Part 1, Semester 2.

Differentiation *

Leonid Pankratov

February 24, 2023

But just as much as it is easy to find the differential [derivative] of a given quantity, so it is difficult
to find the integral of a given differential. Moreover, sometimes we cannot say with certainty whether
the integral of a given quantity can be found or not.
Johann Bernoulli
Science is the Differential Calculus of the mind. Art the Integral Calculus; they may be beautiful
when apart, but are greatest only when combined. Ronald Ross

* File :DIF-F.tex

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Contents
1 Definition and Basic Properties of the Derivative 4

2 Derivatives of Inverse Functions 10

3 Higher-Order Derivatives 13

4 Derivatives of Parametric Functions 14

5 Mean Value Theorem 16

6 Applications of the Mean Value Theorem 23

7 L’Hôspital’s Rule 24

8 Taylor’s Theorem 31

9 An Application of Taylor’s Theorem 32

10 O(f ) and o(f ) 34

11 Remainder in Peano’s Form 34

11.1 Maclaurin’s Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

12 Finding Limits using Maclaurin’s Formula 38

13 Exercises with Solutions 42

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Table of derivatives

1. C ′ = 0
1
2. x′ = 1 12. (cot x)′ = − , x ̸= kπ, k ∈ Z
sin2 x
√
′ 1
3. x = √ , x>0 1
2 x 13. (arcsin x)′ = √ , |x| < 1
1 − x2
4. (ax )′ = ax ln a, a > 0 1
14. (arccos x)′ = − √ , |x| < 1
5. (xα )′ = αxα−1 , x > 0 1 − x2
6. (ex )′ = ex 1
15. (arctan x)′ =
1 1 + x2
7. (loga x)′ = , x, a > 0 1
x ln a 16. (arccotx)′ = −
1 1 + x2
8. (ln x)′ = , x > 0 17. (sinh x)′ = cosh x
x
′
9. (sin x) = cos x 18. (cosh x)′ = sinh x
10. (cos x)′ = − sin x 19. (tanh x)′ =
1
1 cosh2 x
11. (tan x)′ = , 1
cos2 x 20. (cotanhx)′ = − , x ̸= 0
π
x ̸= + kπ, k ∈ Z sinh2 x
2

3
 In this chapter, we discuss basic properties of the derivative of a function and several major theo-
rems, including the Mean Value Theorem and L’Hôpital’s Rule. The applications of derivatives will
be discussed in the next chapter.

1 Definition and Basic Properties of the Derivative

Let G be an open subset of R (i.e., if this subset contains a point, the subset contains a neighborhood
of this point). Consider a function f : G → R. For every a ∈ G, the function

f (x) − f (a)
x−a
is defined on G \ {a}. If we think about x as time variable, and f (x) the distance from the start point,
then the last quotient represents the average velocity.
Since G is an open set, then a is a limit point of G \ {a}, that is a point each neighborhood of
which contains at least one point of G \ {a} different from a. To introduce the instantaneous velocity,
we consider the limit
f (x) − f (a)
lim .
x→a x−a

Definition 1.1 (The Derivative at a Point) Let G be an open subset of R and let a ∈ G. The function
f defined on G is said to have a finite derivative f ′ (x) at a if the limit

f (x) − f (a)
lim
x→a x−a

exists (finite or infinite). This limit is denoted by f ′ (a).

Definition 1.2 (One-sided derivative) A generalization of the concept of a derivative, in which the
ordinary limit is replaced by a one-sided limit. If the following limit exists for a function f of a real
variable x :
f (x) − f (a) f (x) − f (a)
lim or lim
x→a+0 x−a x→a−0 x−a
then it is called the right (respectively, left) derivative of f at the point a.

Notice that if the one-sided derivatives are equal, then the function has an ordinary derivative at a.

Definition 1.3 (The Differentiability) Function f (x) defined in some neighborhood of x0 is referred
to as differentiable at x0 , if its increment at this point can be represented as

∆f (x0 ) ≡ f (x0 + ∆x) − f (x0 ) = A · ∆x + α(∆x)∆x, A ∈ R

4
where
lim α(∆x) = 0,
∆x→0

i.e.
∆f (x0 ) = A∆x + o(∆x), ∆x → 0.

Let’s find a relation between the definitions.

Theorem 1.4 Function f is differentiable at x0 if and only if finite f ′ (x0 ) exists and, moreover,

A = f ′ (x0 ).

Proof of Theorem 1.4. If f is differentiable, we have that

f (x0 + ∆x) − f (x0 ) = A · ∆x + α(∆x)∆x

and
f (x0 + ∆x) − f (x0 )
= A + α(∆x),
∆x
i.e.,
f (x0 + ∆x) − f (x0 )
f ′ (x0 ) = lim = A.
∆x→0 ∆x

Let now
f (x0 + ∆x) − f (x0 )
lim = A.
∆x→0 ∆x
Then
f (x0 + ∆x) − f (x0 )
= A + α(∆x), with lim α(∆x) = 0.
∆x ∆x→0

So ∆f (x0 ) = A∆x + o(∆x) when ∆x → ∞.

So, when we want to prove that a function is differentiable at a point, we can show that the function
has a finite derivative at this point.

Definition 1.5 (The Differentiability on a set) The function f is said to be differentiable on G if f

is differentiable at any point a ∈ int G and has one-side derivatives at boundary point(s), if boundary
point(s) is (are) contained in G.

Example 1.6 In this example we consider some simple functions for which the derivative can be
obtained directly from the definition.

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 1. Let f : R → R be defined by f (x) = x and let a ∈ R. Then

f (x) − f (a) x−a

lim = lim = lim 1 = 1.
x→a x−a x→a x − a x→a

Thus f = x is differentiable at a and f ′ (a) = 1.

2. Let f : R → R be defined by f (x) = x2 and let a ∈ R. Then

f (x) − f (a) x 2 − a2 (x − a)(x + a)

lim = lim = lim =
x→a x−a x→a x − a x→a x−a

= lim (x + a) = 2a.
x→a

Thus f = x2 is differentiable at a and f ′ (a) = 2a.

3. Let f : R → R be defined by f (x) = |x| and let a = 0. Then

f (x) − f (0) |x| x

lim+ = lim+ = lim+ = 1,
x→0 x−0 x→0 x x→0 x

and
f (x) − f (0) |x| −x
lim− = lim− = lim+ = −1.
x→0 x−0 x→0 x x→0 x
This means that the limit
f (x) − f (0)
lim
x→0 x−0
does not exist and f = |x| is not differentiable at 0.

The next result follows directly from the definition of continuity and Definition 1.1.

Theorem 1.7 Let G be an open subset of R and let f be defined on G. If f is differentiable at a ∈ G,

then f is continuous at this point.

Theorem 1.8 Let G be an open subset of R and let f, g : G → R. Suppose that the functions f, g
are differentiable at a ∈ G. Then we have

1. The function f + g is differentiable at a and

(f + g)′ (a) = f ′ (a) + g ′ (a).

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 2. Let C be a constant. The function Cf is differentiable at a and

(C f )′ = C f ′ (a).

3. The product (f g) is differentiable at a and

(f g)′ (a) = f ′ (a) g(a) + f (a)g ′ (a).

4. Suppose now that g(a) ̸= 0. Then the quotient (f /g) is differentiable at a and
 ′
f f ′ (a)g(a) − f (a)g ′ (a)
(a) = .
g (g(a))2
Proof of Theorem 1.8. The proofs of statements (1), (2) are evident and we leave them to the reader.

Consider statement (3). For every x ∈ G \ {a}, we have

(f g)(x) − (f g)(a) f (x)g(x) − f (a)g(x) + f (a)g(x) − f (a)g(a)
= =
x−a x−a
f (x) − f (a) g(x) − g(a)
= g(x) + f (a) .
x−a x−a
By Theorem 1.7, the function g is continuous at a and, therefore,
lim g(x) = g(a). (1.1)
x→a
Then
(f g)(x) − (f g)(a)
lim = f ′ (a) g(a) + f (a)g ′ (a).
x→a x−a
Statement (3) is proved.
Now we turn to statement (4). Since g(a) ̸= 0, then by relation (1.1), there is an open interval I,
a ∈ I such that
g(x) ̸= 0 for all x ∈ I.
We set
f
h= .
g
Then the function h is defined on I and we have
f (x) f (a) f (a) f (a)
− + −
h(x) − h(a) g(x) g(x) g(x) g(a)
=
x−a x−a

1 f (a)
(f (x) − f (a)) + (g(a) − g(x))
g(x) g(x)g(a)
=
x−a
 
1 f (x) − f (a) g(x) − g(a)
= g(a) − f (a) .
g(x)g(a) x−a x−a
Passing to the limit as x → a, we obtain statement (4). This completes the proof of Theorem 1.8.

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Example 1.9 Let f : R → R be defined by f (x) = x2 and let a ∈ R. Using the previous example
and Theorem 1.8, statement (3), compute f ′ (a).

Solution to Example 1.9. In fact, let g : R → R be defined by g(x) = x. Consider f = g · g. Then

f ′ (a) = (gg)′ (a) = g ′ (a)g(a) + g(a)g ′ (a) = 2g ′ (a)g(a) = 2a.

Using the method of mathematical induction, one obtains the derivative of g(x) = xn for n ∈ N:
(xn )′ = n xn−1 . (1.2)
Moreover, if we make use of this fact and Theorem 1.8 we obtain the well known formula for the
derivative of a polynomial p(x) = an xn + ... + a1 x + a0 . It is given by p ′ (x) = nan xn−1 + ... + a1 .
The following result is useful for studying the differentiability of the composition of two functions

f ◦g :A→C

g:A→B f :B→C

x g(x) f (g(x))

A B C

Figure 1: Function composition.

Lemma 1.1 Let G be an open subset of R and f : G → R and let f be differentiable at a. Then
there is a function u : G → R such that

f (x) − f (a) = f ′ (a) + u(x) (x − a) for all x ∈ G with lim u(x) = 0.

 
(1.3)
x→a

Proof of Lemma 1.1. We set


 f (x) − f (a) − f ′ (a),

x ∈ G \ {a};
u(x) = x−a (1.4)

0, x = a.


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The function f is differentiable at a. Then
f (x) − f (a)
lim u(x) = lim − f ′ (a) = f ′ (a) − f ′ (a) = 0.
x→a x→a x−a
Thus, the function u satisfies all the conditions of the lemma.

Theorem 1.10 (Chain Rule) Let f : G1 → R and let g : G2 → R, where G1 , G2 are open subsets
of R with such that f (G1 ) ⊂ G2 . Let f be differentiable at a and g be differentiable at f (a). Then the
composition of functions g ◦ f is differentiable at a and

(g ◦ f )′ (a) = g ′ (f (a))f ′ (a). (1.5)

Proof of Theorem 1.10. We observe that the function f is differentiable at a. Then by Lemma 1.1,
there is a function u defined on G1 such that
f (x) − f (a) = f ′ (a) + u(x) (x − a) ∀x ∈ G1 , lim u(x) = 0.
 
(1.6)
x→a

In a similar way, due to the differentiability of g at f (a), there is a function v defined on G2 with
g(t) − g(f (a)) = g ′ (f (a)) + v(t) t − f (a) ∀t ∈ G2 , lim v(t) = 0.
 

(1.7)
t→f (a)

We make use of (1.7) for t = f (x). We get

g(f (x)) − g(f (a)) = g ′ (f (a)) + v(f (x)) f (x) − f (a) .
  
