2  


2
2
2
2
1
1
1
1
dx



 4 x    4 *  2  4 x    1  2  4 x    1  (2 x  1) 2  1  
2
4
2
2
(2 x  1) 2  1



Let u  2x  1  du  2.dx or dx 
du
2
du 2 1
du
1
1
  2
 tan 1 u  C  tan 1 2 x  1  C
2
1 2 u 1 2
2
u
Homework: Evaluate the following integrals.
1.
4.
7.

dx
x 1 x

dx

e 2 x  25
x.dx
36  x
2
dx
10.  2
2x  2x  5
13.
3x 3  4 x 2  3x
 x 2  1 .dx
sec x tan x.dx
2. 
1  sec 2 x
5.
8.
x
dx
x6  4
e x .dx

4e
cos
x
1

6.

9.

2
4 x .dx
11.

14.
ln 3 x  1
 x ln 2 x  x .dx
1  16 x 2
University of Kufa\Civil Eng. ………………
e x .dx
3.  2 x
e 1
(213)
12.
e x .dx
16  e 2 x
dx
20  8 x  x 2

2( x  1).dx
1  ( x  1) 4
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
5. Hyperbolic Functions:
Definition and Identities:
The hyperbolic cosine and hyperbolic sine
functions are defined by the following equations:
Hyperbolic cosine of x:
Note: when x    e
cosh x 
e x  e x
2
ex
 0 So cosh x 
2
x
when x    e x  0 So cosh x 
e x
2
So D f  (, ) and R f  [1, )
Hyperbolic sine of x:
sinh x 
Note: when x    e  x  0 So sinh x 
e x  e x
2
ex
2
e x
when x    e  0 So sinh x  
2
x
So D f  (, ) and R f  (, )
The notation coshx is often read "kosh x" and sinh x is pronounced as if spelled
"cinch x" or "shine x".
Four additional hyperbolic functions are defined in terms
of cosh x and sinh x as shown below:
tanh x 
sinh x e x  e  x

cosh x e x  e  x
coth x 
cosh x e x  e  x

sinh x e x  e  x
Hyperbolic tangent of x:
D f  (, ) and R f  (1,1)
Hyperbolic cotangent of x:
D f  (, ) \ {0} and R f  (,1)  (1, )
Hyperbolic secant of x:
sech x 
1
2
 x
cosh x e  e  x
D f  (, ) and R f  (0,1]
University of Kufa\Civil Eng. ………………
(214)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
1
2
 x
csch x 
sinh x e  e  x
Hyperbolic cosecant of x:
D f  (, ) \ {0} and R f  (, ) \ {0}
Identities:
Hyperbolic functions
Trigonometric Functions
1 cosh 2 x  sinh 2 x  1
cos 2 x  sin 2 x  1
2 sinh 2x  2 sinh x cosh x
sin 2x  2 sin x cos x
3 cosh 2 x  cosh 2 x  sinh 2 x
cos 2 x  cos 2 x  sin 2 x
4 cosh 2 x 
cosh 2 x  1
2
cos 2 x 
1  cos 2 x
2
5 sinh 2 x 
cosh 2 x  1
2
sin 2 x 
1  cos 2 x
2
tanh 2 x  1  sech2 x
tan 2 x  sec 2 x  1
7 coth 2 x  1  csch2 x
cot 2 x  csc 2 x  1
6
Examples: Prove that:
1. cosh 2 x  sinh 2 x  1
Sol.: left side:
 e x  ex
cosh x  sinh x  
2

2
2
2
  e x  ex
  
2
 



2

e 2 x  2e x e  x  e 2 x e 2 x  2e x e  x  e 2 x

4
4

e 2 x 2e 0 e 2 x e 2 x 2e 0 e 2 x 4e 0






4
4
4
4
4
4
4
 1  right side
o.k.
2. csch2 x =coth2 x-1
2
Sol.: right side:
 e x  ex 
e 2 x  2  e 2 x


