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 Calculus & Analytic Geometry I (MATF 126) 2 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.
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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
Calculus & Analytic Geometry I (MATF 126) 2
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.
Calculus & Analytic Geometry I (MATF 126) 3
Calculus & Analytic Geometry I (MATF 126) 4
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).
Calculus & Analytic Geometry I (MATF 126) 5
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.
Calculus & Analytic Geometry I (MATF 126) 6
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
Calculus & Analytic Geometry I (MATF 126) 7
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
Calculus & Analytic Geometry I (MATF 126) 8
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
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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
Calculus & Analytic Geometry I (MATF 126) 10
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
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Properties of Logarithms
Note: ln lnrrx x
* Use these properties of logarithms to simplify/expand ‘ln’ functions so that differentiation process is simpler.
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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
Calculus & Analytic Geometry I (MATF 126) 13
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
Calculus & Analytic Geometry I (MATF 126) 14
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
Calculus & Analytic Geometry I (MATF 126) 15
The Integral 1du
u
Example:
Calculus & Analytic Geometry I (MATF 126) 16
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
Calculus & Analytic Geometry I (MATF 126) 17
The Integrals of tan x and cot x
Calculus & Analytic Geometry I (MATF 126) 18
Example 7.2.4:
a) sec tan
2 sec
y ydy
y
b) /12
06tan3xdx
Calculus & Analytic Geometry I (MATF 126) 19
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
Calculus & Analytic Geometry I (MATF 126) 20
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
Calculus & Analytic Geometry I (MATF 126) 21
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
Calculus & Analytic Geometry I (MATF 126) 22
Calculus & Analytic Geometry I (MATF 126) 23
Calculus & Analytic Geometry I (MATF 126) 24
Law of Exponents
Calculus & Analytic Geometry I (MATF 126) 25
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
Calculus & Analytic Geometry I (MATF 126) 26
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
Calculus & Analytic Geometry I (MATF 126) 27
Example 7.3.3:
a) 24 x x
y e
b) ln 3y e
c) ln1
y
d) sin 2ln 1ty e t
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Since ex is its own deritive, it is also its own antiderivative.
Calculus & Analytic Geometry I (MATF 126) 29
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
Calculus & Analytic Geometry I (MATF 126) 30
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
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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
Calculus & Analytic Geometry I (MATF 126) 32
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
Calculus & Analytic Geometry I (MATF 126) 33
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.
Calculus & Analytic Geometry I (MATF 126) 34
The graph of y = loga x can be obtained by reflecting the graph of y = ax across the
450 line y = x .
Calculus & Analytic Geometry I (MATF 126) 35
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
Calculus & Analytic Geometry I (MATF 126) 36
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
Calculus & Analytic Geometry I (MATF 126) 37
Calculus & Analytic Geometry I (MATF 126) 38
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
Calculus & Analytic Geometry I (MATF 126) 39
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
Calculus & Analytic Geometry I (MATF 126) 40
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
Calculus & Analytic Geometry I (MATF 126) 41
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
Calculus & Analytic Geometry I (MATF 126) 42
L’Hôpital’s Rule enables us to evaluate limit that involves indeterminate form.
Calculus & Analytic Geometry I (MATF 126) 43
Calculus & Analytic Geometry I (MATF 126) 44
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
Calculus & Analytic Geometry I (MATF 126) 45
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Example 7.5.2: Find the limits :
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
Calculus & Analytic Geometry I (MATF 126) 49
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
Calculus & Analytic Geometry I (MATF 126) 50
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
Calculus & Analytic Geometry I (MATF 126) 51
Example 7.5.3: Find the limits :
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
Calculus & Analytic Geometry I (MATF 126) 52
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
Calculus & Analytic Geometry I (MATF 126) 53
Calculus & Analytic Geometry I (MATF 126) 54
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
Calculus & Analytic Geometry I (MATF 126) 55
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
Calculus & Analytic Geometry I (MATF 126) 56
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
Calculus & Analytic Geometry I (MATF 126) 57
Derivatives of the Inverse Trigonometric Functions
Calculus & Analytic Geometry I (MATF 126) 58
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
Calculus & Analytic Geometry I (MATF 126) 59
Integration Formulas
Calculus & Analytic Geometry I (MATF 126) 60
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
Calculus & Analytic Geometry I (MATF 126) 61
7.7 Hyperbolic Functions
The hyperbolic functions are formed by taking combinations of the two exponential functions ex and e‐x.
Calculus & Analytic Geometry I (MATF 126) 62
Identities for Hyperbolic Functions
Calculus & Analytic Geometry I (MATF 126) 63
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