(1.8)
Then
g(f (x)) − g(f (a)) = g ′ (f (a)) + v(f (x)) f ′ (a) + u(x) (x − a) ∀x ∈ G1 .
  
(1.9)
Now we obtain that
g(f (x)) − g(f (a))  ′
= g (f (a)) + v(f (x)) f ′ (a) + u(x) ∀x ∈ G1 \ {a}.
 
(1.10)
x−a
Taking into account the continuity of the function f at a and the properties of the function v, we have
(using the theorem about composition of limits) that
lim v(f (x)) = 0.
x→a

Therefore
g(f (x)) − g(f (a))
lim = g ′ (f (a))f ′ (a).
x→a x−a
This completes the proof of Theorem 1.10

Example 1.11 Let h : R → R be a function defined as

h(x) = (3x4 + x + 7)15 .

Find h′ (x).

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Solution to Example 1.11. The function h(x) is a polynomial and we can find h′ (x) by expanding
the power and using the previous example. However, Theorem 1.10 suggests a different approach.
We define the functions f, g : R → R as follows:

f (x) = (3x4 + x + 7) and g f (x) = f 15 (x).

Then the function h can be defined as h = g ◦ f . Given a ∈ R from Theorem 1.10 we have that

(g ◦ f )′ (a) = g ′ (f (a))f ′ (a) = 15(3a4 + a + 7)14 (12a3 + 1).

Example 1.12 By iterating the chain rule, generalize the result of Theorem 1.10 to a composition of
more than two functions.

Solution to Example 1.12. For the sake of simplicity, we study the case of three functions. Consider
the functions f : G1 → R, g : G2 → R, and h : G3 → R such that f (G1 ) ⊂ G2 , g(G2 ) ⊂ G3 , the
function f is differentiable at a, g is differentiable at f (a), and h is differentiable at g(f (a)). Then
we have that h ◦ g ◦ f is differentiable at a and

′
h ◦ g ◦ f = h′ g(f (a)) g ′ f (a) f ′ (a).

2 Derivatives of Inverse Functions

Now, when we know how to differentiate a function composition, we proceed with derivatives of the
inverse function. The main result of this section is given by the following theorem.

Theorem 2.1 (Derivative of inverse function) Let the functions f : X → Y and f −1 : Y → X (X

and Y are open subsets) be mutually inverse and continuous at points x0 ∈ X and f (x0 ) = y0 ∈ Y ,
respectively. Then if the function f is differentiable at the point x0 and, in addition, f ′ (x0 ) ̸= 0, then
f −1 is differentiable at the point y0 , and

′ 1
f −1 (y0 ) = . (2.1)
f ′ (x 0)

Proof of Theorem 2.1. Since the functions f and f −1 are mutually inverse, then

f (x) − f (x0 ) and f −1 (y) − f −1 (y0 ), where y = f (x),

are both nonzero if x ̸= x0 . Moreover, it follows from the continuity of the function f at the point x0
and f −1 at the point y0 that
(X ∋ x → x0 ) ⇔ (Y ∋ y → y0 ).

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Applying now the theorem on the limit of a composite function and the arithmetic properties of the
limit, we get
f −1 (y) − f −1 (y0 ) x − x0 1
lim = lim = lim f (x)−f (x0 )
=
Y ∋y→y0 y − y0 X∋x→x0 f (x) − f (x0 ) X∋x→x0
x−x0

1
= .
f ′ (x 0)
This means that the function f −1 : Y → X has a derivative at the point y0 and

′ 1
f −1 (y0 ) = ′ .
f (x0 )
This completes the proof of Theorem 2.1.

Figure 2: Graphs of inverse trigonometric functions.

Remark 1 If it is a priori known that f −1 is differentiable at y0 , from the identity

f −1 ◦ f (x) = x

and the chain rule yields

′
f −1 (y0 ) · f ′ (x0 ) = 1.

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Example 2.2 Let us show (see (13) in the table of derivatives) that

1
arcsin′ y = p |y| < 1.
1 − y2
Solution to Example 2.2. First, we consider the function arcsin x. The sine function sin x is not
one-to-one, and the inverse cannot be defined for all x. Indeed, sin 0 = sin π = 0. We should restrict
the sine in order to have an inverse. The arcsin x is defined as the inverse of sin y for −1 ⩽ x ⩽ 1
and −π/2 ⩽ y ⩽ π/2 so that
sin y = x, arcsin x = sin−1 x = y.
The arcsine-function returns the angle whose sine is a given number. Notice that sin−1 x means the
inverse sin and does not mean 1/ sin.
For example,
π
arcsin 1 = sin−1 1 = = 90◦ .
2
The functions
sin : [−π/2, π/2] → [1, 1] and arcsin : [1, 1] → [−π/2, π/2]
are mutually inverse and continuous. Moreover
sin′ (x) = cos x ̸= 0 if |x| < π/2.
If |x| < π/2, then |y| < 1 for y = sin x. Therefore, by Theorem 2.1
1 1 1 1
arcsin′ y = ′ = =p =p .
sin x cos x 1 − sin2 x 1 − y2
Notice that the sign in front of the radical is taken due to the inequality
cos x > 0 for |x| < π/2.

Example 2.3 Show that

1
arccos′ y = − p |y| < 1.
1 − y2
Solution to Example 2.3. Consider the function arccos x. The arccos x is defined as the inverse of
cos x for −1 ⩽ x ⩽ 1 when cos y = x, 0 ≤ y ≤ π.
arccos x = cos−1 x = y.
Now, as in the previous example, we obtain that
1 1 1 1
arccos′ y = ′
=− = −√ = −p .
cos x sin x 1 − cos2 x 1 − y2
The sign in front of the radical is taken due to the inequality sin x > 0 if 0 < x < π.

12
Example 2.4 Show that
1
arctan′ y = y ∈ R.
1 + y2
Solution to Example 2.4. Consider the function arctan x. The arctan x is defined as the inverse
tan x function for x ∈ R when tan y = x, −π/2 < y < π/2. Then the arctangent is such that
arctan x = tan−1 x = y. The derivative is
1 1 cos2 x
arctan′ y = =  = cos2 x =
tan′ x 1 cos2 x + sin2 x
cos2 x
1 1
= 2
= .
1 + tan x 1 + y2
The desired result is then proved.

3 Higher-Order Derivatives
Suppose that a function f : E → R (E is an open set) is differentiable at every point x ∈ E. Then
the function f ′ : E → R can be defined and its value at x ∈ E is the derivative f ′ of f at this point.
The function f ′ : E → R can also have a derivative (f ′ )′ : E → R. This derivative is said to be
the second derivative of the function f and is usually denoted as
d2 f (x)
f ′′ (x) or .
dx2
In order to generalize this notion, we give the following definition.

Definition 3.1 If the derivative f (n−1) (x) of order (n − 1) of f is defined, then the derivative of order
n is given by

′
f (n) (x) = f (n−1) (x)

The following notation are used in the mathematical literature for the derivative of order n
dn f (x)
f (n) (x) or .
dxn
def
By definition, f (0) (x) = f (x).

Theorem 3.2 (The Leibniz formula) Let u(x) and v(x) be functions with the derivatives up to order
n on a set E. Then Leibniz formula is valid for the n-th derivative of their product. Namely,
n
X
(n)
(uv) = Cnk u(n−k) v (k) , (3.1)
k=0

13
where the binomial coefficients are
n!
Cnk = .
(n − k)!k!

Proof of Theorem 3.2. The proof of (3.1) will be done by the method of mathematical induction.
For n = 1 formula (3.1) is exactly the formula for the derivative of a product of two functions.
If the functions u and v have derivatives up to order (n + 1) then we assume that (3.1) holds for order
n. We differentiate the left-hand and right-hand sides to find
n
X n
X
(n+1)
(uv) = Cnk u (n−k+1) (k)
v + Cnk u(n−k) v (k+1) =
k=0 k=0

n
X
(n+1) (0)
 m
Cn + Cnm−1 u(n+1−m) v (m) + u(0) v (n+1) .

=u v + (3.2)
m=1

Let us make use of the following formula

n! n! n!
Cnm + Cnm−1 = + = +
(n − m)!m! (n − m + 1)!(m − 1)! (n − m)!m!
 
n! · m n! m
+ = 1+ =
(n − m)!m! · (n − m + 1) (n − m)!m! (n − m + 1)
n! n+1 m
= · = Cn+1 (m = 1, 2, ..., n). (3.3)
(n − m)!m! (n + 1 − m)
Now from (3.2) and (3.3), we get
n+1
X
(n+1) m
(uv) = Cn+1 u((n+1)−m) v (m)
m=0

Thus by induction Leibniz’s formula is proved.

4 Derivatives of Parametric Functions

The relationship between the variables x and y can be defined in parametric way using two equations
of the form (
x = x(t),
(4.1)
y = y(t),
where the variable t is called parameter. For example, the functions
(
x = R cos t
y = R sin t, 0 ⩽ t ⩽ 2π,

are parametric equations defining the circle centered at the origin of radius R.

14
Let us find the derivative of a parametric function. To this end we suppose that the functions x = x(t),
y = y(t) are differentiable in the interval α < t < β and x′ (t) ̸= 0. Moreover, we assume that
x = x(t) has an inverse function t = t(x). By the inverse function theorem,

dt 1
= t′x = ′ . (4.2)
dx xt

One can consider the function y(x) as a composite one. Namely,

y (x) = y (t (x)) . (4.3)

Then the derivative of y(x) is given by

1 yt′
yx′ = yt′ · t′x = yt′ · = . (4.4)
x′t x′t

This formula enables to find the derivative of a parametric function y(x). In this case we do not
need the explicit form of this function.
Let us consider some examples.

Example 4.1 Find the derivative of the parametric function

x = t2 and y = t3 .

Solution to Example 4.1. First, we find the derivatives of the functions x, y with respect to the
parameter t. We have
′
′
x′t = t2 = 2t, yt′ = t3 = 3t2 .
Then
dy y′ 3t2 3t
= yx′ = t′ = = (t ̸= 0) .
dx xt 2t 2

Example 4.2 Find the derivative of the parametric function

x = e2t and y = e3t .

Solution to Example 4.2. In this case, as in the previous example, we have that

′
′
x′t = e2t = 2e2t , yt′ = e3t = 3e3t .

Therefore
dy y′ 3e3t 3 3
= yx′ = t′ = 2t = e3t−2t = et .
dx xt 2e 2 2

15
 Now we turn to the high-order derivatives of functions defined parametrically. Namely, let us
consider a function y = f (x) given by
x = x(t), y = y(t). (4.5)
Formula (4.4) implies that the first derivative of y is given by
yt′
y ′ = yx′ = .
x′t
We differentiate this relation with respect to x in order to obtain the second derivative of the function
y(x). We have that  ′ ′  ′ ′
′′ yt ′ yt 1
yxx = ′
· tx = ′
· ′.
xt t xt t xt
Thus
′′ x′ y ′′ − y ′ x′′
yxx = t t ′ 3t t . (4.6)
(xt )

Example 4.3 Consider the function y = f (x) given by the formulas

x(t) = t − sin t and y(t) = 1 − cos t with t ∈ (0, 2π).

