Chapter 2 Some functions and their derivatives

Chapter 2
Some functions and their derivatives
2.1 Derivative of x n for integer n
Recall, from eqn (1.6), for y = f (x),
Also recall that, for integer n,
Hence, if y = x n then
dy
dx = lim
δx→0
(a + b) n = a n + na n−1 b +
f(x + δx) ¡ f(x)
.
δx
n (n ¡ 1)
a n−2 b 2 + ...+ b n .
2!
dy
dx = lim (x + δx) n ¡ x n
δx→0
³
δx
´
x n + nx n−1 δx + n(n−1) x n−2 δx 2 + ...
2
= lim
δx→0 δx
nx n−1 δx + n(n−1) x n−2 δx 2 + ...
2
= lim
δx→0
·
= lim nx n−1 +
δx→0
dy
dx = nxn−1 .
2.2 Trigonometric Functions
See Appendix C for further details.
δx
n (n ¡ 1)
x n−2 δx + ...
2
¸
¡ x n
10

2.2.1 Definition and Graphs of Sine and Cosine
The two basic trigonometric functions are the sine and cosine. Fromanangleθ, 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 < π 2
and, for the right-angled triangle OAP,
sin θ =
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 inradiansisdefined 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:
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By considering the limiting cases of the previously illustrated right-angled triangle
when θ =0and when θ = π/2, we immediately deduce (in accordance with the graphs
just drawn) that
2.2.2 Inverse Trigonometric Functions
sin 0 = 0, (2.1)
cos π 2
= 0, (2.2)
cos 0 = 1, (2.3)
sin π 2
= 1. (2.4)
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

sin : £ ¡ π 2 , π 2
¤
! [¡1, 1] has increasing inverse sin −1 =arcsin:[¡1, 1] ! £ ¡ π 2 , π 2
¤
. By
definition, given any y in [¡1, 1], sin −1 y istheuniqueanglebetween¡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, sotheslope dx
dy
these two points.
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,givenanyy in [¡1, 1], cos −1 (y) = arccos y istheuniqueanglebetween0 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.AnidentityisanequationwhichholdsforALLpermittedvaluesofitsvariables.
See Appendix C for proofs of many trig. identities. Three which will be used in the next
section are
sin(x + y) = sinx cos y +cosx sin y, (2.5)
cos 2x = 1¡ 2sin 2 x (2.6)
cos 2 x +sin 2 x = 1. (2.7)
Appendix C4 also contains the proof of a useful limit:
sin x
lim =1,
x→0 x
(2.8)
i.e. sin x ¼ x for small angles x.
2.2.4 Derivatives of the trigonometric functions
By definition (see section 1.4), the derivative of sin x is
d
(sin x) = lim
dx δx→0
sin(x + δx) ¡ sin x
δx
sin x cos δx +cosx sin δx ¡ sin x
= lim
δx→0 δx
sin x (cos δx ¡ 1) + cos x sin δx
= lim
δx→0 δx
¡2sinx sin 2 (δx/2) + cos x sin δx
= lim
δx→0 δx
·sin δx/2
δx
= ¡ sin x lim
δx→0 2 δx/2
= cosx using (2.8).
¸2
+cosx lim
δx→0
sin δx
δx
using (2.5)
using (2.6)
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)
=
cos 2 x
= cos2 x +sin 2 x
cos 2 x
1
=
using (2.7)
cos 2 x
= sec 2 x.
One similarly deduces that
d
dx (cot x) =¡cosec2 x.
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
dx = ³
1 ´ , provided dx
dx
dy 6=0
which follows directly from the definition of the derivative.
dy
Inverse Sine Set y =sin −1 x and x =siny (with ¡π/2 · y · π/2 and ¡1

