Leif Meilbro
Real Funct ions in One Variable
Exam ples of Elem ent ary Funct ions
Calculus 1c- 2
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Real Funct ions in One Variable – Exam ples of Elem ent ary Funct ions – Calculus 1c- 2
© 2007 Leif Mej lbro & Vent us Publishing ApS
I SBN 978- 87- 7681- 237- 9
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Calculus Analyse c1- 2
Contents
Cont ent s
5
1.
Some Functions known from High School
6
2.
The hyperbolic functions
37
3.
Inverse functions, general
42
4.
The Arcus Functions
46
5.
The Area functions
81
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Preface
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4
Calculus Analyse c1- 2
Preface
Preface
In this volume I present some examples of Elementary Functions, cf. also Calculus 1a, Functions of
One Variable. Since my aim also has been to demonstrate some solution strategy I have as far as
possible structured the examples according to the following form
A Awareness, i.e. a short description of what is the problem.
D Decision, i.e. a reflection over what should be done with the problem.
I Implementation, i.e. where all the calculations are made.
C Control, i.e. a test of the result.
This is an ideal form of a general procedure of solution. It can be used in any situation and it is not
linked to Mathematics alone. I learned it many years ago in the Theory of Telecommunication in a
situation which did not contain Mathematics at all. The student is recommended to use it also in
other disciplines.
One is used to from high school immediately to proceed to I. Implementation. However, examples
and problems at university level are often so complicated that it in general will be a good investment
also to spend some time on the first two points above in order to be absolutely certain of what to do
in a particular case. Note that the first three points, ADI, can always be performed.
This is unfortunately not the case with C Control, because it from now on may be difficult, if possible,
to check one’s solution. It is only an extra securing whenever it is possible, but we cannot include it
always in our solution form above.
I shall on purpose not use the logical signs. These should in general be avoided in Calculus as a
shorthand, because they are often (too often, I would say) misused. Instead of ∧ I shall either write
“and”, or a comma, and instead of ∨ I shall write “or”. The arrows ⇒ and ⇔ are in particular
misunderstood by the students, so they should be totally avoided. Instead, write in a plain language
what you mean or want to do.
It is my hope that these examples, of which many are treated in more ways to show that the solutions
procedures are not unique, may be of some inspiration for the students who have just started their
studies at the universities.
Finally, even if I have tried to write as careful as possible, I doubt that all errors have been removed.
I hope that the reader will forgive me the unavoidable errors.
Leif Mejlbro
17th July 2007
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5
Calculus Analyse c1- 2
1
Some Functions known from High School
Some Functions known from High School
Example 1.1 Differentiate each of the following functions:
1) y = ln(2x),
√
2) y = cos x,
3) y = sin2 x,
4) y = sin x2 ,
5) y = x2 ex ,
6) y =
tan x
.
x
A. Simple differentiations.
D. Determine where the function is defined and where it is differentiable. Then apply some wellknown rules of differentiation.
I. 1) The function y = ln(2x) is defined and differentiable for x > 0. Since
ln(2x) = ln 2 + ln x,
we get
dy
1
=
dx
x
for x > 0.
1
0.5
0
20
40
60
80
100
120
140
160
–0.5
–1
Figure 1: The graph of y = cos
2) The function y = cos
√
√
x, x ≥ 0; different scales on the axes.
x is defined for x ≥ 0 and differentiable for x > 0 with
√
dy
1
= − √ sin x.
dx
2 x
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Calculus Analyse c1- 2
Some Functions known from High School
1
0.8
0.6
0.4
0.2
0
0.2
0.4
Figure 2: The graph of y = cos
0.6
0.8
1
(x) in the neighbourhood of x = 0.
One should also check whether the function is differentiable from the right at x = 0, where
y = 1. The difference quotient is given by
√
1−
ϕ(x) − ϕ(0)
cos( x) − 1
=
=
x−0
x
1
2!
√ 2
( x) + · · · − 1
1
= − + ε(x),
x
2
1
for x → 0+. Therefore, we conclude that the function
2
1
has a half tangent at x = 0+, ϕ′ (0+) = − .
2
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7
Calculus Analyse c1- 2
Some Functions known from High School
3) The function y = sin2 x is defined and differentiable for every x ∈ R, and
dy
= 2 sin x cos x = sin(2x).
dx
1
0.5
–3
–2
–1
2
1
3
–0.5
–1
Figure 3: The graph of y = sin(x2 ).