Find the first and the second derivatives of y(x) with respect to x.
Solution to Example 4.3. By (4.4), we have
yt′ sin t 2 sin(t/2) cos(t/2)
yx′ = ′ = = = cot(t/2).
xt 1 − cos t 2 sin2 (t/2)
Now we differentiate this relation with respect to x to get
 ′  ′
′′ t ′ t 1 1 1
yxx = cot · tx = cot · ′ =− 2 · =
2 t 2 t xt 2 sin (t/2) 1 − cos t
1
=− 4 .
4 sin (t/2)

5 Mean Value Theorem

This section is devoted to the famous Mean Value Theorem which is one of the important tools in
mathematical analysis. It is used in the proof of several fundamental results, such as Fundamental
Theorem of Calculus, L’Hôpital’s rule, the proof of a function being a constant if it has zero derivative,
etc..
We start with the basic definitions of minimum and maximum. Take a ∈ R and δ > 0 and denote
B(a, d) = (a − d, a + d), B+ (a, d) = (a, a + d), B− (a, d) = (a − d, a)
the neighborhoods of point a.

16
Definition 5.1 Let D be a nonempty subset of R and let f : D → R. The function f is said to have
a local (or relative) minimum at c ∈ D if there is δ > 0 such that

f (x) ⩾ f (c) for all x ∈ B(c, d) ∩ D.

In a similar way, the function f is said to have a local (or relative) maximum at a c ∈ D if there
is δ > 0 such that
f (x) ⩽ f (c) for all x ∈ B(c, d) ∩ D.

Theorem 5.2 (Fermat’s Rule) Let I be an open interval and f : I → R. If the function f has a
local minimum (maximum) at c ∈ I and it is differentiable at c, then

f ′ (c) = 0.

y f ′ (b) = 0

y = f (x)

a
b x
f ′ (a) = 0

Figure 3: Illustration of Fermat’s Rule.

Proof of Theorem 5.2. Let us assume that the function f has a local minimum at the point c (the case
of a local maximum is similar). Then there is δ > 0 sufficiently small such that

f (x) ⩾ f (c) for all x ∈ B(c, d) ∩ D.

Since B+ (c, d) is a subset of B(c, d), we have that

f (x) − f (c)
⩾0 for all x ∈ B+ (c, d).
x−c
Now it follows from the differentiability of f at c that

f (x) − f (c) f (x) − f (c)

f ′ (c) = lim = lim+ ⩾ 0.
x→c x−c x→c x−c

17
In a similar way
f (x) − f (c)
⩽0 for all x ∈ B− (c, d)
x−c
and
f (x) − f (c) f (x) − f (c)
f ′ (c) = lim = lim− ⩽ 0.
x→c x−c x→c x−c
Therefore, f ′ (c) = 0.
The proof is similar in the case when f has a local maximum at c. Theorem 5.2 is proved.

Theorem 5.3 (Rolle’s Theorem) Let a, b ∈ R with a < b and f : [a, b] → R. We assume that

1. f is continuous on [a, b];

2. f is differentiable on (a, b);

3. f (a) = f (b).

Then there exists c ∈ (a, b) such that

f ′ (c) = 0. (5.1)

f ′ (c) = 0

y = f (x)

a c b x
f (a) = f (b)

Figure 4: Illustration of Rolle’s Theorem.

Proof of Theorem 5.1. First, we observe that the function f is continuous on the compact set [a, b].
e1 ∈ [a, b] and x
Then by the Extreme Value Theorem, there exist x e2 ∈ [a, b] such that

x1 ) = min {f (x) : x ∈ [a, b]}

f (e x2 ) = max {f (x) : x ∈ [a, b]} .
and f (e

Then
f (e
x1 ) ⩽ f (x) ⩽ f (e
x2 ) for all x ∈ [a, b]. (5.2)

18
 e1 ∈ (a, b) or x
If x e2 ∈ (a, b), then f has a local minimum at x e1 or f has a local maximum at x e2 . By
Theorem 5.2, f ′ (ex1 ) = 0 or f ′ (e
x2 ) = 0, and (5.1) holds for c = xe1 or c = xe2 .
If x
e1 , x
e2 are the endpoints of the interval [a, b], then f (e x1 ) = f (e
x2 ) because f (a) = f (b). By
inequality (5.2), f is a constant function, so f ′ (c) = 0 for any c ∈ (a, b). This completes the proof of
Theorem 5.3.

Example 5.4 (Counterexamples to Theorem 5.3) Let us show that all the three conditions of Rolle’s
theorem are necessary for the theorem to be true.

Counterexample to condition (1). Consider the fractional part f (x) = {x}, on the interval [0, 1].
The derivative of {x} on the open interval (0, 1) is 1 everywhere. In this case, Rolle’s theorem is not
true since f (x) has a discontinuity at x = 1, i.e. f (x) is not continuous everywhere on [0, 1].

Figure 5: Fractional part function.

Figure 6: Absolute value of x (on the left) and the linear function y(x) = x (on the right).

19
Counterexample to condition (2). Consider the absolute value function f (x) = |x| (see Figure 6)
on the interval [−1, 1]. We have shown that f (x) is not differentiable at x = 0. The function f (x) is
continuous on the closed interval [−1, 1], but there is no point inside the interval (−1, 1) where the
derivative equals zero. The Rolle theorem fails because f (x) is not differentiable over (−1, 1).
Counterexample to condition (3). The linear function f (x) = x is continuous on the closed interval
[0, 1] and differentiable on (0, 1). The derivative of the function is 1 everywhere on the interval. The
Rolle’s theorem is not valid because the 3-rd condition is not satisfied (since f (0) ̸= f (1).)

Now we will make use of Rolle’s Theorem to prove the Mean Value Theorem.

Theorem 5.5 (Mean Value Theorem (Lagrange)) Let a, b ∈ R with a < b and f : [a, b] → R. Let
f be continuous on [a, b] and differentiable on (a, b). Then there is c ∈ (a, b) such that
f (b) − f (a)
f ′ (c) = . (5.3)
b−a

y
f (b)
f (c)
f (b) − f (a)
f ′ (c) = y = f (x)
b−a

a c b x
f (a)

Figure 7: Illustration of the Mean Value Theorem.

Proof of Theorem 5.5. Consider the linear function passing through the points (a, f (a)) and (b, f (b)):
f (b) − f (a)
g(x) = · (x − a) + f (a).
b−a
Now we introduce the function
 
f (b) − f (a)
h(x) = f (x) − g(x) = f (x) − (x − a) + f (a) for x ∈ [a, b].
b−a

20
Then  
f (b) − f (a)
h(a) = f (a) − · 0 + f (a) = 0,
b−a
and  
f (b) − f (a)
h(b) = f (b) − (b − a) + f (a) =
b−a
= f (b) − [f (b) − f (a) + f (a)] = 0.
Thus, h(a) = h(b), and the function h satisfies the assumptions of Theorem 5.3. Therefore, there is
c ∈ (a, b) such that h′ (c) = 0. The derivative of the function h(x) is given by

f (b) − f (a)
h′ (x) = f ′ (x) − .
b−a
Then we have that
f (b) − f (a)
f ′ (c) − = 0.
b−a
This means that (5.3) holds true. The proof of Theorem 5.5 is completed.
Theorem 5.5 which is also called the First Mean Value Theorem enables to describe the increment
of a function on an interval in terms of the derivative of this function at an intermediate point of the
interval.
Geometric interpretation. Lagrange’s Theorem has a clear geometrical interpretation. The chord
which goes through the points of the graph corresponding to the ends of the interval a, b has the slope
given by the formula
f (b) − f (a)
k = tan α = .
b−a
Then there is x = c inside [a, b], where the tangent to the graph is parallel to the chord (see Fig. 7).

Corollary 5.6 In the case when the values of f (x) at the endpoints of [a, b] are equal, i.e. f (a) =
f (b), the Lagrange Theorem implies that there exists c ∈ (a, b) such that

f (b) − f (a)
f ′ (c) = = 0,
b−a

which is the statement of Rolle’s theorem. In this way, the last one is a particular case of Lagrange’s
Theorem.

Corollary 5.7 If the derivative f ′ (x) = 0 at all points of [a, b], then f is constant on [a, b]. In fact,
for any two points x1 and x2 in [a, b], there is a point c ∈ (a, b) such that

f (x2 ) − f (x1 ) = f ′ (c) (x2 − x1 ) = 0 · (x2 − x1 ) = 0.

This implies that f (x1 ) = f (x2 ).

21
Remark 2 It is important to note that the fact that f ′ = 0 for x ∈ [a, b] ∪ [c, d] does not imply that f
is (the same) constant on these two intervals. In fact, f might be equal to different constants on these
two intervals.

Example 5.8 Prove that

| sin x| ⩽ |x| for all x ∈ R.

Solution to Example 5.8. Consider the function f (x) = sin x, x ∈ R. Then f ′ (x) = cos x. Now, we
fix x ∈ R and consider two cases.
Case 1: x > 0. By the Mean Value Theorem applied to f on [0, x], there is c ∈ (0, x) such that
sin x − sin 0
= cos c
x−0
and, consequently,
sin x
= | cos c|.
x
Since | cos c| ⩽ 1 we have that | sin x| ⩽ |x| for all x > 0.
Case 2: x < 0. Applying once more the Mean Value Theorem one can show that there is c ∈ (x, 0)
such that
sin 0 − sin x
= cos c.
0−x
Then
sin x
= | cos c| ⩽ 1
x
and one can conclude that | sin x| ⩽ |x| for all x < 0.
Finally, we observe that the equality holds for x = 0 and we get that | sin x| ⩽ |x| for all x ∈ R.

Example 5.9 Prove the inequality

√ (5 + 2x)
1 + 4x < , x > 2. (5.4)
3
√
Solution to Example 5.9. Consider the function f (x) = 1 + 4x for x ⩾ 2. We compute the
derivative of this function:
4 2
f ′ (x) = √ =√ .
2 1 + 4x 1 + 4x
Let us fix x ∈ R such that x > 2 and apply the Mean Value Theorem to f on [2, x]. Since
f (2) = 3, there is c ∈ (2, x) such that
√
1 + 4x − 3 = f ′ (c)(x − 2).

22
We observe that f ′ (2) = 2/3 and f ′ (c) < f ′ (2) for c > 2. Then
√ 2 √ 2x − 4
1 + 4x − 3 < (x − 2) or 1 + 4x < 3 + .
3 3
Now it is clear that (5.4) is valid.
We formulate also a generalization of the Mean Value Theorem.

Theorem 5.10 (Cauchy’s Theorem) Given a, b ∈ R with a < b, suppose the functions f , g are
continuous on [a, b] and differentiable on (a, b). Then there is c ∈ (a, b) such that

[f (b) − f (a)]g ′ (c) = [g(b) − g(a)]f ′ (c). (5.5)

Proof of Theorem 5.10. Let us introduce the function

h(x) = [f (b) − f (a)]g(x) − [g(b) − g(a)]f (x), x ∈ [a, b].

Then
h(a) = f (b)g(a) − f (a)g(b) = h(b),
and the function h verifies all the assumptions of Theorem 5.3. Then there is c ∈ (a, b) such that
h′ (c) = 0. It is clear that

h′ (x) = [f (b) − f (a)]g ′ (x) − [g(b) − g(a)]f ′ (x).

This leads to (5.5) and Theorem 5.10 is proved.

6 Applications of the Mean Value Theorem

In this section, we will assume that a, b ∈ R and a < b.

Lemma 6.1 Let f be continuous on [a, b] and differentiable on (a, b). If f ′ (x) = 0 for all x ∈ (a, b),
then f is constant on [a, b].

