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Calculus & Analytic Geometry I (MATF 126) 1

CHAPTER 7 TRANSCENDENTAL FUNCTIONS

7.1 Inverse Functions and Their Derivatives 7.2 Natural Logarithms 7.3 Exponential Functions 7.5 Indeterminate Forms and L’Hôpital’s Rule 7.6 Inverse Trigonometric Functions 7.7 Hyperbolic Functions


7.1 Inverse Functions and Their Derivatives One‐to‐One Functions Note that some functions f may have same value for different number in its domain. Consider 2( )f x x :

There are distinct points in the same domain that have the same square: f (‐2) = 4 and f (2) =4. Hence 2( )f x x is not one‐to‐one function. A function that has distinct values at distinct elements in its domain is called one‐to‐one. These functions take on any value in their range exactly once.



Inverse Functions Since each output of a one‐to‐one function comes from just one input, the effect of the function can be inverted to send an output back to the input from which it came.

Note: The symbol f ‐1 for the inverse of f is read “f inverse”. The “‐1” in f ‐1 is not an exponent and f ‐1 (x) does not mean 1/ f(x).


Graph of f ‐1 Let f be a one‐to‐one function on [a, b]. If the point (c, d) is on the graph of f, then the point (d, c) is on the graph of f ‐1.

The graph of f ‐1 is symmetric to the graph of f with respect to the line y = x.


Example 7.1.1: In each case, find f ‐1(x) and identify the domain and range of f ‐1 . As a check, show that f ( f ‐1 (x)) = f ‐1 (f(x)) = x.

a) 5( )f x x b) 3

( )2

xf x

x

c) 3

( )x

xf x

d) 2( ) 2 , 1f x x x x


7.2 Natural Logarithms Definition: The Natural Logarithm Function

1

1ln , 0x

x dt xt

Recall:

1

( ) area under the curve ( ) from to

1 1 area under the curve from 1 to , 0

b

a

x

A f x dx y f x x a x b

dt y t t x xt t


t

a) x > 1

b) 0< x < 1

1y

t

1

1ln for 1, ln 0

x

A dt x x xt

y

1 x

1y

t

1

1

1 1= ln

x

xA dt dt x

t t

y

x 1


Note: When x > 1 , ln x > 0 0 < x < 1 , ln x <0

When x = 1, 1

1

1ln1 0 (No area)dt

t i.e ln 1 = 0

Since

1

1

,

1 (ln ) ,

1(ln ) 0

ln is always increasing on 0

1 ln 0

1 0

x

xd dx

dx dx x

dx

dx xx x

x dt xt

dt xt


Since

2

2 2

1 (ln ) ,

1(ln ) , 0

ln is always concave down

0dx

dx x

dx x

dx xx

x

Conclusion: Graph of y = lnx for x > 0 1) 1, ln 0

2) 0 1, ln 0

3) ln 0 when 1

4) ln is always increasing and concave down on 0

x x

x x

x x

x x

lny x

y

x1


Properties of Logarithms

Note: ln lnrrx x

* Use these properties of logarithms to simplify/expand ‘ln’ functions so that differentiation process is simpler.


Example 7.2.1: 1. Express the following in terms of ln 5 and ln 7:

a) ln7 7 b)

1ln35 ln

7

ln25

2. Simplify:

a) lnsec lncos b) ln 8 4 2ln2x c) 3 23ln 1 ln( 1)t t


The Derivatives of y = ln x By the first part of the Fundamental Theorem of Calculus,

1

1 1ln

xd dx dt

dx dx t x

For every positive value of x, we have 1

lnd

xdx x

And the Chain rule extends this formula for positive functions u(x):

ln lnd d du

udx du dx


Example 7.2.2: Find the derivative of y with respect to x, t or :

a) 4ln 1y x

b) ln sin , sin 0y x x

c) 3lny x

d) lny t t

e) ln

1 ln

x xy

x

f) ln ln lny x

g) sin cos

ln1 2ln

y


The Integral 1du

u

Example:


Example 7.2.3:

a) 0

1

3

3 2dx

x

b) /3

0

4sin

1 4cosd

c) 16

2 2 ln

dx

x x


The Integrals of tan x and cot x


Example 7.2.4:

a) sec tan

2 sec

y ydy

y

b) /12

06tan3xdx


Logarithmic Differentiation In some cases, differentiation of complicated functions can be made easier by applying logarithms. The process is called logarithmic differentiation. Example:

Given 2( ) 2 1 3 1 4 1f x x x x , find dy/dx .