4) The function y = sin(x2 ) is defined and differentiable for every x ∈ R and
dy
= cos x2 · 2x = 2x cos x2 .
dx
2.5
2
1.5
1
0.5
–4
–3
–2
0
–1
1
Figure 4: The graph of y = x2 ex .
5) The function y = x2 ex is defined and differentiable for every x ∈ R, and
dy
= 2x ex + x2 ex = x(x + 2) ex .
dx
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8
Calculus Analyse c1- 2
Some Functions known from High School
4
y
2
–8
–6
–4
–2
0
–2
2
6
4
8
x
–4
Figure 5: The graph of y =
6) The function y =
x∈
/ {0} ∪
tan x
.
x
tan x
is defined and differentiable at least when
x
π
+ pπ
2
p∈Z .
However, since both the numerator and the denominator are 0 for x = 0, we shall look closer
at this point.
π
a) When x = 0 and x = + pπ, p ∈ Z, it follows from the rules of differentiation that
2
dy
1 + tan2 x tan x
x + x tan2 x − tan x
=
− 2 =
dx
x
x
x2
x − sin x cos x
x cos2 x + x sin2 x − sin x cos x
=
.
=
2
2
x cos x
x2 cos2 x
b) When x = 0, we get by the continuity that
ϕ(0) = lim
x→0
tan x
1
sin x
= lim
·
= 1 · 1 = 1.
x→0 cos x
x
x
Then consider the difference quotient
ϕ(x) − ϕ(0)
=
x−0
tan x
x
−1
tan x − x
x2 ε(x)
=
=
= ε(x),
x−0
x2
x2
which converges towards 0 for x → 0. Here we have used that the numerator tan x − x is
an odd function, and that the Taylor expansion starts with 0 · x, so the first true term is of
the form c · x3 = x2 ε(x). We therefore conclude that the function is continuously defined
and also differentiable at x = 0 with the derivative ϕ′ (0) = 0, which looks quite reasonable
when we consider the figure.
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Calculus Analyse c1- 2
Some Functions known from High School
Example 1.2 Sketch the graphs of the following functions,
1) y = cos 2x,
2) y = ln(−x),
3) y = ln(ex ),
4) y = e− ln cot x .
Write a programme in MAPLE, by which the graphs are constructed.
A. Drawing of graphs and a MAPLE programme.
D. Determine the domains and reduce the expressions, whenever it is possible.
1
y
–4
–3
–2
–1
0.5
0
1
–0.5
2
3
4
x
–1
Figure 6: The graph of y = cos 2x.
I. 1) Usually there are several possibilities of writing a programme in MAPLE. Personally I prefer
always to describe a function by using a parameter to describe the function. This may seem a
little complicated, but it is actually the best way of doing it. Here, I suggest in the first case
that we use
plot([t,cos(2∗t),t=-4..4],x=-4..4,y=-1.2..1.2,
scaling=constrained,color=black);
2) The function is defined for x < 0. Here we also have several possibilities of the MAPLE
programme. My suggestion is
plot([t,ln(-t),t=-5..0],x=-5..1,y=-2..2,
scaling=constrained,color=black);
3) The function is defined everywhere. By a reduction we get
y = ln (ex ) = x,
so a simple MAPLE programme (which is not unique here either) is
plot([t,t,t=-1..1],scaling=constrained,color=black);
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10
Calculus Analyse c1- 2
Some Functions known from High School
2
1
–5
–3
–4
–2
y
0
–1
1
x
–1
–2
Figure 7: The graph of y = ln(−x), x < 0.
1
0.5
–0.5
–1
0.5
1
–0.5
–1
Figure 8: The graph of y = ln(ex ) = x.
3
2.5
2
y 1.5
1
0.5
0
0.2 0.4 0.6 0.8
1 1.2 1.4
x
Figure 9: The graph of y = e− ln cot x = tan x in the interval 0,
π
.
2
4) The function y = e− ln cos x is defined when cot x > 0, i.e. for
π
.
pπ, pπ +
x∈
2
p∈Z
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11
Calculus Analyse c1- 2
Some Functions known from High School
In this case the expression is reduced to
y = e− ln cot x =
1
= tan x,
cot x
tan x > 0.