Proof of Lemma 6.1. The lemma will be proved by contradiction. To this end we suppose that f is
not constant on [a, b]. Then there exist a1 and b1 such that a ⩽ a1 < b1 ⩽ b, and f (a1 ) ̸= f (b1 ). By
the Mean Value Theorem, there is c ∈ (a1 , b1 ) such that

f (b1 ) − f (a1 )
f ′ (c) = ̸= 0,
b 1 − a1
which is a contradiction. This proves Lemma 6.1.
The next result gives direct criteria for the monotonicity of a function in terms of its derivative.

23
 y

f ′ (c) > 0
y = f (x)

a c b x

Figure 8: Strictly increasing function: for all points c ∈ (a, b) we have f ′ (c) > 0.

Lemma 6.2 Let f be differentiable on (a, b). Then

1. if f ′ (x) > 0 for all x ∈ (a, b), then f is strictly increasing on (a, b);

2. if f ′ (x) < 0 for all x ∈ (a, b), then f is strictly decreasing on (a, b).

Proof of Lemma 6.2. Let us prove statement (1). Fix any x1 , x2 ∈ (a, b) with x1 < x2 . By the Mean
Value Theorem, there is c ∈ (x1 , x2 ) such that

f (x2 ) − f (x1 )
= f ′ (c) > 0.
x2 − x1
This means that f (x1 ) < f (x2 ). Therefore, f is strictly increasing on (a, b).
The proof of statement (2) can be done in a similar way.

Example 6.1 Let n ∈ N and f : (0, +∞) → R be defined by f (x) = xn . Then f ′ (x) = n · xn−1 ,
and f ′ (x) > 0 for all x > 0. This implies that f is strictly increasing. In particular, this proves that
every positive real number has exactly one n-th root.

7 L’Hôspital’s Rule
In this section we are dealing with the results which enable to find the limits by using the derivatives.
The theorems given below are known as L’Hôpital’s Rule.

24
Theorem 7.1 Suppose the functions f , g are continuous on [a, b] and differentiable on (a, b) and, in
addition, f (e
x) = g(e e ∈ [a, b]. Suppose also that there is δ > 0 such that g ′ (x) ̸= 0 for
x) = 0, where x
all x ∈ B(e e, where B(e
x; δ) ∩ [a, b], x ̸= x x − δ, x
x; δ) = (e e + δ). Then it holds

f ′ (x) f (x)
if lim =ℓ =⇒ lim = ℓ. (7.1)
x g ′ (x)
x→e x g(x)
x→e

Proof of Theorem 7.1. Let {xk } ⊂ [a, b] that converges to x e and such that xk ̸= x
e for every k. By
Cauchy’s Theorem, for each k, there is a sequence {ck }, with ck between xk and xe, such that

x)]g ′ (ck ) = [g(xk ) − g(e

[f (xk ) − f (e x)]f ′ (ck ).

Since f (e x) = 0, and g ′ (ck ) ̸= 0 for k sufficiently large,

x) = g(e

f (xk ) f ′ (ck )
= ′ .
g(xk ) g (ck )

Taking into account that g ′ (x) ̸= 0 for x near x x) = 0, we also have that g(xk ) ̸= 0 for k
e and g(e
sufficiently large. By the Squeeze Theorem, the sequence {ck } converges to x
e. Therefore,

f (xk ) f ′ (ck ) f ′ (x)

lim = lim ′ = lim ′ = ℓ.
k→∞ g(xk ) k→∞ g (ck ) x→ex g (x)

Consequently relation (7.1) follows from the Sequential Characterization of Limits Theorem. Theo-
rem 7.1 is proved.

Example 7.2 Prove that

2x + sin x
lim = 1.
x→0 x2 + 3x
Solution to Example 7.2. First we observe that the conditions of Theorem 7.1 are valid. We introduce
the functions
f (x) = 2x + sin x, g(x) = x2 + 3x, and choose x e = 0.
For the sake of definiteness, we take [a, b] = [−1, 1]. Then the functions f and g are continuous
on [a, b] and differentiable on (a, b). In addition f (x)/g(x) is well defined on [−1, 1] \ {0}. We set
δ = 7/3. Then g ′ (x) = 2x + 3 ̸= 0 for x ∈ B(e x; δ) ∩ [−1, 1]. Finally we obtain the limit of the
quotient of derivatives to obtain

f ′ (x) 2 + cos x limx→0 (2 + cos x) 2+1

lim = lim = = = 1.
x g ′ (x)
x→e x→0 2x + 3 limx→0 (2x + 3) 3
Now Theorem 7.1 implies the desired result
2x + sin x
lim = 1.
x→0 x2 + 3x

25
Example 7.3 Calculate the limit

3x3 − 2x2 + 4x − 5
lim .
x→1 4x4 − 2x − 2
Solution to Example 7.3. As in the previous example we apply Theorem 7.1. We introduce the
functions
f (x) = 3x3 − 2x2 + 4x − 5 and g(x) = 4x4 − 2x − 2.
It is clear that f (1) = g(1) = 0. Moreover,

f ′ (x) = 9x2 − 4x + 4 and g ′ (x) = 16x3 − 2.

Since the derivative g ′ (1) = 16 − 2 = 14 ̸= 0 and it is a continuous function, then we have that
g ′ (x) ̸= 0 for x near the point 1. It is easy to calculate that

9x2 − 4x + 4 9
lim 3
= .
x→1 16x − 2 14
9
The desired limit is also .
14
If the derivatives of f , g satisfy the assumptions of Theorem 7.1 we can use L’Hospital’s rule to
determine first the limit of f ′ (x)/g ′ (x) and then apply this rule once more in order to find the limit of
f (x)/g(x).

Example 7.4 Calculate the limit

x2
lim .
x→0 1 − cos x

Solution to Example 7.4. In this example f (x) = x2 and g(x) = 1 − cos x. It is clear that these
functions and all their derivatives are continuous. In addition f (0) = g(0) = 0, g ′ (x) = sin x,
g ′ (0) = 0, f ′ (0) = 0 = g ′ (0) and g ′′ (x) = cos x ̸= 0 for x in a neighborhood of 0. Moreover,

f ′′ (x) 2
lim = lim = 2.
x→0 g ′′ (x) x→0 cos x

Applying now L’Hospital’s rule we obtain that

f ′ (x) f ′′ (x) 2
lim = lim ′′ = lim = 2.
x→0 g ′ (x) x→0 g (x) x→0 cos x

Applying L’Hospital’s rule for the second time we have that

x2 f (x) f ′ (x)
lim = lim = lim ′ = 2.
x→0 1 − cos x x→0 g(x) x→0 g (x)

Finally, the desired limit equals 2.

26
Example 7.5 Consider the functions g(x) = x + 3x2 and f : R → R defined by

2 1
 x sin , if x ̸= 0,


x
f (x) =

 0, if x = 0.


Find the limit

f (x) x2 sin x1
lim = lim .
x→0 g(x) x→0 x + 3x2

Solution to Example 7.5. Taking into account the definition of derivative at the point x = 0 one can
see that the function f is differential and its derivative is given by

 2x sin 1 − cos 1 , if x ̸= 0,

′ x x
f (x) =

0, if x = 0.

One can observe that limx→0 f ′ (x) does not exist. Then we conclude that L’Hospital’s rule cannot be
applied. However
x2 sin x1
lim does exist.
x→0 x + 3x2

This fact can be seen from the following relations

x2 sin x1 x sin x1 limx→0 (x sin x1 )

lim = lim = = 0.
x→0 x + 3x2 x→0 1 + 3x limx→0 (1 + 3x)

Theorem 7.6 Let a, b ∈ R, a < b, and x

e ∈ (a, b). Consider the functions f, g : (a, b) \ {e
x} → R
which are differentiable on (a, b) \ {e
x} and assume

lim f (x) = lim g(x) = ∞.

x→e
x x→e
x

Let us suppose that there exists δ > 0 such that g ′ (x) ̸= 0 for all x ∈ B(e
x; δ) ∩ (a, b), x ̸= x
e.
Then, if ℓ ∈ R and
f ′ (x)
lim = ℓ, (7.2)
x g ′ (x)
x→e

we have that
f (x)
lim = ℓ. (7.3)
x g(x)
x→e

27
Proof of Theorem 7.6. It is given that

lim f (x) = lim g(x) = ∞.

x→e
x x→e
x

Then one can choose δ > 0 such that f (x) ̸= 0 and g(x) ̸= 0 for all x ∈ B(e
x; δ) ∩ (a, b). Let us prove
that
f (x)
lim+ = ℓ.
x→ex g(x)

The case x → x̃ is similar.

Fix any ε > 0. We have to choose δ0 > 0 such that

f (x)
−ℓ <ε
g(x)

whenever x ∈ B+ (ex; δ0 ) ∩ (a, b), where B+ (e

x; δ0 ) = (e
x, x
e + δ0 ). It follows from (7.2) that we can
find K > 0 and a positive δ1 < δ such that

f ′ (x) f ′ (x) ε
⩽K and −ℓ < , (7.4)
g ′ (x) ′
g (x) 2

whenever x ∈ B+ (e x; δ1 ) ∩ (a, b), x ̸= x

e.
Fix α ∈ B+ (e
x; δ1 ) ∩ (a, b) (in particular, α > x
e). We know that

lim f (x) = ∞.
x→e
x

Then we can find δ2 > 0, δ2 < min{δ1 , α − x e} and f (x) ̸= f (α) for x ∈ B+ (e x; δ2 ) ∩ (a, b) =
B+ (e ′
x; δ2 ). Moreover, for such x, taking into account that g (z) ̸= 0 if x < z < α, Rolle’s Theorem
implies that g(x) ̸= g(α). Therefore, for all x ∈ B+ (e
x; δ2 ),
g(α)
f (x) f (x) − f (α) 1 − g(x)
= · . (7.5)
g(x) g(x) − g(α) 1 − f (α)
f (x)

Let us introduce the function

g(α)
1− g(x)
Hα (x) = f (α)
for x ∈ B+ (e
x; δ2 ).
1− f (x)

Since
lim f (x) = lim g(x) = ∞,
x→e
x x→e
x

then we have that

lim Hα (x) = 1.
x+
x→e

This means that there exists γ : 0 < γ < δ2 such that

ε
|Hα (x) − 1| < whenever x ∈ B+ (e
x; γ).
2K

28
Now it follows from Cauchy’s Theorem applied on [x, α] that for any x ∈ B+ (e
x; γ),
[f (x) − f (α)]g ′ (c) = [g(x) − g(α)]f ′ (c),
for some c ∈ (x, α) ∩ B(e
x; δ1 ). For this c, from (7.5), we have that
f (x) f ′ (c)
= ′ Hα (x).
g(x) g (c)
Since c ∈ B(e
x; δ1 ), from (7.4) we obtain that, for x ∈ B+ (e
x; γ),
f ′ (c) f ′ (c) f ′ (c)
 
f (x)
− ℓ = ′ Hα (x) − ℓ = ′ (Hα (x) − 1) + −ℓ
g(x) g (c) g (c) g ′ (c)

f ′ (c) f ′ (c) ε ε
⩽ ′
|Hα (x) − 1| + ′
−ℓ <K + = ε.
g (c) g (c) 2K 2
Now we set δ0 = γ. This completes the proof of Theorem 7.6.