Solution:

2

Apply ln on both sides:

ln ( ) ln 2 1 3 1 4 1

1ln ( ) ln 2 1 ln 3 1 2ln 4 1

2Differentiate both sides with respect to

1ln ( ) ln 2 1 ln 3 1 2ln 4 1

2

f x x x x

f x x x x

x

d df x x x x

dx dx


2

1 1 1 2( ) (2) (3) (4)

( ) 2 1 3 1 4 1

( ) 2 3 8

( ) 2 1 3 1 4 1

2 3 8( ) ( )

2 1 3 1 4 1

2 3 8 2 1 3 1 4 1

2 1 3 1 4 1

f xf x x x x

f x

f x x x x

f x f xx x x

x x xx x x

Example 7.2.5: Find dy/dx for the following functions:

a) 1

( 1)y

t t

b) tan 2 1y

c)

3 2

1 2

1 2 3

x x xy

x x


7.3 Exponential Functions Recall: 1. ln x is always increasing with domain 0, and range , .

2. ln x is one‐to‐one , hence its inverse exists. Notation: If ( ) lnf x x then,

1( ) natural exponential function in

x

f x x

e

1

( ) ln : D f 0,

R f ,

( ) : D f ,

R f 0,

x

f x x

f x e




Law of Exponents


Example 7.3.1: Solve for t:

a) 0.01 1000te b) (ln2) 1

2te

c) 2 2 1( ) xx te e e

d) 80 1te

Example 7.3.2: Solve for y in terms of t or x. a) ln(1 2 )y t

b) 2ln( 1) ln( 1) ln(sin )y y x


The Derivative and Integral of ex We calculate its derivative using the inverse relationships and Chain Rule:

Inverse relationship

Differentiate both

ln

ln 1

11

sides

Chain Rule

Solve for the de

riva tiv e

x

x

x

x

x x

e x

de

dx

de

e dx

de e

dx


Example 7.3.3:

a) 24 x x

y e

b) ln 3y e

c) ln1

y

d) sin 2ln 1ty e t


Since ex is its own deritive, it is also its own antiderivative.


Example 7.3.4:

a) 22 3x xe e dx

b) ln16

/4

0

xe dx

c) 43 tt e dt

d) /2cot 2

/41 csce d

e) 2 2ln

02 cosx xxe e dx


The General Exponential Function ax

When a = e, the definition gives

ax = e x lna = e x lne = ex(1) = ex The Derivative of au


Proof:

ln

ln ln

ln

x x a

x a

x

d da e

dx dxd

e x adx

a a

Example 7.3.5: Differentiate:

a) 22

sy b) 1 ey t

b) lny d)

1xy x

e) sinxy x f) lnlnx

y x


The Integral of au If 1a , so that ln1 0 then,

Example 7.3.6:

a) 0

25 d

b)

4

1

2 x

dx

c) ln2

1

2 x

dxx


Logarithms with Base a

If a is any positive number other than 1, the function ax is one‐to‐one function and has a nonzero derivative at every point.

It therefore has a differentiable inverse.

We call the inverse the logarithm of x with base a and denote it by loga x.


The graph of y = loga x can be obtained by reflecting the graph of y = ax across the

450 line y = x .


Example 7.3.7: Simplify the expression:

a) 2log 32 b) log 7 c) 3log 1/9

d) 2525log 3x e) sin4log 2

xe x


The function loga x is actually just a numerical multiple of ln x. To see this, we let y = loga x then take the natural log of both sides.

log

log

log

Inverse =x

ln ln Take natural log on both sides

ln ln

ln Solve for

ln

a

a

a

xy

xy

y

y x

a a

a x a

a x

y a x

xy y

a



Derivatives and Integrals Involving loga x

To find the derivatives or integrals involving base a logarithms, we convert them to natural logarithms.

If u is a positive differentiable function of x, then

ln 1 1 1log ln

ln ln ln lna

d d u d duu u

dx dx a adx a u dx


Example 7.3.8: Differentiate: a) 3log 1 ln3y b) 3 9log logy r r c) 8 23log logy t

d) ln5

57

log3 2

xy

x

e)

2 2

2log2 1

x ey

x


Example 7.3.9: Integrate:

a) ln2 1

1

ex b)

42

1

log xdx

x c) 10

10

1/10

log 10xdx

x

d) 28log

dx

x x e) 1

1xe

dtt


7.5 Indeterminate Forms and L’Hôpital’s Rule Indeterminate Form 0/0 If the continuous function f(x) and g(x) are both zero at x = a, then

( )lim

( )x a

f x

g x

cannot be found by substituting x = a. The substitution produces 0/0, a meaningless expression which we cannot evaluate. We use 0/0 as a notation for an expression known as an indeterminate form. Other indeterminate forms are:

0/ , 0, ,0 and 1


L’Hôpital’s Rule enables us to evaluate limit that involves indeterminate form.