π
we can e.g. use the following MAPLE programme,
2
plot([t,tan(t),t=0..Pi/2-.1],x=0..Pi/2,y=0..3,
scaling=constrained,color=black);
In the interval 0,
π
. (This can
2
also be removed by the command discont=true). Notice that one can get more of the graph
by changing y=0..3 to e.g. y=0..4.
where we have cheated a little in order not to be troubled by the vertical line x =
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Calculus Analyse c1- 2
Some Functions known from High School
Example 1.3 Reduce the following expressions:
1) y = cos2 x + sin2 x,
2) y = cos2 x − sin2 x,
3) y = e− ln x ,
4) y = ln (x ex ) − ln x.
A. Reduction of simple mathematical expressions.
D. Determine the domain and then apply high school mathematics.
I. 1) It is obvious that the expression is defined for every x ∈ R and that
y = cos2 x + sin2 x = 1.
Remark. It is very important for an engineering student to know that (x, y) = (cos t, sin t),
t ∈ R, is a parametric description of the unit circle, run through infinitely often. The most
important movements are the straight movements and the circular movements. ♦
2) It should also be well-known that
y = cos2 x − sin2 x = cos 2x,
for every x ∈ R.
3) The function y = e− ln x is defined when ln x is defined, i.e. for x > 0. In this case we get
y = e− ln x =
1
,
x
for x > 0.
Remark. The trap is of course that one should believe that the function is defined if only
x = 0. This is not true because we have not defined the logarithm of a negative number. ♦
4) The function y = ln(ex ) − ln x is defined for x > 0. In that case we have
y = ln (x ex ) − ln x = ln x + ln (ex ) − ln x = x,
for x > 0.
Example 1.4 Prove the following two formulæ of the derivative of the function
y = tan x.
A. Prove that
d(tan x)
1
=
= 1 + tan2 x.
dx
cos2 x
D. Use the known rules of differentiation on tan x =
sin x
.
cos x
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13
Calculus Analyse c1- 2
Some Functions known from High School
I. When this is done we get for x =
d(tan x)
dx
=
d
dx
sin x ·
π
+ pπ, p ∈ Z, that
2
1
cos x
= cos x ·
− sin x
1
+ sin x · −
cos x
cos2 x
= 1 + tan2 x
cos2 x + sin2 x
1
sin2 x
=
=
.
= 1+
2
cos x
cos2 x
cos2 x
Example 1.5 For the two angles u and v we introduce the vectors
e = (cos u, sin u)
and
f = (cos v, sin v).
We can express the scalar product of e and f in two ways: Either by means of coordinates, or by taking
the length of e multiplied by the length (positive or negative) of the projection of f onto the direction
defined by e. Apply this to prove an addition formula for trigonometric functions.
A. Derivation of an addition formula.
D. Consider a scalar product of two unit vectors in two different ways.
1
0.5
–1
–0.5
0.5
1
–0.5
–1
Figure 10: The vectors e and f with the angle v − u between them.
I. In rectangular coordinates the scalar product of the two vectors is given by
e·f
= (cos u, sin u) · (cos v, sin v)
= cos u · cos v + sin u · sin v.
The angle between the vectors e and f calculated from e towards v is given by v − u, hence the
projection of the unit vector f onto the direction given by e is cos(v − u).
By an identification we therefore get
cos(v − u) = cos u · cos v + sin u · sin v.
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14
Calculus Analyse c1- 2
Some Functions known from High School
If we in this formula replace u by −u, we get
cos(v + u) = cos u · cos v − sin u · sin v.
Example 1.6 For two angles u and v we introduce the vectors
e = (cos u, sin u)
and
f = (cos v, sin v).
Prove an addition formula for trigonometric functions by first calculating the scalar product ˆ
e · f in
two ways: Either by means of rectangular coordinates, or by taking the product of the length of one of
the vectors and the signed length of the projection of the second vector onto the direction of the first
one.
A. A trigonometric addition formula.
D. First find the coordinates of the vector ˆ
e. Then use the description above to calculate the scalar
product in two different ways and identify the coordinates.
1
0.5
–1
–0.5
0.5
1
–0.5
–1
Figure 11: The vectors e and f , and the vector ˆ
e, which is obtained by turning e the angle
π
.