Example 7.7 Find the limit

ln x2
lim 1 .
x→0 1 + √ 3 2
x
√
3
Solution to Example 7.7. In this example f (x) = ln x2 , g(x) = 1 + 1/ x2 , and x e = 0. As for the
interval containing zero one can take any interval (a, b) such that 0 ∈ (a, b). The functions f , g are
differentiable and g ′ (x) ̸= 0 for all x ̸= 0. Moreover,
lim f (x) = lim g(x) = ∞.
x→e
x x→e
x

Let us consider the quotient of the derivatives. We have

√
(ln x2 )′ 2/x
3
x5 √
3
lim 1 ′
= lim 2 1 = −3 lim = −3 lim x2 = 0.
x→0 (1 + √ 3 2)
x
x→0 − √
3
3 x 5
x→0 x x→0

Then from Theorem 7.6 we have that

ln x2
lim 1 = 0.
x→0 1 + √ 3 2
x

The next theorem can be proved in the same way as Theorem 7.1.

Theorem 7.8 Suppose that f and g are differentiable on (a, ∞), g ′ (x) ̸= 0 for all x ∈ (a, ∞), and

lim f (x) = lim g(x) = 0.

x→∞ x→∞

If ℓ ∈ R, then
f ′ (x) f (x)
lim =ℓ =⇒ lim = ℓ. (7.6)
x→∞ g ′ (x) x→∞ g(x)

29
Example 7.9 Find the limit
1
lim π .
x→∞
x − arctan x
2
Solution to Example 7.9. We divide the numerator and the denominator by x and use Theorem 7.8.
The limit of the quotient of the derivatives is given by
1/x −1/x2 x2 + 1
lim = lim −1 = lim = 1.
x→∞ ( π − arctan x) x→∞ x→∞ x2
2 x2 +1

Now Theorem 7.8 implies that the desired limit is also 1.

The next theorem can be proved by the arguments similar to ones from the proof of Theorem 7.6.

Theorem 7.10 Let f and g be differentiable on (a, ∞), g ′ (x) ̸= 0 for all x ∈ (a, ∞) and

lim f (x) = lim g(x) = ∞.

x→∞ x→∞

If ℓ ∈ R, then
f ′ (x) f (x)
lim = ℓ, ⇒ lim = ℓ. (7.7)
x→∞ g ′ (x) x→∞ g(x)
ln x
Example 7.11 Find the limit lim .
x→∞ x

Solution to Example 7.11. In this example we set f (x) = ln x and g(x) = x. These functions satisfy
the conditions of Theorem 7.10. Then we have that
f ′ (x) 1/x 1
lim = lim = lim =0
x→∞ g ′ (x) x→∞ 1 x→∞ x

and the desired limit is also zero.

Example 7.12 Show that the following limit cannot be calculated by L’Hospital’s rule and find it :
x + sin x
lim .
x→∞ x
Solution to Example 7.12. Here
f (x) = x + sin x, g(x) = x and f ′ (x) = 1 + cos x, g ′ (x) = 1.
def def

It is clear that
1 + cos x
lim = lim (1 + cos x) does not exist!!!
x→∞ 1 x→∞
However, we can easily find the desired limit. In fact,
 
x + sin x sin x
lim = lim 1 + = 1.
x→∞ x x→∞ x

30
8 Taylor’s Theorem
In this section, we will discuss the results which enable to approximate differentiable functions by
polynomials.

Theorem 8.1 (Taylor’s Theorem. Remainder in Lagrange’s form) Let n ∈ N. Suppose f : [a, b] →
R is a function such that its n-th derivative f (n) is a continuous function on [a, b] and (n + 1)-th
derivative f (n+1) exist for all x ∈ (a, b). Let x0 ∈ [a, b]. Then for any x ∈ [a, b] with x ̸= x0 , there is
θ ∈ (x0 , x) (or θ ∈ (x, x0 )) such that

f (n+1) (θ)
f (x) = Pn (x) + (x − x0 )n+1 ,
(n + 1)!

where
n
X f (k) (x0 )
Pn (x) = (x − x0 )k .
k=0
k!

Proof of Theorem 8.1. Consider x0 ∈ [a, b] and fix x ̸= x0 . Since x − x0 ̸= 0, then there is λ ∈ R
such that
λ
f (x) = Pn (x) + (x − x0 )n+1 . (8.1)
(n + 1)!
Our goal is to prove that
λ = f (n+1) (θ), for θ ∈ (x0 , x).
Let us introduce the function
n
X f (k) (t) λ
g(t) = f (x) − (x − t)k − (x − t)n+1 . (8.2)
k=0
k! (n + 1)!

Now it follows from (8.1) that

n
X f (k) (x0 ) λ
g(x0 ) = f (x) − (x − x0 )k − (x − x0 )n+1
k=0
k! (n + 1)!

λ
= f (x) − Pn (x) − (x − x0 )n+1 = 0,
(n + 1)!
and
n
X f (k) (x) λ
g(x) = f (x) − (x − x)k − (x − x)n+1
k=0
k! (n + 1)!

n
X f (k) (x) λ
= f (x) − f (x) − (x − x)k − (x − x)n+1
k=1
k! (n + 1)!

= f (x) − f (x) = 0.

31
It follows from Rolle’s theorem that there is θ ∈ (x0 , x) such that g ′ (θ) = 0. Now taking into
account that x is fixed, the independent variable is denoted by t, and applying the product rule for the
derivatives we obtain that
n 
f (k+1) (θ) f (k) (θ)
X 
′ ′ k k−1
g (θ) = −f (θ) + − (x − θ) + (x − θ)
k=1
k! (k − 1)!
λ λ f (n+1) (θ)
+ (x − θ)n = · (x − θ)n − · (x − θ)n = 0.
n! n! n!
From this relation we conclude that λ = f (n+1) (θ). This completes the proof of Theorem 8.1.

Definition 8.2 The polynomial Pn (x) in Theorem 8.1 is called the n-th Taylor polynomial of the
function f at the point x0 .

Example 8.3 Estimate the error of approximation the function f (x) = sin x with its Taylor polyno-
mial at x0 = 0 on the interval [−1, 1].

Solution to Example 8.3. We observe that

f ′ (x) = cos x, f ′′ (x) = − sin x, f ′′′ (x) = − cos x.

Then it is easy to see that

x3
P3 (x) = x − .
3!
Moreover, for any θ ∈ R,
|f (4) (θ)| = | sin θ| ⩽ 1.
Thus, for x ∈ [−1, 1] and θ ∈ (x, 0) we obtain that

|f (4) (θ)| 4 1
|sin x − P3 (x)| = |x| ⩽ ≃ 0, 042.
4! 4!

9 An Application of Taylor’s Theorem

Theorem 9.1 Let n be an even positive natural number. Suppose f (n) exists and continuous on (a, b).
Consider x0 ∈ (a, b) satisfying

f ′ (x0 ) = ... = f (n−1) (x0 ) = 0 and f (n) (x0 ) ̸= 0.

Then

32
 1. f (n) (x0 ) > 0 if and only if f has a local minimum at x0 ;

2. f (n) (x0 ) < 0 if and only if f has a local maximum at x0 .

Proof of Theorem 9.1. Consider statement (1), i.e. suppose f (n) (x0 ) > 0. Since f (n) (x0 ) > 0 and
f (n) is continuous at the point x0 , then there is δ > 0 such that
f (n) (t) > 0 for all t ∈ B(x0 ; δ) ⊂ (a, b).
Consider x ∈ B(x0 ; δ). By Taylor’s theorem and the assumptions on f (n) one can conclude that there
is c ∈ (x0 , x) such that
f (n) (c)
f (x) = f (x0 ) + · (x − x0 )n .
n!
Since n is even and c ∈ B(x0 ; δ), f (x) ⩾ f (x0 ). This means that the function f has a local minimum
at x0 .
Now let us assume that the function f has a local minimum at the point x0 . Then there is δ > 0
such that
f (x) ⩾ f (x0 ) for all x ∈ B(x0 ; δ) ⊂ (a, b).
Consider a sequence {xk } ⊂ (a, b) which converges to x0 and such that xk ̸= x0 , ∀k ∈ N. Due to
Taylor’s theorem, there is a sequence {ck }, with ck ∈ (xk , x0 ) (or ck ∈ (x0 , xk )), ∀k ∈ N, satisfying
f (n) (ck )
f (xk ) = f (x0 ) + (xk − x0 )n
n!
We observe that xk ∈ B(x0 ; δ) for k sufficiently large. Then f (xk ) ⩾ f (x0 ) and
f (n) (ck )
f (xk ) − f (x0 ) = · (xk − x0 )n ⩾ 0.
n!
One can conclude that f (n) (ck ) ⩾ 0 for these values of k. The sequence {ck } converges to x0 and
consequently
f (n) (x0 ) = lim f (n) (ck ) ⩾ 0.
k→∞
The proof of statement (2) of the theorem can be done using by similar arguments.

Example 9.2 Consider the function f defined on R by

f (x) = x2 cos x.

Then it is easy to see that

f ′ (x) = 2x cos x − x2 sin x, f ′′ (x) = 2 cos x − 4x sin x − x2 cos x,

and
f (0) = f ′ (0) = 0, f ′′ (0) = 2 > 0.

Theorem 9.1 implies that the function f has a local minimum at 0.

It is important to notice that since f (0) = 0 and f (π) < 0, then 0 is not a global minimum.

33
 In what follows we are going to explain how to find the limits with the help of Taylor’s and
Maclaurin’s Theorems. We start by introducing the big O and small o notations, and then we will
formulate Taylor’s and Maclaurin’s theorems with the remainder in Peano’s form.

10 O(f ) and o(f )

Let us define the symbols O(f ) and o(f ).
Big O is a mathematical notation providing a convenient way of describing how fast a function is
growing, when its argument goes to some particular value or infinity.
Suppose the function g(x) satisfies the condition g(x) ̸= 0 in a deleted neighborhood of a point
x0 . Then if
f (x)
lim = 1,
x→x0 g(x)

we will say that f is equivalent to g as x → x0 . We write f (x) ∼ g(x) as x → x0 .

Suppose that there is a constant C > 0 such that in a deleted neighborhood of a point x0 ,
f (x)
⩽ C.
g(x)
Then we will say that the function f (x) is big O of g(x) as x → x0 , and write
f (x) = O(g(x)) as x → x0 .
If
f (x)
lim = 0,
x→x0 g(x)
the function is said to be little o of g(x) as x → x0 , and we write
f (x) = o(g(x)) as x → x0 . (10.1)

Example 10.1 Let us give some examples when (10.1) holds:

1
x2 = o(x), cos x sin2 x = o(x), tan3 x sin = o(x), x → 0.
x
In particular, if g(x) = 1 then the formula f (x) = o(1) as x → x0 means that f (x) → 0 as
x → x0 .

11 Remainder in Peano’s Form

Suppose that the n-th derivative of the function f , i.e., f (n) (x0 ), exists at the point x0 . Then in a
neighborhood B(x0 ) of x0 we have
n
X f (k) (x0 )
f (x) = (x − x0 )k + Rn (f, x) = Pn (x) + Rn (f, x). (11.1)
k=0
k!

34
Here the polynomial Pn (x) is known as the n-th Taylor polynomial of f at x0 and Rn (f, x) is the
remainder in Taylor’s formula.

◦
Lemma 11.1 Suppose that there exist f (n) (x0 ) and f ′ in some deleted neighborhood B (x0 ) of the
◦
point x0 . Then for x ∈ B (x0 ),
(Rn (f, x))′ = Rn−1 (f ′ , x). (11.2)

Proof of Lemma 11.1. From (11.1), we have that

n
!′
X f (k) (x0 )
(Rn (f, x))′ = f (x) − (x − x0 )k = Rn−1 (f ′ , x)
k=0
k!

and Lemma 11.1 is proved.

