Class Notes For Math 1220-004
Chapter 6
Section 6.1: The Natural Logarithm Function
Question: We know that:
1
Dx ( x2 ) = x = x 1
2
D (x) = 1 = x 0
x
1
1
Dx (− ) = 2 = x -2
x
x
But whose derivative is
1
= x -1 ?
x
Definition
The natural logarithm function is defined by
Z
ln x =
1
Geometric Interpretation:
1
x
1
dt
t
Derivative
It follows directly from the definition of the log function:
0
(ln x) =
1
x
Properties:
If a and b are positive numbers and r is any rational number, then:
(1) ln 1 = 0
(2) ln ab = ln a + ln b
(3) ln
a
= ln a − ln b
b
(4) ln ar = r ln a
Proof:
2
Example 1
Dx (ln (3 sin x −
Example 3
√
2x + 7))
Dx (
Example 2
Show that: Dx ln |x| =
1 + sin x2
3
(x + 2) 4
)
Example 4 Calculate:
R
(i) tan xdx
1
x
(ii)
3
R
π
4
0
tan xdx
Example 6 Calculate:
Z
2t2
t+1
dt
+ 4t + 3
Section 6.2: Inverse Functions and Their Derivatives
Notation: We use y = f −1 (x) to denote the inverse of y = f (x)
4
Inverse Functions
If f (x) and f −1 (x) are inverse functions:
• f (x) must be one-to-one, i.e., the inverse exists when we can get back
to an x given a y. The horizontal line test may be used.
• If (a, b) is on f (x), then (b, a) is on f −1 (x).
• f (f −1 (x)) = f −1 (f (x)) = x.
• The domain of f (x) becomes the range of f −1 (x).
• The range of f (x) becomes the domain of f (x).
Theorem
If f is strictly monotonic on its domain, then f has an inverse.
5
Example 1
Rx
Show that g(x) = 1 (cos2 t + 4t)dt has an inverse on (−∞, +∞), whereas
h(x) = x2 − 3 doesn’t.
Example 2
Find the inverse of f (x) =
x3 + 2
and show that f (f −1 (x)) = x.
x3 + 1
Review: Three steps to find the inverse
• Step 1:
• Step 2:
• Step 3:
6
Here comes the main theorem of this lecture.
Theorem
let f be differentiable and strictly monotonic on some certain domain. If f 0 (x) 6= 0 at a certain x in I, then f −1 is also differentiable at the
corresponding point y = f (x) in the range of f and
(f −1 )0 (y) =
1
f (x)
Which is often written symbolically as
dx
1
=
dy
dy
dx
Interpretation (A way to memorize the theorem):
Example 3
Let f (x) = 3x5 + x − 2, find f −1 (2).
7
Section 6.3: The natural Exponential Function
Definition: The inverse of ln x is called the natural exponential function,denoted by exp x.
Thus, we have the following relation:
x=exp y ⇐⇒ y=ln x
Since y = ln x and y = exp x are inverses of each other, it can be observed
immediately that :
• ln(exp x) = x
• exp(ln x) = x
Definition: The letter e denotes the unique
number such that ln e = 1
Fact:e ≈
Key Observation:
er = exp(ln er ) = exp(r ln e) = expr
From the above analysis, we know that ln x and ex are inverse functions of
each other, we have:
(1) eln x = x
(2) ln(ex ) = x
Properties
(1)ea eb = ea+b
ea
= ea−b
eb
d
(3) ex = ex
dx
(2)
8
Examples
(4) Evaluate the following integrals
ln v 2014 −ln
(1) simplify e
(2) Find Dx (ex
(3) Find
2 cos(√x)
sin3 v
ln v 801
(i)
)
(ii)
dy
if 2y = exy − 2 tan x
dx
R1
0
R
x4 e−7x
5 +1
dx
2ex
dx
ex − 1
(5) let f (x) =
lnx
for x ∈
1 + (lnx)2
(0, ∞), find:
(a) limx→0+ f (x)and limx→∞ f (x)
(b) the maximum and minimum
values of f(x)
9
Section 6.4: General Exponential and Logarithmic Functions
1. Observations: ax =
2. Definition: For a > 0 and any real number x.
ax = exlna
3. Definition of the inverse function:
y = loga x ⇐⇒ x = ay
4.Properties of Exponents:
If a > 0, b > 0, and x and y are real numbers,
(i) ax ay = ax+y
(ii)
ax
= ax−y
ay
(iii) (ax )y = axy
(iv) (ab)x = ax bx
a
ax
(v) ( )x = x
b
b
5. Derivatives and Integrals
(1) Dx ax = ax lna
(2)
R
ax dx = (
(3) Dx loga x =
1
)ax + C
lna
1
xlna
10
Examples
(3)
(1) If y = xx , find
R
R
34x−2 dx and (4x − 2)3 dx
dy
dx
√
R 2 x
√ dx
(4)
x
(2) differentiate:
√
(x2 + 3)sin
2x
√
2
+ (sin 2x)x +3
11
Section 6.5: Exponential Growth and Decay
Motivation: We consider a population model, which is the simplest one.
Assumptions
(1) Population as a function of time t is differentiable.