2
I. When e = (cos u, sin u), we get
ˆ
e = (− sin u, cos u),
(interchange the coordinates and then change the sign on the first coordinate). Then the inner
product becomes
ˆ
e · f = (− sin u, cos u) · (cos v, sin v) = cos u · sin v − sin u · cos v.
π
On the other hand, the signed angle between ˆ
e and f is given by v − u − , i.e. the projection of
2
the unit vector f onto the line determined by the direction ˆ
e is
cos v − u −
π
2
= cos
π
− (v − u) = sin(v − u).
2
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15
Calculus Analyse c1- 2
Some Functions known from High School
When these two expressions are identified we get
sin(v − u) = sin v · cos u − cos v · sin u.
Finally, when u is replaced by −u we get
sin(u + v) = sin u · cos v + cos u · sin v,
and we have proved the addition formula.
Example 1.7 Prove that
tan
v
sin v
=
,
2
1 + cos v
v = π + 2pπ.
A. Proof of a trigonometric formula.
D. Express e.g. the right hand side by half of the angle and reduce.
I. First note that both sides of the equality sign is defined, if and only if
v = π + 2pπ,
p ∈ Z.
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16
Calculus Analyse c1- 2
Some Functions known from High School
Then calculate the right hand side by changing variable to the half angle,
v
v
v
v
v
2 sin cos
2 sin · cos
sin
sin v
2
2
2
2
2 = tan v .
=
=
=
v
v
v
v
1 + cos v
2
1 + cos2 − sin2
2 cos2
cos
2
2
2
2
Example 1.8 Sketch in the same coordinate system the functions f (x) = a x for a = 2, a = 3 and
a = 4. It should in particular be indicated when some graph lies above another one.
A. Graph sketches of exponentials.
D. Write a suitable MAPLE programme.
16
14
12
10
8
6
4
2
–2
–1
1
2
x
Figure 12: The graphs of f (x) = ax for a = 2, 3 and 4. Different scales on the axes.
I. Here I have used the following MAPLE programme:
plot({2^x,3^x,4^x},x=-2..2,color=black);
Every graph goes through (0, 1). To the left of the y-axis we have
4x < 3x < 2x ,
for x < 0,
and to the right of the y-axis we have instead
2x < 3x < 4x ,
for x > 0.
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17
Calculus Analyse c1- 2
Some Functions known from High School
1
Example 1.9 Sketch in the same coordinate system the functions f (x) = x α , x ≥ 0, for α = ,
2
α = 2 and α = 3. Indicate in particular when some graph lies above another one.
A. Graph sketches of power functions.
D. Write a suitable MAPLE programme.
8
6
4
2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
x
Figure 13: The graphs of f (x) = xα , x ≥ 0, for α =
1
, 2 and 3. Different scales on the axes.
2
I. The following MAPLE programme has been applied:
plot({sqrt(x),x^2,x^3},x=0..2,color=black);
Every graph goes through (0, 0) and (1, 1). In the interval ]0, 1[ we have
√
x3 < x2 < x,
for x ∈ ]0, 1[,
and when x > 1, we have instead
√
for x > 1.
x < x2 < x3 ,
Example 1.10 Given three positive numbers a, r and s such that
ar+s = 128,
ar−s = 8,
ars = 1024.
Find the numbers a, r and s.
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18
Calculus Analyse c1- 2
Some Functions known from High School
A. Three nonlinear equations in three unknowns.
D. Apply the logarithm on all three equations and solve the new equations.
I. Since 128 = 27 and 8 = 23 and 1024 = 210 , we get by taking the logarithm of the three given
equations that
⎧
⎨ (r + s) ln a = 7 ln 2,
(r − s) ln a = 3 ln 2,
⎩
rs ln a = 10 ln 2.
It follows from the first two equations by an addition and a subtraction, etc. that
r ln a = 5 ln 2,
s ln a = 2 ln 2.
When these expressions are inserted into the last equation we get
10 ln 2 = rs ln a = s(r ln a) = s · 5 ln 2,
i.e. s = 2;
10 ln 2 = rs ln a = r(s ln a) = r · 2 ln 2. i.e. r = 5.
Finally, it follows e.g. from r ln a = 5 ln 2 and r = 5 that ln a = ln 2, thus a = 2. Hence we have
found that
a = 2,
r = 5,
s = 2.