The following form of Taylor’s Theorem with less restrictive hypotheses on the function f is
known as Taylor’s Theorem with Peano’s Form of Remainder.

Theorem 11.1 (Taylor’s formula with Peano’s form of reminder) If f is a function such that its
n-th derivative f (n) (x0 ) at x0 exists then
n
X f (k) (x0 )
f (x) = · (x − x0 )k + o ((x − x0 )n ) as x → x0 . (11.3)
k=0
k!

Proof of Theorem 11.1. The proof of the theorem is done by the method of mathematical induction.
For n = 1 the statement of the theorem holds true. In fact, in this case the function f is differen-
tiable at x0 . Then

f (x) − f (x0 ) = f ′ (x0 )(x − x0 ) + o ((x − x0 )) as x → x0 .

Suppose that the statement of the theorem is valid for (n − 1), n ⩾ 1. Let us prove that (11.3) holds
true. We assume for the sake of definiteness that x > x0 and make use of Lagrange’s theorem and
Lemma 11.1. Then

Rn (f, x) = Rn (f, x) − Rn (f, x0 ) = Rn−1 (f ′ , ξ)(x − x0 ), x0 < ξ < x. (11.4)

By the induction assumption,

Rn−1 (f ′ , ξ) = o (ξ − x0 )n−1 = o (x − x0 )n−1

as x → x0 . (11.5)

Then from (11.4), (11.5) we obtain that

Rn (f, x) = o ((x − x0 )n ) as x → x0 .

This completes the proof of Theorem 11.1.

35
 ◦
Theorem 11.2 (Uniqueness of Taylor’s expansion) Suppose that for x ∈ B (x0 ), some function f
is represented by two Taylor’s expansions

f (x) = a0 + a1 (x − x0 ) + ... + an (x − x0 )n + o ((x − x0 )n ) as x → x0 , (11.6)

and
f (x) = b0 + b1 (x − x0 ) + ... + bn (x − x0 )n + o ((x − x0 )n ) as x → x0 . (11.7)

Then
a0 = b0 , a1 = b1 , ... , an = bn .

Proof of Theorem 11.2. We subtract term by term representation (11.7) from (11.6) and obtain

c0 + c1 (x − x0 ) + ... + cn (x − x0 )n = o ((x − x0 )n ) as x → x0 , (11.8)

where c0 = (a0 − b0 ) , c1 = (a1 − b1 ) , ..., cn = (an − bn ). Therefore, in order to prove the theorem
we have to show that
c0 = c1 = ... = cn = 0. (11.9)
Let us pass to the limit in (11.8) as x → x0 . Obviously, c0 = 0. We now divide (11.8) by (x − x0 )
and obtain
c1 + ... + cn · (x − x0 )n−1 = o (x − x0 )n−1


as x → x0 . (11.10)
Passing to the limit in (11.10) as x → x0 , we get c1 = 0. In a similar way we prove that

c2 = ... = cn = 0.

This completes the proof of Theorem 11.2.

11.1 Maclaurin’s Formulae

Definition 11.3 Taylor’s formula (11.3) for x0 = 0 is called Maclaurin’s formula with Peano’s form
of remainder, and it takes the form
n
X f (k) (0) k
f (x) = · x + o (xn ) as x → 0. (11.1)
k=0
k!

Now let us present examples of Maclaurin formulae for some common functions.

36
 Maclaurin’s Formulae for the Elementary Functions
a) exponential function
x2 xn
ex = 1 + x + + ··· + + o(xn )
2! n!
or n
X xk
ex = + o(xn );
k=0
k!
b) hyperbolic functions

x3 x5 x2n+1
sinh x = x + + + ··· + + o(x2n+2 )
3! 5! (2n + 1)!
or n
X x2k+1
sinh x = + o(x2n+2 );
k=0
(2k + 1)!
2
x x4 x2n
cosh x = 1 + + + ··· + + o(x2n+1 )
2! 4! (2n)!
or n
X x2k
cosh x = + o(x2n+1 );
k=0
(2k)!
c) trigonometric functions

x3 x5 (−1)n x2n+1
sin x = x − + + ··· + + o(x2n+2 )
3! 5! (2n + 1)!
or n
X (−1)k x2k+1
sin x = + o(x2n+2 );
k=0
(2k + 1)!
2
x x4 (−1)n x2n
cos x = 1 − + + ··· + + o(x2n+1 )
2! 4! (2n)!
or n
X (−1)k x2k
cos x = + o(x2n+1 );
k=0
(2k)!
d) power function
n  
p
X p
(1 + x) = xk + o(xn ),
k
k=0

where    
p p(p − 1)(p − 2) . . . (p − k + 1) p
= and = 1;
k k! 0
In particular,
1
= 1 − x + x2 + · · · + (−1)n xn + o(xn )
1+x

37
or n
1 X
= (−1)k xk + o(xn );
1 + x k=0
1
= 1 + x + x2 + · · · + xn + o(xn )
1−x
or n
1 X
= xk + o(xn );
1 − x k=0
e) logarithmic function

x2 x 3 (−1)n−1 xn
ln(1 + x) = x − + + ··· + + o(xn )
2 3 n
or n
X (−1)k−1 xk
ln(1 + x) = + o(xn )
k=1
k
and n
X xk
ln(1 − x) = − + o(xn ).
k=1
k

12 Finding Limits using Maclaurin’s Formula

In this section we consider three important examples that will give us a general recipe for computing
more complicated limits.

Example 12.1 Suppose that f is n times differentiable and g is m times differentiable, and f (n) (0) ̸=
0, g (m) (0) ̸= 0. Find a limit

f (x)
lim , where f (0) = g(0) = 0
x→0 g(x)

and
f (i) (0) = 0, g (j) (0) = 0, i = 1, n − 1, j = 1, n − 1.

Solution to Example 12.1. We represent the functions by their Maclaurin’s formulae

f (x) = axn + o(xn )(a ̸= 0) and g(x) = bxm + o(xm )(b ̸= 0).

There are three possibilities:

If m = n, then
f (x) axn + o(xn ) a
lim = lim n = . (12.1)
x→0 g(x) x→0 bx + o(xn ) b

38
If n > m, then
f (x)
lim = 0. (12.2)
x→0 g(x)

If n < m, then
f (x)
lim = ∞. (12.3)
x→0 g(x)

Example 12.2 Taylor’s formula is often used to calculate the limits of the form

 g(x)
lim f (x) ,
x→x0

where
f (x) > 0, lim f (x) = 1, lim g(x) = ∞.
x→x0 x→x0

Solution to Example 12.2. Consider first the case x0 = 0. Suppose that the functions f and g can be
represented as
1
f (x) = 1 + axk + o(xk ) and g(x) = , as x → 0,
bxk + o(xk )

where a ̸= 0, b ̸= 0, k ∈ N. Since
 1 axk + o(xk ) a
lim 1 + axk + o(xk ) axk +o(xk ) = e and

lim = ,
x→0 x→0 bxk + o(xk ) b

then g(x)  1
= lim 1 + axk + o(xk ) bxk +o(xk ) =
 
lim f (x)
x→0 x→0

  1  axk +o(xk )
k k axk +o(xk ) bxk +o(xk )
= lim 1 + ax + o(x ) =
x→0

1
 
 
axk +o(xk ) 
( )
axk +o(xk ) ln  1+axk +o(xk )



a
= lim e bxk +o(xk ) = eb . (12.4)
x→0

If the functions f, g have the representations

1 + axk + o(xk )
f (x) = = 1 + (a − a1 )xk + o(xk ), as x → 0
1 + a1 xk + o(xk )

and
1
g(x) = , as x → 0,
bxk + o(xk )

39
where a ̸= 0, a1 ̸= 0, b ̸= 0, k ∈ N, then
 g(x) (a−a1 )
lim f (x) =e b . (12.5)
x→0

Suppose now that

1
f (x) = 1 + axn + o(xn ) and g(x) = , as x → 0,
bxm + o(xm )

where a ̸= 0, b ̸= 0, m, n ∈ N. Then, from (12.4), we have

g(x)
= e0 = 1

lim f (x) for n > m. (12.6)
x→0

If m > n and (m − n) is an odd number then

 g(x)
lim f (x) does not exist.
x→0

If m > n and (m − n) is an even number then

(
 g(x) +∞, ab > 0,
lim f (x) =
x→0 0, ab < 0.

 g(x)
Remark 3 Notice that in order to find the limit of f (x) as x → 0, one can find the limit of the
logarithm of this function limx→0 g(x) ln f (x), and then compute

g(x)
= elimx→0 [g(x) ln f (x)] .

lim f (x)
x→0

Remark 4 If we calculate the limits with Taylor’s formula at some point x0 ̸= 0, one can set

t = x − x0

and reduce our problem to the calculation of the limit at t = 0. The case when x → ∞ can be reduced
to the case t = 0 by the change of variable x = 1/t.

Example 12.3 Find the limit

√
1 + 2 tan x − ex + x2 N (x)
lim = lim . (12.7)
x→0 arcsin x − sin x x→0 D(x)

40
Solution to Example 12.3. We observe that the numerator and the denominator in (12.7) go to zero
as x → 0, i.e.,
lim N (x) = lim D(x) = 0.
x→0 x→0

For the function D(x) we make use of the expansions

x3 x3
sin x = x − + o(x3 ) and arcsin x = x + + o(x3 ) as x → 0,
6 6
and Maclaurin’s formula becomes
x3 x3 x3
D(x) = arcsin x − sin x = x + −x+ + o(x3 ) = + o(x3 ) as x → 0.
6 6 3
This means that we have to represent N (x) by Maclaurin’s formula with o(x3 ). To this end we make
use of the formulae
√ 1 1 1
1 + t = 1 + t − t2 + t3 + o(t3 ) as t → 0, (12.8)
2 8 16
x3
tan x = x + + o(x3 ) as x → 0. (12.9)
3
Then, from (12.8), (12.9) we get
√ 1 1 1
1 + 2 tan x = 1 + (2 tan x) − (2 tan x)2 + (2 tan x)3 + o (tan x)3


2 8 16
3 2 3
x x x
=1+x+ − + + o(x3 )
3 2 2
x2 5 3
=1+x− + x + o(x3 ) as x → 0. (12.10)
2 6
Recall Maclaurin’s formula for the exponential ex

x 2 x3
ex = 1 + x + + + o(x3 ) as x → 0. (12.11)
2 6
Then we can obtain the representation of the function N (x). Expansions (12.10) and (12.11) yield
√ 2
N (x) = 1 + 2 tan x − ex + x2 = x3 + o(x3 ) as x → 0.
3
In this way the fraction becomes

2x3 /3 + o(x3 )
as x → 0.
x3 /3 + o(x3 )

This means that the limit (12.7) is given by

√
1 + 2 tan x − ex + x2
lim = 2.
x→0 arcsin x − sin x

41
13 Exercises with Solutions

Exercise 13.1 Find the derivatives of sin x and cos x using the definition of the derivative.

Solution to Exercise 13.1. (1) Consider

sin(x + ∆x) − sin x sin x cos ∆x + cos x sin ∆x − sin x

= =
∆x ∆x
1 − cos ∆x sin ∆x
= − sin x · + cos x · .
∆x ∆x
Since  
2 ∆x 2 ∆x ∆x
1 − cos ∆x = 1 − cos − sin = 2 sin2 ,
2 2 2
then as ∆x → 0,

sin(x + ∆x) − sin x sin2 ∆x

2 sin ∆x
= −2 sin x · + cos x · −→ cos x
∆x ∆x ∆x
and
(sin x)′ = cos x.