(2) The increase dy in population is proportional to the short time increment
dt and the size of the population at that instant, which is y(t). (dy =
k ∗ dt ∗ y(t))
The associate differential equation is:
dy
= ky(t)
dt
We now solve this ordinary differential equation.
We begin by rewriting the ordinary differential equation as:
dy
= kdt
y
12
Example 1. A bacterial population grows at a rate that is proportional to
its size. Initially, it is 10,000 and after ten days it is 20,000. What is the
population after 25 days?
Example 2 (Theorem). Calculate
1
x
lim (1 + x)
x→0
Example 3. Calculate the following integrals and compare results with Example 2.
(a)
lim (1 + x)100000
x→0
(b)
1
lim 1 x
x→0
13
3
Example 4. Using results of example 2, calculate the following.
(a)
1
lim (1 + 5x) x
x→0
(b)
lim (
n→∞
n+2 n
)
n
14
Section 6.8: The Inverse Trigonometric Functions and Their Derivatives
1. Motivation: Recall that for a function to have an inverse, the function
has to be one-to-one, that is, it has to pass the horizontal line test on some
certain domain. Especially, we know that if a function is monotonic on some
domain, then it has an inverse.
Part I: Basic Relations
2. Definition: Inverse Sine and Inverse Cosine (Review)
π π
• x = sin−1 y ⇐⇒ y = sinx, x ∈ [− , ]
2 2
• x = cos−1 y ⇐⇒ y = cosx, x ∈ [0, π]
π π
• x = tan−1 y ⇐⇒ y = tanx, x ∈ (− , )
2 2
• x = sec−1 y ⇐⇒ y = secx, x ∈ [0, π], x 6= 0
3. Reciprocal Relations:
1
• sec−1 y = cos−1 ( )
y
1
• cot−1 y = tan−1 ( )
y
1
• csc−1 y = sin−1 ( )
y
4. Some useful identities:
√
• sin(cos−1 x) = 1 − x2
√
• cos(sin−1 x) = 1 − x2
√
• sec(tan−1 x) = 1 + x2
• tan(sec−1 x) = sgn(x) ∗
√
x2 − 1, for |x| ≥ 1
15
Examples:
(a) sin−1 (sin
3π
)=
2
π
(b) sin−1 (sin( )) =
4
(c) sec−1 (−1) =
1
(d) tan(2 tan−1 ( )) =
3
Part II: Derivatives
1. Derivatives of the basic functions: (Review)
• Dx (sinx) =
• Dx (cosx) =
• Dx (tanx) =
• Dx (secx) =
• Dx (cscx) =
• Dx (cotx) =
2. Derivatives of the inverse functions (New)
(i) Dx (sin−1 x) = √
1
, x ∈ (−1, 1)
1 − x2
(ii) Dx (cos−1 x) = − √
1
, x ∈ (−1, 1)
1 − x2
1
1 + x2
1
√
(iv) Dx (sec−1 x) =
, |x| > 1
|x| x2 − 1
(iii) Dx (tan−1 x) =
16
Examples:
(1) Dx (xsin−1 (x3 + 2))
√
(2) Dx (e
x tan−1 (x
+ 2sinx))
Part III: Integrals
New relations:
(1)
(2)
(3)
R
√
1
x
dx = sin−1 ( ) + C
a
a2 − x2
R
a2
R
1
x
1
dx = tan−1 ( ) + C
2
+x
a
a
1
1
|x|
√
dx = sec−1 ( ) + C
2
2
a
a
x x −a
Examples:
(a)
Z
2
√
2
dx
√
x x2 − 1
17
(b)
Z
(c)
Z
ex
4 + 9e2x
1
dx
x2 + 8x + 18
Section 6.9: The Hyperbolic Functions and Their Derivatives
Part I: Basic Relations
1. Definition:
sinh x =
ex − e−x
2
cosh x =
ex + e−x
2
tanh x =
sinh x
cosh x
coth x =
cosh x
sinh x
sech x =
1
cosh x
csch x =
1
sinh x
18
Example 1: Show that cosh2 x − sinh2 x = 1
Part II: Derivatives
(i) Dx sinh x=
(ii) Dx cosh x=
(iii) Dx tanh x=
(iv) Dx coth x = −csch2 x
(v) Dx sech x = −sech x tanh x
(vi) Dx csch x = −csch x coth x
Examples:
(1)
(2)
√
Dx cosh3 (ln( 3x + 2))
Z
tanh xdx
19
Part III: Inverse Hyperbolic Functions
1. Basic Relations (You need to know how to derive these relations):
√
(1) sinh−1 x = ln(x + x2 + 1)
√
(2) cosh−1 x = ln(x + x2 − 1), x ≥ 1
1 1+x
ln
, −1 < x < 1
2 1−x
√
1 + 1 − x2
x = ln(
), 0 < x ≤ 1
x
(3) tanh−1 x =
(4) sech−1
2. Derivatives of the inverse hyperbolic functions:
(1) Dx sinh−1 x = √
1
x2
(2) Dx cosh−1 x = √
(3) Dx tanh−1 x =
+1
1
x2
−1
, x>1
1
, −1 < x < 1
1 − x2
1
(4) Dx sech−1 x = − √
,
x 1 − x2
0<x<1
20
Example: Show that Dx sinh−1 x = √
1
x2
21
+1
by two different methods.