C. Check. When a = 2 and r = 5 and s = 2, we get by insertion into the original equations that
ar+s
ar−s
ars
= 25+2
= 25−2
= 25·2
= 27
= 23
= 210
= 128,
=
8,
= 1024.
We see that all three equations are fulfilled.
Example 1.11 Given three positive numbers a and b and r such that
(ab)r = 3,
a−r =
1
,
2
1
a r = 16.
Find the numbers a and b and r.
A. Three nonlinear equations in three unknowns, which all must be positive.
D. Take the logarithm and solve the new equations.
I. By taking the logarithm of the three equations we get
⎧
ln 3 = r ln(ab) = r ln a + r ln b,
⎪
⎪
⎨
1
ln
= −r ln a,
2
⎪
⎪
1
⎩
ln 16 =
ln a,
r
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19
Calculus Analyse c1- 2
Some Functions known from High School
which we rewrite as
⎧
⎪
⎨ r ln a + r ln b = ln 3,
r ln a = ln 2,
⎪
⎩ 1 ln a = ln 16 = 4 ln 2.
r
When we divide the latter equation into the former equation, we get
r ln a
1
ln 2
= .
= r2 =
1
4 ln 2
4
ln a
r
Since r > 0 according to the assumptions, we get
r=
1
,
2
hence by an insertion into the second equation,
1
ln a = ln 2, hvoraf a = 4.
2
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Calculus Analyse c1- 2
Some Functions known from High School
Finally we get from the first equation that
ln 2 +
1
ln b = ln 3,
2
hence
9
ln b = 2{ln 3 − ln 2} = ln ,
4
dvs. b =
9
.
4
As a conclusion the solution is
a = 4,
b=
9
,
4
r=
1
.
2
C. Test. By insertion of these values into the original equations we get
(ab)r =
4·
9
4
1
1
2
1
= 9 2 = 3,
a−r = 4− 2 = 2−1 =
1
a r = 42 = 16.
1
,
2
It is seen that all three original equations are fulfilled.
Remark. It can be proved that if one does not require that a and b and r are all positive, then we
get another solution,
a=
1
,
4
b=
4
,
9
1
r=− .
2
The proof of this claim is left to the reader. ♦.
Example 1.12 Prove the three power rules
ar+s = ar · as ,
(ab)r = ar · br ,
s
(ar ) = ars ,
by assuming the logarithmic rules.
A. We shall prove three rules for power functions, where we assume that the rules of logarithm hold,
and that the function ln : R+ → R is continuous and strictly increasing.
D. Set up the rules of logarithm and derive the power rules.
I. We can use the following three rules,
I ln(a, b) = ln a + ln b, for a, b > 0.
a
II ln
= ln a − ln b, for a, b > 0.
b
III ln (ar ) = r ln a, for a > 0 and r ∈ R.
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21
Calculus Analyse c1- 2
Some Functions known from High School
1) By applying rule III twice and at the last equality using rule I we get
ln ar+s
= (r + s) ln a = r ln a + s ln a
= ln (ar ) + ln (as ) = ln (ar · as ) .
Since ln is one-to-one, we get by the exponential that
ar+s = ar · as .
2) Analogously we get by applying the rules III, I, III, I that
ln ((ab) r) = r · ln(ab) = r{ln a + ln b} = r ln a + r ln b
= ln (ar ) + ln (br ) = ln (ar · br ) .
Since ln is one-to-one we get by the exponential that
(ab)r = ar br .
3) Finally we get by using rule III at every equality sign that
s
ln ((ar ) ) = s ln (ar ) = rs ln a = ln (ars ) .
Since ln is one-to-one we get by the exponential that
s
(ar ) = ars .
Example 1.13 Given a positive number a and a natural number n. Then a n can be defined in two
ways: Either as the product of a with itself n times, or by an = en ln a . Explain why the two definitions
give the same result.
A. Two apparently different definitions should give the same.
D. Use the rules of the logarithm.
I. Since
ln(ab) = ln a + ln b
for a, b > 0,
we get for b = a,
ln a2 = ln(a · a) = 2 ln a,
hence the claim is true for n = 1 and for n = 2, where a > 0 is any number. This indicates that
we should try
Induction. Assume that
ln (an ) = n ln a
for some n ∈ N.
This assumption has been proved to be true for n = 1 and for n = 2.