1

1
coth x-1   x
x 
e 2 x  2  e 2 x
e e 
2

e 2 x  2  e 2 x  e 2 x  2  e 2 x
e 2 x  2  e 2 x
e2x
4
2 

 x
2 x
x 
2e
e e 
= csch2 x = left side
University of Kufa\Civil Eng. ………………
(215)
2
o.k.
……………………. Mathematics \1st class
Chapter Seven
3. cosh 2 x 
Transcendental Functions
cosh 2 x  1
2


cosh 2 x  1 e 2 x  e 2 x 2  1

2
2
Sol.: right side:

e 2 x  e 2 x  2 e 2 x  2  e 2 x

4
4
 e x  e x
 
2

2

 =cosh2 x= left side

o.k.
4. cosh x + sinh x = ex
Sol.: left side:
e x  e x e x  e x
cosh x + sinh x=

2
2



1 x
2e x
e  ex  e x  ex 
2
2
 e x  right side
o.k.
Examples: Solve the following equations:
1. 5 cosh x  3 sinh x  5
Sol.: 5
e x  e x
e x  e x
3
5
2
2
5e x  5e  x  3e x  3e  x  10  2e x  8e  x  10  e x  4e  x  5
4
e2x  4
 e  x 5 
5
e
ex
x
 e 2 x  4  5e x  e 2 x  5e x  4  0
 (e x  4)(e x  1)  0
 either (e x  4)  0  e x  4  x  ln 4
or (e x  1)  0  e x  1  x  ln 1  0
2. 3 cosh x  2 sinh x  10
Sol.: 3
e x  e x
e x  e x
2
 10
2
2
3e x  3e  x  2e x  2e  x  20  e x  5e  x  20

e 2 x  5  20e x  e 2 x  20e x  5  0
 (20)  (20) 2  4(1)(5) 20  400  20 20  380
e 


2(1)
2
2
x
 either
e x  19.74  x  ln 19.74  2.98
University of Kufa\Civil Eng. ………………
(216)
……………………. Mathematics \1st class
Chapter Seven
or
Transcendental Functions
e  0.254  x  ln 0.254  1.373
x
Derivatives of Hyperbolic Function:
If u is any function of x, then:
Derivative of hyperbolic functions
Derivative of trigonometric functions
1
d
du
sinh u  cosh u.
dx
dx
d
du
sin u  cos u.
dx
dx
2
d
du
cosh u  sinh u.
dx
dx
d
du
cos u   sin u.
dx
dx
3
d
du
tanh u  sec h 2 u.
dx
dx
d
du
tan u  sec 2 u.
dx
dx
4
d
du
coth u   csc h 2u.
dx
dx
d
du
cot u   csc 2 u.
dx
dx
5
d
du
sec hu   sec hu tanh u.
dx
dx
d
du
sec u  sec u tan u.
dx
dx
6
d
du
csc hu   csc hu coth u.
dx
dx
d
du
csc u   csc u cot u.
dx
dx
Examples: Prove that:
1.
d
sinh x  cosh x
dx
Sol.:
2.
d
d  e x  e  x  e x  (e  x ) e x  e  x

sinh x  

 cosh x o.k.
dx
dx 
2 
2
2
d
csc hx   csc hx coth x
dx
Sol.:
d
d  1  sinh x * (0)  1 * cosh x
csc hu 


dx
dx  sinh x 
sinh 2 x

Examples: Find
 cosh x
 1 cosh x

.
  csc hx coth x
2
sinh x sinh x sinh x
o.k.
dy
of the following:
dx
1. y  sinh 3x
Sol.:
dy
 cosh 3x * 3  3 cosh 3x
dx
2. y  tanh( 1  x 3 )
University of Kufa\Civil Eng. ………………
(217)
……………………. Mathematics \1st class
Chapter Seven
dy
 sec h 2 (1  x 3 ) * 3x 2  3 x 2 sec h 2 (1  x 3 )
Sol.:
dx
3. y  coth
Sol.:
Transcendental Functions
1
x
dy
 1   1 1
1
  csc h 2   *  2   2 csc h 2  
dx
 x  x  x
 x
4. y  x sec hx 2
Sol.:


dy
 x  sec hx 2 . tanh x 2 .2 x  sec hx 2
dx
 2 x 2 sec hx 2 . tanh x 2  sec hx 2
5. y  csc h 2 ( x 2  1)
Sol.:

dy
 2 csc h( x 2  1)  csc h( x 2  1) coth( x 2  1) * 2 x
dx

 4 x csc h 2 ( x 2  1) coth( x 2  1)
6. y  ln tanh 2 x
dy sec h 2 2 x * 2


Sol.:
dx
tanh 2 x

2
1
2
cosh 2 x
cosh 2 2 x 
*
2
sinh 2 x
cosh 2 x sinh 2 x
cosh 2 x
2*2
4

 4 csc h 4 x
2 cosh 2 x. sinh 2 x sinh 4 x
7. y  sinh x x
Sol.: ln y  x ln sinh x
1 dy
cosh x
.  x.
 ln sinh x
y dx
sinh x
dy
 cosh x

x
 y x.
 ln sinh x   sinh x   x coth x  ln sinh x 
dx
 sinh x

Integrals of Hyperbolic Function:
If u is any function of x, then:
1.  sinh u.du  cosh u  C
2.  cosh u.du  sinh u  C
3.  sec h 2u.du  tanh u  C
4.  csc h 2u.du   coth u  C
5.  sec hu. tanh u.du   sec hu  C
6.  csc hu. coth u.du   csc hu  C
University of Kufa\Civil Eng. ………………
(218)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Examples: Evaluate the following integrals:
1.  coth 5 x.dx  
cosh 5 x
.dx
sinh 5 x
  cosh 5 x.dx 
Let u  sinh 5x  du  cosh 5x.5dx
1 du
u. 5
du
5
1
1
ln | u | C  ln | sinh 5 x | C
5
5

ln 2
e x  ex
2.  4e sinh x.dx   4e
.dx   2e 2 x  2 .dx
2
0
0
0
ln 2

ln 2
x
x


ln 2


 
 e 2 x  2 x 0  e 2 ln 2  2 ln 2  e0  2 * 0

 e ln 2  2 ln 2  1  4  2 ln 2  1  3  2 ln 2
2
cosh 2 x  1
1  sinh 2 x

.dx  
 x
3.  sinh x.dx  
2
2 2
0
0
0
1
1
1
2
1  sinh 2   sinh 0
 sinh 2 1
 
 1  
 0  
  0.40672
2  2
4
2
  2

4.  tanh 3x. sec h 2 3x.dx
Sol.: let u  tanh 3x  du  3 sec h 2 3x.dx
 u.
du
3

 sec h 2 3 x.dx 

 csc h 2 x.dx  du
du
u2
tanh 2 3x

C 
C
3 2*3
6
5.  e coth x . csc h 2 x.dx
Sol.: let u  coth x  du   csc h 2 x.dx
e
u
(du )  e u  C  e coth x  C
csc h 2 x
.dx
6. 
x
Sol.: let u 2  x  2u.du  dx
csc h 2u
2
 u .2u.du  2 csc h u.du  2 coth u  C  2 coth x  C
University of Kufa\Civil Eng. ………………
(219)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Homework:
I. Verify the following identities:
1. cosh x  sinh x  e  x
2. sinh(  x)   sinh x
3. cosh(  x)  cosh x
4. sinh( x  y )  sinh x cosh y  cosh x sinh y
5. cosh( x  y)  cosh x cosh y  sinh x sinh y
6. tanh( x  y) 
tanh x  tanh y
1  tanh x tanh y
7. (cosh x  sinh x) n  cosh nx  sinh nx for every positive integer n.
II. Find dy/dx of the following functions:
sec hx 2
x2  1
1. y  cosh 4 x 2  3
2. y 
4. y  tan 1 (tanh x)
5. y  e3 x sec hx
6. y  sec h5 x
8. y  ln(csc hx  coth x)
9. y  sinh 2 x
7. y 
1
tanh x  1
3. y  x csc he 4 x
10. sinh xy  ye x
11. x 2 tanh y  ln y
12. y  ( x 2  1) sec h(ln x)
(Hint: Express the terms of exponentials and
simplify before differentiating.)
III. Evaluate the following integrals:
1.