Repeatedly differentiating (2.9), we see that
d 2 y
dx = dy
2 dx = y, and more generally dn y
= y for all positive integers n.
dxn It follows (from Taylor’s Theorem - which you haven’t met yet!) that a function satisfies
condition (2.9) if and only if
y(x) =y(0)
·1+x ¢¢¢¸
+ x2
2! + x3
3! + = y(0)
∞X
n=0
x n
n!
for all real x, (2.10)
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
dx = y(0) d
dx
·
= y(0)
= y(0)
·1+x + x2
2! + x3
3! + ¢¢¢¸
¸
¢¢¢
0+1+ 2x
2! + 3x2 + 4x3
3! 4!
·1+x ¢¢¢¸
+ x2
2! + x3
3! + = y(x).
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, ofwhichmorelateroninthis 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)]
dx
= exp(x) (for all real x), (2.11)
exp (0) = 1. (2.12)
Theuniquefunctionexp(x) satisfying these is given explicitly by
This definition:
exp(x) =1+x + x2
2! + x3
3! + ¢¢¢= ∞
X
n=0
x n
n! . (2.13)
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,
exp (¡x) =
1
exp (x)
Given exp (x) > 0 it follows from (2.11) that d [exp(x)] > 0 and so
dx
exp (x) is a strictly increasing function, i.e. x>y) exp(x) > exp(y). (2.14)
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From 2.13 we can see that
and so, given exp (¡x) = 1
exp(x) ,
So, the graph of the function exp(x) looks like:
exp(x) ! +1 as x ! +1, (2.15)
exp(x) ! 0 as x !¡1. (2.16)
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, y2 (0, 1),
y =exp(x) if and only if x =lny. (2.17)
Taking x =0in (2.17) and using exp (0) = 1, weseethat
Similarly it follows from (2.15), (2.16) and (2.17) that
ln 1 = 0. (2.18)
ln x ! +1 as x ! +1, (2.19)
ln x ! ¡1 as x ! 0. (2.20)
Properties (2.18)—(2.20) all show up clearly on the graph of ln:
19

2.3.3 Definition of an Arbitrary Real Power of a Positive Real
Number
We know
so we see
exp(x 1 + x 2 )=exp(x 1 )exp(x 2 ) (2.21)
exp(x 1 + x 2 + ¢¢¢+ x n )=exp(x 1 )exp(x 2 ) ...exp(x n ), (2.22)
where n is any positive integer and x 1 ,x 2 ,...,x n any real numbers. Taking x 1 = x 2 =
¢¢¢= x n =lnx, wherex>0, and using (2.17), we get
exp(n ln x) =[exp(lnx)] n = x n ,
for any positive integer n.
We can define an arbitrary real power x α of a positive real number x by
equivalent [by (2.17)] to
x α =exp(α ln x) (x>0, α real), (2.23)
ln(x α )=α ln x (x >0, α real). (2.24)
The definition (2.23) satisfies the standard rules for multiplication and division of
powers:
for all real α and β.
x α x β = x α+β ,
x α
x β = x α−β ,
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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 =exp1, (2.25)
from which
Using (2.23),
i.e.
2.3.5 Derivatives
The exponential function
ln e =1. (2.26)
e x =exp(x ln e) =exp(x)
exp x = e x for all real x. (2.27)
By definition.
d
[exp(x)] = exp(x). (2.28)
dx
The logarithmic function
y =lnx, andsox =exp(y) . Then
d
dx [ln(x)] = 1 dx
dy
=
1
exp (y) = 1 x
(for x>0). (2.29)
Arbitrary real powers
From the above definition of x α we have
d
dx (xα ) = d (exp (α ln x))
dx
= exp(α ln x)£ α £ 1 x
(using the chain rule)
= αx α 1 x
= αx α−1 .
Thus
d
dx (xα )=αx α−1 (α real, x>0), (2.30)
NBthereisasimilartrickfor
d
dx (ax )= d (exp (x ln a)) = ...
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”).
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2.4.1 Definitions
cosh x = 1 2 (ex + e −x ), (2.31)
sinh x = 1 2 (ex ¡ e −x ), (2.32)
tanh x = sinh x
cosh x = ex ¡ e −x 1 ¡ e−2x
=
e x + e−x 1+e = e2x ¡ 1
−2x e 2x +1 , (2.33)
coth x = 1
tanh x = cosh x
sinh x = ex + e −x 1+e−2x
=
e x ¡ e−x 1 ¡ e = e2x +1
−2x e 2x ¡ 1
(x 6= 0),(2.34)
sech x = 1
cosh x = 2
,
e x + e−x (2.35)
cosech x = 1
sinh x = 2
e x ¡ e −x (x 6= 0). (2.36)
2.4.2 The Fundamental Identity
Corresponding to the well-known identity cos 2 x +sin 2 x =1for the trigonometric functions,
we have for the hyperbolic functions the identity
cosh 2 x ¡ sinh 2 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
4 (ex +e −x ) 2 ¡ 1 4 (ex ¡e −x ) 2 = 1 4 (e2x +e −2x +2e x e −x )¡ 1 4 (e2x +e −2x ¡2e x e −x )=e x e −x = e x−x = e 0 =1.
2.4.3 Further Identities
1 ¡ tanh 2 x = sech 2 x, (2.38)
coth 2 x ¡ 1 = cosech 2 x, (2.39)
sinh(x + y) = sinhx cosh y +coshx sinh y, (2.40)
sinh(x ¡ y) = sinhx cosh y ¡ cosh x sinh y, (2.41)
cosh(x + y) = coshx cosh y +sinhx sinh y, (2.42)
cosh(x ¡ y) = coshx cosh y ¡ sinh x sinh y, (2.43)
sinh 2x = 2sinhx cosh x, (2.44)
cosh 2x = sinh 2 x +cosh 2 x, (2.45)
cosh 2 x = 1 (cosh 2x +1),
2
(2.46)
sinh 2 x = 1 (cosh 2x ¡ 1),
2
(2.47)
sinh x cosh y = 1 [sinh (x + y)+sinh(x ¡ y)] ,
2
(2.48)
cosh x cosh y = 1 [cosh (x + y)+cosh(x ¡ y)] ,
2
(2.49)
sinh x sinh y = 1 [cosh (x + y) ¡ cosh (x ¡ y)] .
2
(2.50)
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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 = 0,
sinh x ! +1 as x ! +1,
sinh x ! ¡1 as x !¡1,
sinh (¡x) = ¡ sinh (x) (i.e. it is an ODD function).
24

