Download Chapter 2 Some functions and their derivatives

Chapter 2
Some functions and their derivatives
2.1
Derivative of xn for integer n
Recall, from eqn (1.6), for y = f (x),
f (x + δx) ¡ f (x)
dy
= lim
.
dx δx→0
δx
Also recall that, for integer n,
(a + b)n = an + nan−1 b +
n (n ¡ 1) n−2 2
a b + . . . + bn .
2!
Hence, if y = xn then
dy
=
dx
(x + δx)n ¡ xn
δx→0
δx
³
lim
xn + nxn−1 δx +
=
lim
δx→0
n−1
=
n(n−1) n−2
x δx2
2
lim
nx
δx +
δx
n(n−1) n−2
x δx2
2
´
+ . . . ¡ xn
+ ...
δx
·
¸
n
(n
¡ 1) n−2
n−1
= lim nx
x δx + . . .
+
δx→0
2
δx→0
dy
= nxn−1 .
dx
2.2
Trigonometric Functions
See Appendix C for further details.
10
2.2.1
Definition and Graphs of Sine and Cosine
The two basic trigonometric functions are the sine and cosine. From an angle θ, we
may define cos θ and sin θ geometrically. Given the point P in the (x, y) plane such that
OP is of length 1 and makes an anticlockwise angle θ with the positive x axis, then
cos θ = x-coordinate of P.
sin θ = y-coordinate of P.
In the figure, 0 < θ <
sin θ =
π
2
and, for the right-angled triangle OAP,
length of side opposite to θ
,
length of hypotenuse
cos θ =
length of side adjactent to θ
.
length of hypotenuse
In calculus, and THROUGHOUT THIS COURSE, all angles will be measured
in RADIANS. In the picture, the angle θ in radians is defined to be the length of the
arc from the point (1, 0) to P of the circle of radius 1 centre O.
Our geometrical definitions of cos θ and sin θ make sense not just for 0 < θ < π/2 as
in the picture but for all real values of θ of either sign. The graphs of these two functions
in the range (¡3π, 3π) are:
11
By considering the limiting cases of the previously illustrated right-angled triangle
when θ = 0 and when θ = π/2, we immediately deduce (in accordance with the graphs
just drawn) that
sin 0
π
cos
2
cos 0
π
sin
2
2.2.2
= 0,
(2.1)
= 0,
(2.2)
= 1,
(2.3)
= 1.
(2.4)
Inverse Trigonometric Functions
From the form of the graph of the sine function, we see that it cannot have an inverse
as it stands, since each horizontal line y = c (with ¡1 < c < 1) cuts the graph in
infinitely many points. We restrict it to an interval on which it is an increasing function,
and the conventional choice for that interval is [¡π/2, π/2]. The increasing restriction
12
£
£
¤
¤
sin : ¡ π2 , π2 ! [¡1, 1] has increasing inverse sin−1 = arcsin : [¡1, 1] ! ¡ π2 , π2 . By
definition, given any y in [¡1, 1], sin−1 y is the unique angle between ¡π/2 and π/2 whose
sine is y.
NB. It is totally different from (sin y)−1 =cosec y.
NB The graph of sin−1 y has vertical tangents at y = §1, so the slope
these two points.
dx
dy
is infinite at
Question: Over what restriction will cos have an inverse, arccos?
Answer: We restrict cos to be either increasing or decreasing. It is usual to choose the
decreasing restriction cos : [0, π] ! [¡1, 1], with inverse arccos : [¡1, 1] ! [0, π].
13
As for arcsin, given any y in [¡1, 1], cos−1 (y) = arccos y is the unique angle between 0 and π
whose cosine is y.
tan and arctan
14
The function tan is increasing over many intervals, but we conventionally take the interval
containing x = 0: tan : (¡π/2, π/2) ! R with inverse arctan : R ! (¡π/2, π/2).
15
2.2.3
Some identities and a useful limit
NB. An identity is an equation which holds for ALL permitted values of its variables.
See Appendix C for proofs of many trig. identities. Three which will be used in the next
section are
sin(x + y) = sin x cos y + cos x sin y,
cos 2x = 1 ¡ 2 sin2 x
cos2 x + sin2 x = 1.