Example 7.5.1: Find the limits :

a) 3

23

4 15lim

12t

t t

t t

b) 2

0

sinlimt

t

t c)

0

3 1lim

2 1

x

xx

d) sin

0

3 1lim

e)

2

0lim

ln secx

x

x f)

1

1lim

ln sinx

x

x x




a)

2

ln 1lim

logx

x

x

b)

2

0

ln 2lim

lnx

x x

x

c) lim ln2 ln 1

xx x

d) 0

lim ln lnsinx

x x

e) 22

1lim sin

4xx

x

f) 2lim 3x

x x x


Indeterminate Powers

Example 8: Apply L’Hôpital’s Rule to show that 1/

0lim 1

x

xx e

.

Solution: Apply natural log on both sides;

1/ 1ln ( ) ln 1 ln 1

xf x x x

x

L’Hôpital’s Rule now applies to give

0 0

0

1

ln 1 0lim ln ( ) lim

0

1/(1 ) lim

1 1

( )

x x

x

xf x

x

x

f x e e


Example 9: Find 1/lim x

xx

.

Solution: Apply natural log on both sides;

1/ 1ln ( ) ln lnxf x x x

x

L’Hôpital’s Rule now applies to give

0

lnlimln ( ) lim

1/ lim

1 0

( ) 1

x x

x

xf x

xx

f x e



a) 1/ 1

1lim x

xx

b) 1/

lim lnx e

x ex

c) 1/

0lim

xx

xe x

d) 0

lim x

xx

e) 2

0lim lnx

x x

f) 20

lim lnx

x x


7.6 Inverse Trigonometric Functions Note: Trigonometric functions are not one‐to‐one.

Eg: y = sin x

Restrict the domain so that f (x) = sin x is one‐to‐one. i.e Domain: /2, /2

Range : 1,1

Now f (x) = sin x is one‐to‐one and f ‐1(x) exists, denoted by sin‐1x or arc sin x.

Note : 1 1sin

sinx

x



Consider:

1

1 1

1

sin sin 3/2 ? of the form : sin sin if Dom sin i.e 1,1sin sin 3/2 3/2

if

x x xx

1

3/2 1,1Hence, sin sin 3/2 is undefined.

Examples:

1

1

1

cot cot /4

Compare with definition:

cot cot 0

since cot /4 /4 0,

cot /4 is undefined

cot cot /4 is also undefined

y x x y y x


1

1

11 4

1

1sin , find y

2

Note: sin sin

/2, /2 , 1 1

1 1sin sin , y ,

2 2

1Since sin y

2

y

x y x y

y x

y y Q Q

y Q

Example: Use reference triangle to find:

a) 1 1sin

2

b) 1 1sin

2

c)

1 3sin

2

Solution for (a):

2 1

3

1

306

1sin

2 6

y


Example 7.6.1: 1. Evaluate the following expressions:

a) 1 3cos

2

b) 1cos cos3 c)

1 1sin sin

2

2. a) Suppose . Find . b) Find an alternative form for in terms of x.

1sin 2/5 cos and tan

1cot cos /4x


Derivatives of the Inverse Trigonometric Functions


Example 7.6.2: Find the derivatives:

a) 1 1cosy

x

b) 1sec 5y s c) 1tan lny x

d) 2 11 secy s s e) 2 1ln 4 tan2

xy x x


Integration Formulas


Example 7.6.3: Integrate the following:

a) 29 3

dx

x b)

3 2

4

20 9 4

ds

s c) 23 3 25

dx

x x

d)

/4

21

4

1 ln

e dt

t t

e) 22

dx

x x f)

2 6 10

dy

y y

g) 1

2

tan

1

xdx

x


7.7 Hyperbolic Functions

The hyperbolic functions are formed by taking combinations of the two exponential functions ex and e‐x.


Identities for Hyperbolic Functions


Example 7.7.1: 1. Given sinh x = 4/3. Find the values of the remaining 5 hyperbola functions.

2. Rewrite the following in terms of exponential and simplify: a) sinh 2lnx

b) ln cosh sinh ln cosh sinhx x x x


Derivatives of Hyperbolic Functions


Example 7.7.2: Differentiate the following:

a) 1sinh 2 1

2y x b) ln coshy z c) 21

lnsinh coth2

y v v


Integral Formulas for Hyperbola Functions


Example 7.7.3: Integrate the following:

a) sinh5

xdx b)

csc ln coth lnh t tdt

t c) ln2

0tanh2x dx

d) /2

02sinh sin cos d

e)

ln102

04sinh

2

xdx

CHAPTER TRANSCENDENTAL FUNCTIONSkdirectory1213.weebly.com/uploads/8/1/0/3/8103000/...Calculus &...

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