tanh 0 = 0,
tanh x ! +1 as x ! +1,
tanh x ! ¡1 as x !¡1,
tanh (¡x) = ¡ 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. Notethatsinhand
tanh are both increasing functions and that the graph of tanh has horizontal asymptotes
y = §1 approached as x !§1respectively.
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 !¡1,
coth x ! +1 as x ! 0 from above , coth x !¡1as x ! 0, from below,
sech x ! 0 as x !§1,
cosech x ! 0 as x !§1,
sech 0=1
cosech x ! +1 as x ! 0 from above, cosech x !¡1as x ! 0 from below.
2.4.6 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). Fory ¸ 0, x¸ 1, wehavey =cosh −1 x ,
x =coshy. 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 =coshy,
that is
x = 1 ¡
e y + e −y¢ .
2
Multiplying through by 2e y gives
a quadratic for e y ,withroots
2xe y = e 2y +1
e 2y ¡ 2xe y +1 = 0
(e y ) 2 ¡ 2xe y +1 = 0
e y = 2x § p 4x 2 ¡ 4
= x § p x
2
2 ¡ 1.
³
so, y = ln x § p ´
x 2 ¡ 1 .
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But we want y>0, so we choose the + sign, i.e.
³
y =cosh −1 x =ln x + p ´
x 2 ¡ 1
for x ¸ 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,
sinh −1 x =ln
³x + p ´
1+x 2 . (2.52)
tanh
tanh : R ! (¡1, 1) is an increasing function, so there is no need for restriction. It is easy
to show that
tanh −1 x = 1 µ 1+x
2 ln for ¡ 1

Inverse Cosh Recall y =cosh −1 x , x =coshy with y ¸ 0, x¸ 1. So,sinh y ¸ 0
and hence
Then
i.e.
sinh 2 y = cosh 2 y ¡ 1
q
sinh y = + cosh 2 y ¡ 1=+ p x 2 ¡ 1.
dy
dx = 1 dx
dy
d
dx (cosh−1 x)=
= 1
sinh y =+ 1
p
x2 ¡ 1
1
p for x>1.
x2 ¡ 1
You can also prove this by differentiating eqn (2.51) directly.
Inverse Sinh
Similarly
d
dx (sinh−1 x)=
1
p
1+x
2 .
Inverse Tanh
y =tanh −1 x ) x =tanhy, sowehave
so
dx
dy = sech2 x =1¡ tanh 2 x =1¡ y 2 ,
d
dx (tanh−1 x)= 1 dx
dy
= 1 for ¡ 1