sinh
x
x
.dx
2.

cosh ln x
.dx
x
3.
1
 coth
2
3x
.dx
4.  sinh x cosh x.dx
5.  tanh 3x sec h3x.dx
6.  sinh x cosh x .dx
sec h 2 x
.dx
7. 
1  2 tanh x
e sinh x
.dx
8. 
sec hx
9.  cosh 2 3x.dx
2
10.
cosh(ln x)
1 x .dx
11.

sin 1 (cosh x) sinh x
1  cosh x
2
.dx 12.
 4e
x
cosh x.dx
13.  tanh x ln cosh x.dx
14.

cosh x  1.dx
(Hint: multiply by
University of Kufa\Civil Eng. ………………
cosh x  1
)
cosh x  1
(220)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
5. Inverse of Hyperbolic Functions:
All hyperbolic functions have inverses, they are:
Inverse of hyperbolic functions
Their domains
1
sinh 1 x  ln( x  x 2  1)
( ,  )
2
cosh 1 x  ln( x  x 2  1)
[1, )
3
tanh 1 x 
1 1  x 
ln 

2 1 x 
(1,1)
4
coth 1 x 
1  x  1
ln 

2  x 1
(, ) \ [1,1]
5
1  1  x2
sec h x  ln 

x





(0,1]
6
1
1  x2
csc h 1 x  ln  
x
|x|





(, ) \ {0}
1
Example: Prove that: sinh 1 x  ln( x  x 2  1) .
Sol.: let y  sinh 1 x 
x
x  sinh y 
e y  e y
2

e y  2 x  e  y  0 multiply both sides by ey

e 2 y  2 xe y  1  0 
ey 
2x  4x 2  4
 x  x2  1
2
Since ey is never negative, we must discard the minus sign.
 y  ln( x  x 2 1)
That is sinh 1 x  ln( x  x 2  1)
Derivatives and Integrals:
a. Derivatives
If u is any function of x, then
d
du dx
sinh 1 u 
dx
1 u2
d
du dx
2. cosh 1 u  2
dx
u 1
d
du dx
3. tanh 1 u 
dx
1 u2
1.
u 1
University of Kufa\Civil Eng. ………………
u 1
(221)
……………………. Mathematics \1st class
Chapter Seven
d
du dx
4. coth 1 u 
dx
1 u2
d
 du dx
5. sec h 1u 
dx
u 1 u2
d
 du dx
6. csc h 1u 
dx
u 1 u2
Transcendental Functions
u 1
0  u 1
u0
b. Integrals:
If u is any function of x, then
1.

2.

du
1 u
 sinh 1 u  C
2
du
u 1
2
 cosh 1 u  C
tanh 1 x  C if
du

3. 
1  u 2 coth 1 x  C if
4.
u
5.
u
du
1 u
2
du
1 u
2
u 1
u 1
  sec h 1 u  C
  csc h 1 u  C
Useful identities
1
1. sec h 1 x  cosh 1  
 x
1
2. csc h 1 x  sinh 1  
 x
1
3. coth 1 x  tanh 1  
 x
Example: Show that
d
cosh 1 x 
dx
1
x2  1
.
Sol.: Let y  cosh 1 x   x  cosh y
And by implicit differentiation:
1  sinh y.