(2) In a similar way,

cos(x + ∆x) − cos x cos x cos ∆x − sin x sin ∆x − cos x

= =
∆x ∆x
1 − cos ∆x sin ∆x
= − cos x · − sin x · −→ − sin x, as ∆x → 0.
∆x ∆x
Thus
(cos x)′ = − sin x.

Exercise 13.2 Find the derivative of f (x) = |x| using the chain rule.
√ √
Solution to Exercise 13.2. By definition, |x| = x2 . Let us consider the function f (x) = x2 as a
composite, i.e.
√
f = g and g = x2 .
Applying now Theorem 1.10, we obtain that

′ √
′ 1 x x
|x| = x2 = √ · 2x = √ = .
2 x2 x2 |x|

42
Exercise 13.3 Compute the derivative of function y(x) = 2x ln |x|, x ̸= 0.

Solution to Exercise 13.3. By Theorem 1.8 (3), we have

′
′ h i
y ′ = 2x · ln |x| + 2x · ln |x| = see Table of Derivatives, statement 4
 
x x 1 x x 1
= 2 ln 2 · ln |x| + 2 · · = 2 ln 2 · ln |x| + .
|x| |x| x

Exercise 13.4 Find the derivative of y(x) = sinh x · cosh x.

Solution to Exercise 13.4. By the definition of hyperbolic functions sinh, cosh,

ex − e−x ex + e−x
sinh x = and cosh x = .
2 2
Then by Theorem 1.8 (3),

′
′
y ′ = sinh x cosh x + cosh x sinh x

e2x + 2 + e−2x + e2x − 2 + e−2x

= cosh2 x + sinh2 x =
4

e2x + e−2x
= = cosh(2x).
2

x

Exercise 13.5 Find the derivative of y(x) = ln ln .
2
Solution to Exercise 13.5. By the chain rule,
1 2 1 1
y′ = x · · = x.
ln x 2 x ln
2 2

Exercise 13.6 Find the derivative of y(x) = ln | sin x|.

Solution to Exercise 13.6. First we notice that x cannot be a multiple of π, that is x ̸= πk, k ∈ Z.
Using the chain rule, we obtain
1
′ 1 sin x
y′ = · | sin x| = · · cos x = cot x,
| sin x| | sin x| | sin x|
where sin x ̸= 0 or x ̸= πk (k ∈ Z).

43
 √

Exercise 13.7 Find the derivative of y(x) = ln x2 + x4 + 1 .

Solution to Exercise 13.7. By the chain rule,

1 √
′
y′ = √ · x2 + x4 + 1
x2 + x4 + 1

4x3
 
1 1
= √ · 2x + · √ =
x2 + x4 + 1 2 x4 + 1
√
x2 x2 + x4 + 1
 
1 2x
= √ · 2x · 1 + √ = √ · √
x2 + x4 + 1 x4 + 1 x2 + x4 + 1 x4 + 1

2x
=√ .
x4 + 1

In the next Exercises (13.8)-(13.11) we illustrate how the Inverse Function Theorem 2.1 can be
used for computing the derivatives.

√
Exercise 13.8 Find the derivative of the function y(x) = n
x.
√
Solution to Exercise 13.8. Let us find the inverse function for y = n x. To this end, we represent x
in terms of y, i.e.,
√ √
n
y = f (x) = n x ⇒ y n = n x ⇒ x = φ (y) = y n .

From the Inverse Function Theorem 2.1

√
′ 1
n
x = f ′ (x) = .
φ′ (y) y=f (x)

Since
φ′ (y) = (y n )′ = n y n−1 ,
√
we plug in y = n
x instead of y to get
√ 1 1
( n x)′ = √ n−1
= √
n
, x > 0,
n( x)
n
n xn−1
i.e.,
√
′ 1
n
x = √
n
(x > 0) .
n xn−1

Exercise 13.9 Find the derivative of y(x) = ln x.

44
Solution to Exercise 13.9. The functions ln x and ex are mutually inverse. Then

x = ϕ (y) = ey , x > 0, y ∈ R.

Let us recall that eln x = x. Then the derivative of ln x can be found from the derivative of exponential
function, namely,
1 1 1 1
(ln x)′ = ′ = y ′ = ln x = .
ϕ (y(x)) (e ) y=ln x e x
Thus
1
(ln x)′ = .
x

√
3
Exercise 13.10 Find the derivative of y(x) = x + 1.

Solution to Exercise 13.10. Let us find the inverse function for our function y, which is monotonically
increasing. We have √
y = 3 x + 1 ⇒ y 3 = x + 1 ⇒ x = y 3 − 1.
Let us find the derivative y ′ . We have
√ ′ 1 1
3
x + 1 = y ′ (x) = = =
x′ (y) y=y(x) (y 3 − 1)′ y=y(x)

1 1
= = 2 , x ̸= −1.
3y 2
p
y=y(x) 3
3 (x + 1)

 
1
Exercise 13.11 Find the derivative of y(x) = arctan .
x

Solution to Exercise 13.11. The inverse function x = x(y) to the function y = y(x) can be found in
a following way:
1 1 1
y = arctan ⇒ = tan y ⇒ x = , where x ̸= 0.
x x tan y

Now the derivative of y = f (x) is given by

 ′
1 1 1
arctan = y ′ (x) = ′ = ′
x x (y) y=y(x) 1 y=y(x)
tan y

1 1 tan2 y
= · =− .
− tan12 y cos2 y y=y(x) 1 + tan2 y y=y(x)

45
The desired derivative is
′ 1 2
tan2 arctan x1



1 x 1
arctan =− 2 1

=− 2 =− .
x 1 + tan arctan x 1+ 1 1 + x2


x

Let us now use the chain rule and Statement 15 from the Table of Derivatives to compute the same
derivative:  ′  ′
1 1 1 1 1 1
arctan = 1 · =− 2 · 1 = − .
x 1 + x2 x x 1 + x2 1 + x2

In the Exercises 13.12-13.16 we are dealing with the higher order derivatives.

Exercise 13.12 Find the derivatives of the n-th order, n = 1, ..., 5 of the function

y(x) = (2x + 1)3 (x − 1) .

Solution to Exercise 13.12. It is clear that

y(x) = (2x + 1)3 (x − 1) = (8x3 + 12x2 + 6x + 1) · (x − 1)

= 8x4 + 12x3 + 6x2 + x − 8x3 − 12x2 − 6x − 1 =

= 8x4 + 4x3 − 6x2 − 5x − 1.

Now, by (1.2), we can find the derivatives of the n-th order (n = 1, ..., 5):

y ′ = (8x4 + 4x3 − 6x2 − 5x − 1)′ = 32x3 + 12x2 − 12x − 5,

′
y ′′ = (y ′ ) = (32x3 + 12x2 − 12x − 5)′ = 96x2 + 24x − 12,

′
y (3) = (y ′′ ) = (96x2 + 24x − 12)′ = 192x + 24,

′
y (4) = y (3) = (192x + 24)′ = 192,

′
y (5) = y (4) = (192)′ = 0.

Exercise 13.13 Find the n-th order derivative of y(x) = ln x.

46
Solution to Exercise 13.13. In this exercise we make use of the method of mathematical induction.
We start by computing several derivatives of y = y(x) to make a guess. It follows from (1.2) that
1
y ′ = (ln x)′ =
,
x
 ′
′′ ′ ′ 1
′ 1
y = (y ) = = x−1 = −x−2 = − 2 ,
x x
 ′
′′ ′ 1 2
y (3)
= (y ) = − 2 = 2x−3 = 3 ,
x x
 ′
′′′ ′ 2 6
y (4)
= (y ) = = −6x−4 = − ,
x3 x4
 ′
(4) ′ 6 24
(5)
= − 4 = 24x−5 = 5 .


y = y
x x

Our hypothesis is that the derivative of the n-th order has the form

(−1)n−1 (n − 1)!
y (n) = . (13.1)
xn
As it was shown above, for n = 1 the formula (13.1) holds true. Now we differentiate the n-th
derivative to find the derivative of the (n + 1)th order. From (13.1) and (1.2) we obtain that

′
y (n+1) = (−1)n−1 (n − 1)! · x−n = (−1)n−1 (n − 1)! · (−n)x−[n+1] =

(−1)n n!
= ,
x[n+1]
and (13.1) is established.

Exercise 13.14 Find the n-th order derivative of y = x ln x at the point x = 1.

Solution to Exercise 13.14. We make use of the formula for the derivative of the product and formula
(1.2) to have
1
y ′ = (x ln x)′ = 1 · ln x + x · = ln x + 1,
x
1
y ′′ = (ln x + 1)′ = = x−1 ,
x

′ 1
y (3) = x−1 = −1 · x−2 = − ,
x2

′ 2!
y (4) = −1 · x2 = 2! x−3 = 3,
x

′ 3!
y (5) = 2! x−3 = 2! · (−3) · x−4 = −3! x−4 = − .
x4

47
Thus, our guess is that the n-th derivative, for n ⩾ 2, is given by

(n) (−1)n (n − 2)!

y = .
x(n−1)
This result can be proved by the method of mathematical induction (prove it!) and if x = 1 we obtain
that
(−1)n (n − 2)!
y (n) (1) = = (−1)n (n − 2)!.
1(n−1)

Exercise 13.15 Find the n-th order derivative of the function y = sin x.

Solution to Exercise 13.15. In this exercise we apply the method of mathematical induction. To
guess a form for the nth derivative, we compute several derivatives of y = sin x:
 π
y ′ = (sin x)′ = cos x = sin x + ,
2
 π
y ′′ = (cos x)′ = − sin x = sin x + 2 · ,
2

(3) ′
 π
y = (− sin x) = − cos x = sin x + 3 · ,
2
 π
y (4) = (− cos x)′ = sin x = sin x + 4 · .
2
Our hypothesis is that the derivative of the n-th order has the form

(n) (n)
 nπ 
y = (sin x) = sin x + .
2
This result can be proved by the method of mathematical induction.

1
Exercise 13.16 Find the n-th order derivative of the function y(x) = .
x
Solution to Exercise 13.16. As in the previous exercises the proof of the final result is based on the
method of mathematical induction. To this end we have to find the first derivatives of the function

48
y = y(x). Here we make use of the formula (1.2). We have
 ′
′ 1 (−1)
y = = −x−2 = 2 ,
x x x
′
(−1)2 · 2

′′ 1 −2 ′

−3
y = − 2 = −1 · x x = −1 · (−2) · x = ,
x x x3
!′
2
(−1) · 2 2 −3 ′
y ′′′ = = (−1)3 · 2 · 3 · x−4


3
= (−1) · 2 · x x
x
x
3
(−1) · 2 · 3
= ,
x4 !′
(−1)3 · 2 · 3 3 −4 ′
y (4) = (−1)4 · 4! · x−5


= 4
= (−1) · 2 · 3 · x
x
x
(−1)4 4!
= .
x5
Our hypothesis is that the derivative of the n-th order has the form

(n) (−1)n n!
y = .
xn+1
The proof can be done by the method of mathematical induction, as before.
In the Exercises 13.17-13.19 we are dealing with the derivatives of parametric functions.

Exercise 13.17 Consider the parametric equations defining an ellipse:

x(t) = a cos t, y(t) = b sin t,

where a, b are the semi-axes of the ellipse and t is a parameter. Find the first three derivatives of
y = y(x) with respect to x.