Appendix C4 also contains the proof of a useful limit:
sin x
lim
= 1,
x→0 x
i.e. sin x ¼ x for small angles x.
2.2.4
(2.5)
(2.6)
(2.7)
(2.8)
Derivatives of the trigonometric functions
By definition (see section 1.4), the derivative of sin x is
d
(sin x) =
dx
=
=
=
=
=
sin(x + δx) ¡ sin x
δx→0
δx
sin x cos δx + cos x sin δx ¡ sin x
lim
using (2.5)
δx→0
δx
sin x (cos δx ¡ 1) + cos x sin δx
lim
δx→0
δx
2
¡2 sin x sin (δx/2) + cos x sin δx
using (2.6)
lim
δx→0
δx
·
¸2
δx sin δx/2
sin δx
¡ sin x lim
+ cos x lim
δx→0 2
δx→0
δx/2
δx
cos x
using (2.8).
lim
This is eqn B.4 of Appendix B. There is a similar proof that
d
(cos x) = ¡ sin x.
dx
Derivatives of other trig functions may be derived from those of sin and cos. e.g,
using the quotient rule:
µ
¶
d
d sin x
(tan x) =
dx
dx cos x
cos x (cos x) ¡ sin x(¡ sin x)
=
cos2 x
2
cos x + sin2 x
=
cos2 x
1
using (2.7)
=
cos2 x
= sec2 x.
One similarly deduces that
d
(cot x) = ¡cosec2 x.
dx
The remaining derivatives are all given in Appendix B.
16
2.2.5
Derivatives of the inverse trigonometric functions
Derivatives of inverse functions To find the derivatives of inverse functions, we use
the result
dy
1
dx
= ³ ´ , provided
6= 0
dx
dx
dy
dy
which follows directly from the definition of the derivative.
y · π/2 and ¡1 < x < 1
Inverse Sine Set y = sin−1 x and x = sin y (with ¡π/2 · p
p
2
2
) cos y ¸ 0). Using sin y + cos y = 1, we have cos y = + 1 ¡ sin2 y = + 1 ¡ x2 .
Hence
d
dy
(sin−1 x) =
dx
dx
1
= ³ ´
dx
dy
= ³
=
1
´
d
(sin y)
dy
1
cos y
and so
d
1
(sin−1 x) = p
(¡1 < x < 1).
dx
1 ¡ x2
Note that this ! +1 as x ! §1.
Inverse Cosine Using an exactly similar argument we find that
d
1
(¡1 < x < 1),
(cos−1 x) = ¡ p
dx
1 ¡ x2
which also ! ¡1 as x ! §1.
Inverse Tan x = tan y )
dx
dy
= sec2 y = 1 + tan2 y = 1 + x2 , so
1
1
d
for any real x.
(tan−1 x) = dx =
dx
1 + x2
dy
2.3
2.3.1
The Exponential and Logarithm Functions
The Exponential Function
Considering the unique function y(x) (defined for all real x) satisfying the condition
dy
= y(x) for all real x,
dx
[NB this is a differential equation cf MATH1970 next semester.]
17
(2.9)
Repeatedly differentiating (2.9), we see that
d2 y
dn y
dy
=
y,
and
more
generally
=
= y for all positive integers n.
dx2
dx
dxn
It follows (from Taylor’s Theorem - which you haven’t met yet!) that a function satisfies
condition (2.9) if and only if
¸
·
∞
X
xn
x2 x3
+
+ ¢ ¢ ¢ = y(0)
for all real x,
(2.10)
y(x) = y(0) 1 + x +
2!
3!
n!
n=0
where n! = n(n ¡ 1)(n ¡ 2) . . . 3.2.1 is the product of the first n positive integers (called
“factorial n”) and y(0) is the value of y(x) at x = 0.
For now, you can satisfy yourself that (2.10) “works” by differentiating it:
·
¸
dy
d
x2 x3
= y(0)
1+x+
+
+¢¢¢
dx
dx
2!
3!
¸
·
2x 3x2 4x3
+
+
¢¢¢
= y(0) 0 + 1 +
2!
3!
4!
¸
·
x2 x3
+
+ ¢ ¢ ¢ = y(x).
= y(0) 1 + x +
2!
3!