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22
Calculus Analyse c1- 2
Some Functions known from High School
Then we get for the successor,
ln an+1 = ln (an · a) = ln (an ) + ln a = n ln a + ln a = (n + 1) ln a,
and we conclude that the formula follows by induction.
We have now proved that
ln (an ) = n ln a = ln en ln a .
Since ln : R+ → R is bijective, we have
an = a · · · a = en ln a ,
and it follows that the two definitions agree for every n ∈ N.
Example 1.14 Investigate in each of the following cases if the claim is correct or wrong:
1) (xy)z = xy z ,
2) xy = ey ln x ,
3) ln(a − b) =
ln a
,
ln b
4) xy+z = xy + xz ,
5) sin(x + y) = sin x + sin y,
6) (a + b)2 = a2 + b2 ,
7) sin v = 2 cos2
v
− 1,
2
8)
xα dx =
xα+1
,
α+1
9)
2x dx =
2x+1
,
x+1
10)
1
dx = − cot x.
sin2 x
A. The formulation of this example is on purpose very sloppy, because this is more or less how the
students’ exercises are formulated without any “proof”. We shall find out if some given “formulæ’
are correct or not. The solutions will not be too meticulous, because that would demand a lot
more.
D. If one of the formulæ is correct, it should of course be proved, and its domain should be specified.
If some formula is wrong, one should give a counterexample. This part is a little tricky because the
formula may be right for carefully chosen x, y and z. In particular, one cannot give some general
guidelines for how to do it.
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23
Calculus Analyse c1- 2
Some Functions known from High School
I. 1) The claim is in general wrong. If we e.g. choose x = y = z = 2, we get
(xy)z = (2 · 2)2 = 16,
xy z = 2 · 22 = 8 = 16 = (xy)z .
Notice however that the formula is correct if z = 1, or if x = 1 (or x = 0).
2) The claim is true for x > 0.
3) The claim is wrong. First notice that its potential domain is given by 0 < b < a. If we here
e.g. choose b = a − 1 for any a > 1, we get
ln(a − b) = ln 1 = 0 and
ln a
= 0, when b = 1.
ln b
4) The claim is wrong. Choosing e.g. x = y = z = 1, we get
xy+z = 11+1 = 1 and xy + xz = 11 + 11 = 2 = 1 = xy+z .
It follows by continuity that the formula is not correct in some open domain containing (1, 1, 1).
π
5) The claim is wrong. If we choose x = y = we get
2
sin(x + y) = sin π = 0 and
sin x + sin y = 1 + 1 = 2 = 0 = sin(x + y).
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Calculus Analyse c1- 2
Some Functions known from High School
6) The claim is wrong when both a = 0 and b = 0. In fact,
(a + b)2 = a2 + b2 + 2ab,
and we see that the additional term 2ab = 0, when a = 0 and b = 0.
Remark. A very frequent error made by the students is to put (a + b)2 equal to a2 + b2 , which
is not correct. ♦
7) The claim is wrong. The left hand side sin v is an odd function = 0, and the right hand side is
an even function = 0. The only function, which is both odd and even is 0.
8) This claim is correct for α = −1. In fact,
d
dx
xα+1
α+1
α+1 α
· x = xα .
α+1
=
When α = −1, the right hand side is not defined (never divide by 0). It is well-known that
x−1 dx =
1
dx = ln |x|,
x
for x = 0.
9) The claim is wrong which can be seen by at small test,
d
dx
2x+1
x+1
=
=
=
d
dx
e(x+1) ln 2
x+1
e(x+1) ln 2
ln 2 · e(x+1) ln 2
−
x+1
(x + 1)2
x+1
2
{(x + 1) ln 2 − 1} = 2x .
(x + 1)2
10) The claim is correct for x = pπ, p ∈ Z. In fact, we get by a test
d
d
{− cot x} = −
dx
dx
cos x
sin x
=−
− sin2 − cos2 x
sin2 x
=
1
.
sin2 x
Example 1.15 Prove that
1) cos 2x = 2 cos2 x − 1 = 1 − 2 sin2 x,
2) sin 2x = 2 sin x cos x.
A. Two simple applications of the rules of calculations.
D. Apply the rules with y = x and the sign +.
I. We shall also need the trigonometric fundamental equation
cos2 x + sin2 x = 1.
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