Solution to Exercise 13.17. Consider the first derivative of the function y with respect to x. We have

′ yt′ (b sin t)′ b cos t b

y = yx′ = ′ = ′ = = − cot t.
xt (a cos t) −a sin t a

In a similar way we can find the second derivative y ′′ , namely,

′
′ ′ b b 1



(y ) − cot t − − 2
y ′′ = yxx
′′
= ′t =x a t
= a sin t
=
xt (a cos t)′t (−a sin t)

b 1 b 3
=− 2 3 = − 2 csc t,
a sin t a

49
where
1
csc t = . (13.2)
sin t

(3)
Finally, yxxx is given by

′
′′ ′ − ab2 csc3 t t − ab2 · 3csc2 t · (csc t)′t


(3) (3) (yxx )t
y = yxxx = = =
x′t (a cos t)′t (−a sin t)

3b csc2 t · (− cot t) · csc t 3b csc3 t cot t 3b

=− 3
· = − 3
· = − 3 csc4 t cot t.
a (− sin t) a sin t a

Notice that by the chain rule

1
(csc3 t)′t = 3(csc2 t) · (csc t)′t = −3 csc2 t · · cos t =
sin2 t
= −3 csc2 t · cot t · csc t = −3 csc3 t · cot t.

Exercise 13.18 Consider the function y defined parametrically by

x = 1 + t2 , y = t − t3 .

Find the first three derivatives of the function y = y(x) with respect to x at t = 1.

Solution to Exercise 13.18. The first derivative of the function y with respect to x is
′
yt′ (t − t3 )t 1 − 3t2 1 3t
y ′ = yx′ = ′
= ′ = = − .
xt (1 + t2 )t 2t 2t 2

In a similar way we compute the second derivative y ′′ , namely,

′ ′ 3t ′
1 1
· (−t−2 ) − 32 1 3


′′ ′′ (y )
x t 2t
− 2 2 2t2
+ 2 1 3
y = yxx = ′ = ′ = =− =− 3
− .
xt 2
(1 + t ) 2t 2t 4t 4t

′′′
Finally, yxxx is given by

′
′′ ′ − 4t13 − 4t3 − 14 · (−3t−4 ) − 34 · (−t−2 )
′′′ ′′′ (yxx )t t
y = yxxx = = = =
x′t (1 + t2 )′t 2t
3
4t4
+ 4t32 3 3
= = 5 + 3.
2t 8t 8t

50
At the point where t = 1 we have that

yx′ = −1,
t=1

′′
yxx = −1,
t=1

′′′ 3 3 3 3 3
yxxx = 5
+ 3
= + = .
t=1 8·1 8·1 8 8 4

Exercise 13.19 Consider the parametric function y

x(t) = 1 + sin t, y(t) = t − cos t.

Find the third derivative of y = y(x) with respect to x at t = 0.

Solution to Exercise 13.19. The first derivative of the function y with respect to x is

yt′ (t − cos t)′t 1 + sin t

y ′ = yx′ = ′
= ′ = .
xt (1 + sin t)t cos t

The second derivative y ′′ is

1+sin t ′ cos t·cos t−(1+sin t)·(− sin t)
(yx′ )′t


′′ ′′ cos t cos2 t
y = yxx = ′ = = =
xt (1 + sin t)′ cos t

cos2 t + sin t + sin2 t 1 + sin t

= = .
cos3 t cos3 t

(3)
Finally, yxxx is given by
′′ ′ 1+sin t ′


′′′ (yxx )t 3
cos t t
y (3) = yxxx = =
x′t (1 + sin t)′t

cos4 t + 3 cos2 t sin t + 3 cos2 t sin2 t

=
cos7 t

cos2 t cos2 t + 3 sin t + 3 sin2 t

=
cos7 t

cos2 t + 3 sin t + 3 sin2 t

=
cos5 t

51
 ′′′
and at the point corresponding to t = 0 yxxx equals

′′′ cos2 0 + 3 sin 0 + 3 sin2 0 1+0+0

yxxx = 5
= = 1.
t=0 cos 0 1

The forthcoming Exercises 13.20-13.23 are devoted to the applications of the Mean Value Theo-
rem.

Exercise 13.20 Consider f (x) = x2 + 2x. Find all ξ ∈ [−2, 0] such that f ′ (ξ) = 0.

Solution to Exercise 13.20. Let us show that the function f (x) satisfies three conditions of Rolle’s
theorem.

1. First, we observe that f (x) is a quadratic function and, therefore, it is continuous in the interval
[−2, 0].

2. Since f (x) is quadratic, then it is differentiable over the open interval (−2, 0).

3. Finally, we have to check the third condition of Rolle’s theorem. In fact

f (−2) = (−2)2 + 2 · (−2) = 0 and f (0) = 02 + 2 · 0 = 0.

Thus f (−2) = f (0) = 0 and the third condition is also satisfied.

This enables to use Rolle’s theorem. Let us differentiate the function:

′
f ′ (x) = x2 + 2x = 2x + 2.

Now in order to calculate ξ, we have to find the solution to the equation f ′ (ξ) = 0, i.e.,

f ′ (ξ) = 2ξ + 2 = 0.

Therefore, we obtain that f ′ (ξ) = 0 for ξ = −1.

Exercise 13.21 Given f (x) = x2 − 6x + 5, find all ξ ∈ [2, 4] such that f ′ (ξ) = 0.

Solution to Exercise 13.21. Let us show that the function f (x) satisfies three conditions of Rolle’s
theorem.

1. First, we observe that f (x) is a quadratic function and, therefore, it is continuous in the interval
[2, 4].

2. Since f (x) is quadratic, then it is differentiable over the open interval (2, 4).

52
 3. Finally, we have to check the third condition of Rolle’s theorem. In fact

f (2) = 22 − 6 · 2 + 5 = −3 and f (4) = 42 − 6 · 4 + 5 = −3.

Thus f (2) = f (4) = −3 and the third condition is also satisfied.

Then
′
f ′ (x) = x2 − 6x + 5 = 2x − 6.
The next step is to find the solution to the equation f ′ (ξ) = 2ξ − 6 = 0 which yields ξ = 3.

Exercise 13.22 Let the function f = f (x) be given by

f (x) = x2 − 3x + 5, x ∈ [1, 4].

Show that all the conditions of Lagrange’s theorem are fulfilled and find ξ given by the theorem.

Solution to Exercise 13.22. Let us show that the function f (x) satisfies the conditions of Lagrange’s
theorem.

1. First, we observe that f (x) is a quadratic function and, therefore, it is continuous in the interval
[1, 4].

2. Since f (x) is quadratic, then it is differentiable over the open interval (1, 4). Moreover, the
derivative of f is given by

′
f ′ (x) = x2 − 3x + 5 = 2x − 3.

Lagrange’s theorem states that there exists ξ such that

f (b) − f (a) (42 − 3 · 4 + 5) − (12 − 3 · 1 + 5)

f ′ (ξ) = ⇒ 2ξ − 3 = .
b−a 4−1
Thus
9−3
2ξ − 3 = = 2 ⇒ 2ξ = 5.
3
The solution to this equation is ξ = 2, 5 ∈ (1, 4).

Exercise 13.23 Let the function be given by

√
f (x) = x+4 on the interval [0, 5].

Show that all the conditions of Lagrange’s theorem are fulfilled and find ξ satisfying the conditions of
this theorem.

53
Solution to Exercise 13.23. Let us show that the function f (x) satisfies the conditions of Lagrange’s
theorem.

1. First, we observe that f (x) is continuous in [0, 5].

2. f is differential on (0, 5), and the derivative of f is given by

√ ′ 1
′
f (x) = x+4 = √ .
2 x+4

This enables to apply Lagrange’s theorem and find the desired point ξ. We have
√ √
′ f (b) − f (a) 1 5+4− 0+4
f (ξ) = ⇒ √ =
b−a 2 ξ+4 5−0
1 1
⇒ √ =
2 ξ+4 5
and
p 5 25
ξ+4= or ξ + 4 = .
2 4
Thus the desired point is the point ξ = 2.25 ∈ (0, 5).

The remaining exercises concern finding limits with the help of Maclaurin’s expansion.

Exercise 13.24 Find the limit :

1 x2
earctan x − +
lim  1−
x 2 .
x→0 1+x
ln − 2x
1−x

Solution to Exercise 13.24. We make use of the following Maclaurin formulae

x 2 x3 x 2 x3
ln(1 + x) = x − + + o(x3 ), ln(1 − x) = −x − − + o(x3 ).
2 3 2 3
Then we can find the Maclaurin formula for the denominator of the fraction
2x3
 
1+x
ln − 2x = ln(1 + x) − ln(1 − x) − 2x = + o(x3 ) as x → 0.
1−x 3

Now we have to represent the numerator of our fraction by the Maclaurin formula with o(x3 ). We
know that
x3
arctan x = x − + o(x3 ) as x → 0 (13.3)
3
and, in addition,
t2 t3
et = 1 + t + + + o(t3 ) as t → 0. (13.4)
2 6
54
We plug (13.3) in (13.4) in order to have
x3 x2 x3
 
arctan x
e =1+ x− + + + o(x3 ) =
3 2 6
x2 x3
=1+x+ − + o(x3 ) as x → 0. (13.5)
2 6
Now we make use of the following Maclaurin formula
1
= 1 + x + x2 + x3 + o(x3 ) as x → 0.
1−x
Then we can obtain the Maclaurin formula for the numerator of the fraction. Namely,
1 x2 7
earctan x − + = − x3 + o(x3 ) as x → 0.
1−x 2 6
In this way, our fraction can be represented in the following form
− 67 x3 + o(x3 )
2x3
as x → 0.
3
+ o(x3 )
7
This means that the desired limit is equal to − .
4

Exercise 13.25 Find the limit

cot x3
lim [f (x)]g(x) = lim cos(xex ) − ln(1 − x) − x .
x→0 x→0

Solution to Exercise 13.25. First, we see that

1 1
g(x) = cot x3 = 3
= 3 as x → 0.
tan x x + o(x3 )
This means that we have to represent the function
f (x) = cos(xex ) − ln(1 − x) − x
by the Maclaurin formula with o(x3 ). To this end we make use of the following relations :
xex = x + x2 + o(x2 ), as x → 0,
t2
cos t = 1 − + o(t3 ) as t → 0,
2
x2 x3
− ln(1 − x) = x + + + o(x3 ) as x → 0.
2 3
Then we obtain the Maclaurin formula for the function f (x)
2
f (x) = 1 − x3 + o(x3 ) as x → 0.
3
Now we have that   3 1 3
g(x) 2 3 3
x +o(x )
[f (x)] = 1 − x + o(x ) as x → 0
3
and the desired limit is e−2/3 .

55
Exercise 13.26 Find the limit
h√ x i1/ sin2 (x−2)
lim 3 − x + ln .
x→2 2
Solution to Exercise 13.26. We set x − 2 = t. Then we obtain
h√ 1/ sin2 t
x i1/ sin2 (x−2) √
 
t
lim 3 − x + ln = lim 1 − t + ln 1 + .
x→2 2 t→0 2

Let us recall the corresponding Maclaurin’s formulae

√ t t2
sin2 t = t2 + o(t2 ); 1−t=1− − + o(t2 ) as t → 0
2 8
and
t t2
 
t
ln 1 + = − + o(t2 ) as t → 0.
2 2 8
Now we have that
1/ sin2 t 1/(t2 +o(t2 ))
√ t2
  
t 2
1 − t + ln 1 + = 1 − + o(t ) ,
2 4

and the desired limit is equal to e−1/4 .

56