It turns out that the sum of infinitely many terms is nevertheless finite in value for any
real x. [It is an example of a Taylor series, of which more later on in this course.]
Definition of the exponential function We define the exponential function exp(x)
to be the unique function (defined for all real x) satisfying the two conditions
d
[exp(x)] = exp(x) (for all real x),
dx
exp (0) = 1.
(2.11)
(2.12)
The unique function exp(x) satisfying these is given explicitly by
exp(x) = 1 + x +
X xn
x2 x3
+
+¢¢¢ =
.
2!
3!
n!
n=0
∞
(2.13)
This definition:
² allows the numerical evaluation of exp (x) for all real x.
² allows one to show (see Appendix D.1) that:
exp (x + y) = exp (x) exp (y) ,
exp (x) > 0,
1
exp (¡x) =
exp (x)
Given exp (x) > 0 it follows from (2.11) that
d
[exp(x)]
dx
> 0 and so
exp (x) is a strictly increasing function, i.e. x > y ) exp(x) > exp(y).
18
(2.14)
From 2.13 we can see that
exp(x) ! +1 as x ! +1,
and so, given exp (¡x) =
(2.15)
1
,
exp(x)
exp(x) ! 0 as x ! ¡1.
(2.16)
So, the graph of the function exp(x) looks like:
2.3.2
The Logarithmic Function
Since exp : R ! (0, 1) is a strictly increasing function, it must have a (unique) strictly
increasing inverse ln : (0, 1) ! R called the logarithmic function or natural logarithm.
For x 2 R, y 2 (0, 1),
y = exp(x) if and only if x = ln y.
(2.17)
Taking x = 0 in (2.17) and using exp (0) = 1, we see that
ln 1 = 0.
(2.18)
Similarly it follows from (2.15), (2.16) and (2.17) that
ln x ! +1 as x ! +1,
ln x ! ¡1 as x ! 0.
Properties (2.18)—(2.20) all show up clearly on the graph of ln:
19
(2.19)
(2.20)
2.3.3
Definition of an Arbitrary Real Power of a Positive Real
Number
We know
exp(x1 + x2 ) = exp(x1 ) exp(x2 )
(2.21)
exp(x1 + x2 + ¢ ¢ ¢ + xn ) = exp(x1 ) exp(x2 ) . . . exp(xn ),
(2.22)
so we see
where n is any positive integer and x1 , x2 , . . . , xn any real numbers. Taking x1 = x2 =
¢ ¢ ¢ = xn = ln x, where x > 0, and using (2.17), we get
exp(n ln x) = [exp(ln x)]n = xn ,
for any positive integer n.
We can define an arbitrary real power xα of a positive real number x by
xα = exp (α ln x)
(x > 0, α real),
(2.23)
equivalent [by (2.17)] to
ln(xα ) = α ln x
(x > 0, α real).
(2.24)
The definition (2.23) satisfies the standard rules for multiplication and division of
powers:
xα xβ = xα+β ,
xα
= xα−β ,
xβ
for all real α and β.
20
2.3.4
The Number e, and the Exponential Function as a Power
We define the positive real number e (whose value is approximately 2.718) by
e = exp 1,
(2.25)
ln e = 1.
(2.26)
from which
Using (2.23),
ex = exp (x ln e) = exp (x)
i.e.
exp x = ex for all real x.
2.3.5
(2.27)
Derivatives
The exponential function By definition.
d
[exp(x)] = exp(x).
dx
(2.28)
The logarithmic function y = ln x, and so x = exp (y) . Then
1
1
1
d
[ln(x)] = dx =
=
dx
exp (y)
x
dy
(for x > 0).
(2.29)
Arbitrary real powers From the above definition of xα we have
d
d α
(x ) =
(exp (α ln x))
dx
dx
= exp (α ln x)£ α £
= αxα
1
x
(using the chain rule)
1
x
= αxα−1 .
Thus
d α
(x ) = αxα−1
dx
NB there is a similar trick for
(α real, x > 0),
(2.30)
d x
d
(a ) =
(exp (x ln a)) = . . .
dx
dx
2.4
The Hyperbolic Functions
These are constructed from the exponential function, but obey identities closely analogous
to those satisfied by the corresponding trigonometric functions (obtained by dropping the
final “h”).