dy
dx
dy
1


dx sinh y
1
cosh 2 y  1

1
x2  1
University of Kufa\Civil Eng. ………………
o.k.
(222)
……………………. Mathematics \1st class
Chapter Seven
Transcendental Functions
Examples: Find dy/dx of the following functions:
1. y  sinh 1 3x
Sol.:
dy
3
3


dx
1  (3x) 2
1  9x2
2. y  cosh 1 e x
ex
dy
Sol.:

dx
e2x  1
x
3. y  2 tanh 1  tan 

2
 x 1
2 sec 2   *
dy
2 2

Sol.:
dx
 x
1  tan 2  
2
1
1
 x
 x
cos 2  
cos 2  
1
1
2 
2



 sec x
 x  cos x
2 x 
2 x 
2 x 
sin   cos    sin   cos 2. 
2
2
2
 2
1
 x
 x
cos 2  
cos 2  
2
2
1
4. y  coth 1  
 x
Sol.:
dy
 1 x2
 1 x2
1
1


 2

2
2
x  1 x  1 1  x2
dx 1  1 x
x2
5. y  sec h 1 (cos x)
Sol.:
dy
 ( sin x)
sin x
1



 sec x
2
dx cos x 1  cos x cos x sin x cos x
Examples: Evaluate the following integrals:
1
1.

0
1
2.dx
1  4x2

0
2.dx
1  (2 x) 2
Let u  2 x  du  2.dx , at x  0  u  0
at x  1  u  2
2

0
du
1 u2

 
2

 sinh 1 u 0  sinh 1 2  sinh 1 0  1.4436
University of Kufa\Civil Eng. ………………
(223)
……………………. Mathematics \1st class
Chapter Seven
dx

2.  2
9 x  25 
Transcendental Functions
dx
1
dx


 9
 25  3 x  2
25 x 2  1
  1
 25

 5 
3x
3.dx
 du 
5
5
Let u 
 dx 
5.du
3
1
1
 3x 
tanh 1 u  C 
tanh 1    C

1 5.du 3 1
du
 1 du
 15
15
 5
  2


2
2


25 u  1 15 u  1 15 1  u
  1 coth 1 u  C   1 coth 1  3x   C
 15
15
 5
3.

dx
4x  9
2
dx

dx

4

9 x 2  1
9

 2 x  2 
9   1
 3 

2
dx
1
3 1
 2x 
3
 
*  cosh 1    C
3  2 x 2
2 2
 3 

1
 
 3 
4.

sin x.dx
1  cos 2 x
Let u  cos x  du   sin x.dx
 sin x.dx  du

5.
 du
1 u
  sinh 1 u  C   sinh 1 cos x   C
2
4 tanh 1 x
 1  x 2 .dx
Let u  tanh 1 x  du 
 4u.du 
dx
1  x2


2
4u 2
 C  2u 2  C  2 tanh 1 x  C
2
1
e coth x
.dx
6. 
1  x2
Let u  coth 1 x  du 
 e .du  e
u
7. .
u
1
 C  ecoth
x
dx
1  x2
C
cos x.dx
 sin x
1  sin 2 x
Let u  sin x  du  cos x.dx
University of Kufa\Civil Eng. ………………
(224)
……………………. Mathematics \1st class
Chapter Seven
du
u
Transcendental Functions
1 u
  csc h 1 u  C   csc h 1 sin x  C
2
Homework:
1. Find dy/dx of the following:
a. y  sinh 1 5x
b. y  sinh 1 e x
c. y  cosh 1 x
d. y  tanh 1 ( x 2  1)
e. y  tanh 1 sin 3x
1
f. y  x sinh 1  
h. y  ln cosh 1 4 x
i. y  cosh 1 ln 4 x 
g. y 
1
1
sinh
x
2
 x
2. Evaluate the following integrals:

a.

d.
g.

j.

m.
dx
81  16 x 2
e x .dx
e
16
2x
dx
5e
2x
dx
9  4x
x
2
dx
b.

e.
 5  3x
h.

k.

16 x 2  9
2dx
2
x.dx
25 x  36
2
x.dx
9  4x
2
dx
c.
 49  4 x
f.
x
i.

l.
x
2
dx
9  x4
dx
25 x 2  36
dx
9  4x2
dx
4x2  9
University of Kufa\Civil Eng. ………………
(225)
……………………. Mathematics \1st class