21
2.4.1
Definitions
cosh x =
sinh x =
sinh x
cosh x
cosh x
1
=
coth x =
tanh x
sinh x
1
sech x =
cosh x
1
cosech x =
sinh x
tanh x =
2.4.2
=
=
=
=
1 x
(e + e−x ),
2
1 x
(e ¡ e−x ),
2
ex ¡ e−x
1 ¡ e−2x
e2x ¡ 1
,
=
=
ex + e−x
1 + e−2x
e2x + 1
ex + e−x
1 + e−2x
e2x + 1
=
=
ex ¡ e−x
1 ¡ e−2x
e2x ¡ 1
2
,
x
e + e−x
2
(x 6= 0).
x
e ¡ e−x
(2.31)
(2.32)
(2.33)
(x 6= 0),(2.34)
(2.35)
(2.36)
The Fundamental Identity
Corresponding to the well-known identity cos2 x + sin2 x = 1 for the trigonometric functions, we have for the hyperbolic functions the identity
cosh2 x ¡ sinh2 x = 1
(2.37)
(holding for all real x). NB the minus sign! To prove (2.37), just substitute the definitions
(2.31) and (2.32) into the LHS, to get
1 x −x 2 1 x −x 2 1 2x −2x
1
(e +e ) ¡ (e ¡e ) = (e +e +2ex e−x )¡ (e2x +e−2x ¡2ex e−x ) = ex e−x = ex−x = e0 = 1.
4
4
4
4
2.4.3
Further Identities
1 ¡ tanh2 x
coth2 x ¡ 1
sinh(x + y)
sinh(x ¡ y)
cosh(x + y)
cosh(x ¡ y)
sinh 2x
cosh 2x
=
=
=
=
=
=
=
=
cosh2 x =
sinh2 x =
sinh x cosh y =
cosh x cosh y =
sinh x sinh y =
sech2 x,
cosech2 x,
sinh x cosh y + cosh x sinh y,
sinh x cosh y ¡ cosh x sinh y,
cosh x cosh y + sinh x sinh y,
cosh x cosh y ¡ sinh x sinh y,
2 sinh x cosh x,
sinh2 x + cosh2 x,
1
(cosh 2x + 1),
2
1
(cosh 2x ¡ 1),
2
1
[sinh (x + y) + sinh (x ¡ y)] ,
2
1
[cosh (x + y) + cosh (x ¡ y)] ,
2
1
[cosh (x + y) ¡ cosh (x ¡ y)] .
2
22
(2.38)
(2.39)
(2.40)
(2.41)
(2.42)
(2.43)
(2.44)
(2.45)
(2.46)
(2.47)
(2.48)
(2.49)
(2.50)
Each of these has a trigonometric counterpart (see Appendix C), but there are some
important differences of sign, which should be carefully noted. The identities are all easy
to prove from the definitions above.
2.4.4
The Graphs of cosh, sinh and tanh
Using the definitions of the hyperbolic functions, and our knowledge of the exponential function, we can sketch the following graphs and deduce some properties for the
hyperbolic functions.
cosh 0 = 1,
cosh x ! +1 as x ! §1,
cosh (¡x) = cosh (x) (i.e. it is an EVEN function).
23
sinh 0
sinh x
sinh x
sinh (¡x)
=
!
!
=
0,
+1 as x ! +1,
¡1 as x ! ¡1,
¡ sinh (x) (i.e. it is an ODD function).
24
tanh 0
tanh x
tanh x
tanh (¡x)
=
!
!
=
0,
+1 as x ! +1,
¡1 as x ! ¡1,
¡ tanh (x) (i.e. it is an ODD function).
Like all even functions, cosh has a a graph which is symmetrical about Oy, i.e. invariant under reflection in Oy. It has a minimum value of 1 at x = 0, where its derivative
sinh x is zero. On the other hand, the graphs of the functions sinh and tanh, like those of
all odd functions, are invariant under rotation through 180◦ about O. Note that sinh and
tanh are both increasing functions and that the graph of tanh has horizontal asymptotes
y = §1 approached as x ! §1 respectively.
2.4.5
Symmetry and Asymptotic Properties of other hyperbolic
functions
Similarly, we can deduce the following properties of coth, sech and cosech:
coth(¡x) = ¡ coth x, sech (¡x) = sechx, cosech (¡x) = ¡cosechx.
coth x ! 1 as x ! +1,
coth x ! +1 as x ! 0 from above ,
sech x ! 0
cosech x ! 0
sech 0 = 1
cosech x ! +1 as x ! 0 from above,
2.4.6
coth x ! ¡1 as x ! ¡1,
coth x ! ¡1 as x ! 0, from below,
as x ! §1,
as x ! §1,
cosech x ! ¡1 as x ! 0 from below.
Inverse Hyperbolic Functions
cosh
We restrict cosh x to the interval [0, 1), on which it is an increasing function with
increasing inverse cosh−1 : [1, 1) ! [0, 1). For y ¸ 0, x ¸ 1, we have y = cosh−1 x ,
x = cosh y. cosh−1 x may be expressed in terms of logarithms as follows.
If y = cosh−1 x (x ¸ 1) then y is the positive solution of the equation x = cosh y,
that is
¢
1¡ y
x=
e + e−y .
2
y
Multiplying through by 2e gives
2xey = e2y + 1
e2y ¡ 2xey + 1 = 0
(ey )2 ¡ 2xey + 1 = 0
a quadratic for ey , with roots
p
p
4x2 ¡ 4
ey =
= x § x2 ¡ 1.
´
³ 2p
so, y = ln x § x2 ¡ 1 .
2x §
25
But we want y > 0, so we choose the + sign, i.e.
³
´
p
for x ¸ 1.
y = cosh−1 x = ln x + x2 ¡ 1
(2.51)
sinh
sinh: R ! R is an increasing function, so there is no need for restriction. By similar
arguments to those used for cosh, one shows that, for any x,
´
³
p
(2.52)
sinh−1 x = ln x + 1 + x2 .
tanh
tanh : R ! (¡1, 1) is an increasing function, so there is no need for restriction. It is easy
to show that
µ
¶
1+x
1
−1
tanh x = ln
for ¡ 1 < x < 1.
(2.53)
2
1¡x
2.4.7
Derivatives
The Derivatives of the Hyperbolic Functions We know:
d x
d −x
(e ) = ex ,
(e ) = ¡e−x
dx
dx
Hence
·
¸
¢
d 1¡ x
d
−x
(cosh x) =
e +e
=
dx
dx 2
·
¸
¢
d
d 1¡ x
−x
(sinh x) =
e ¡e
=
dx
dx 2
¢
1¡ x
e ¡ e−x = sinh x,
2
¢
1¡ x
e + e−x = cosh x.
2
By the quotient rule we deduce that
µ
¶
d
1
d sinh x
cosh x cosh x ¡ sinh x sinh x
=
= sech2 x,
(tanh x) =
=
2
2
dx
dx cosh x
cosh x
cosh x
d
1
d
sinh x
(sech x) =
(cosh x)−1 = ¡ (cosh x)−2 sinh x = ¡
= ¡sech x tanh x,
dx
dx
cosh x cosh x
and one may similarly prove (provided x 6= 0) that
d
(coth x) = ¡cosech2 x,
dx
d
(cosech x) = ¡cosech x coth x.
dx
(see Appendix B.)
Derivatives of Inverse Hyperbolic Functions
26
Inverse Cosh Recall y = cosh−1 x , x = cosh y with y ¸ 0, x ¸ 1. So, sinh y ¸ 0
and hence
Then
i.e.
sinh2 y = cosh2 y ¡ 1
q
p
sinh y = + cosh2 y ¡ 1 = + x2 ¡ 1.
1
1
1
dy
= dx =
= +p
2
dx
sinh y
x ¡1
dy
d
1
for x > 1.
(cosh−1 x) = p 2
dx
x ¡1
You can also prove this by differentiating eqn (2.51) directly.
Inverse Sinh Similarly
d
1
.
(sinh−1 x) = p
dx
1 + x2
Inverse Tanh y = tanh−1 x ) x = tanh y, so we have
dx
= sech2 x = 1 ¡ tanh2 x = 1 ¡ y 2 ,
dy
so
d
1
1
(tanh−1 x) = dx =
for ¡ 1 < x < 1.
dx
1 ¡ x2
dy
27