Mathematics for Business

8. Differentiation

Dr. Edwin Setiawan Nugraha

President University, October 28, 2019

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Outline

1 Introduction
2 Definition
3 The Rules of Differentiation
4 Exercises
5 Derivative of Trigonometric Function
6 Chain Rule
7 Hight Order Derivative
8 Maximum and Minimum
9 Exercises

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Background

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Definition of derivative

Definition
Derivative of a function f is another function f 0 whose value at any
number x
f (x + h) − f (x)
f 0 (x) = lim (1)
h→0 h

Example: Find f 0 (x) if f (x) = x 2 .

Answer
f (x + h) − f (x)
f 0 (x) = lim
h→0 h
(x + h)2 − x 2
= lim
h→0 h
2xh + h2
= lim
h→0 h
= 2x

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Rules of Differentiation

Theorem (Power Rule)

If f (x) = x n , n is a real number, then f 0 (x) = nx n−1

Theorem (Constant Multiple Rule)

If k is a constant, and f is differentiable function, then

f 0 (kx) = kf 0 (x)

Theorem (Sum Rule)

If f and g are differential function, then

(f + g )0 (x) = f 0 (x) + g 0 (x)

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Rules of Differentiation

Theorem (Difference Rule)

If f and g are differential function, then

(f − g )0 (x) = f 0 (x) − g 0 (x)

Theorem (Product Rule)

If f and g are differential function, then

(f ·g )0 (x) = f 0 (x)g (x) + f (x)g 0 (x)

Theorem (Quotient Rule)

If f and g are differential function, then
 0
f f 0 (x)g (x) − f (x)g 0 (x)
(x) =
g g (x)2
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Example and Exercise

For no.1-5, find derivative for the following function

1 a.f (x) = x 10 , b.f (x) = 20x −5 c.f (x) = x 4 + x 2
√ x +1
2 a.f (x) = x 8 − x, b.f (x) = (3x − 1)(4x + 9) c.f (x) =
x −1
3 4 3 2 2
a.f (x) = x + x + x + x + π; b.f (x) = (x − x)(x − x + 5) 2

a 1
4 a.f (x) = − 2 ; b.f (x) = (x 2 − x)(x 2 − x + 5)
x x
5 if f (0) = 4, f 0 (0) = −1, g (0) = −3, g 0 = 5, find (f ·g )0 (0)
6 Find the equation of tangent line to y = x 2 + 2x + 2 at (1, 1)

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Derivative of Trigonometric Function

Theorem (Quotient Rule)

The function f (x) = sin x and g (x) = cos x are both differentiable, and

f 0 (x) = cos x and g 0 = − sin x

Using the rule of derivative, find derivative

f (x) = tan x, g (x) = cot x, h(x) = sec x, k(x) = csc x

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Chain Rule

The Chain Rule

If g is differentiable at x, and f is differentiable at g (x). Then the
composite function F = f ◦g defined by F = f (g (x)) is differentiable at x
and F 0 is given by
F 0 = f 0 (g (x))·g 0 (x)
In Leibnize notation, if y = f (u) and u=g(x) are both differentiable
functions, then
dy dy du
=
dx du dx

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Example
Find derivative functions below

1.f (x) = (3x 6 + 8)5 , 2.g (x) = sin (2x + 3)

Answer
1. f (x) = (3x 6 + 8)5 , let u = 3x 6 + 8, so y = u 5
dy du
= 5u 4 , = 18x 5
du dx
dy du
f 0 (x) = ·
du dx
= 5u 4 ·18x 5
= 90x 5 (3x 6 + 8)4

2. g (x) = sin (2x + 3), let u = 2x + 3, so y = sin u

dy du
= cos u, =2
du dx
dy du
f 0 (x) = ·
du dx
= cos u·2
= 2 cos (2x + 3)

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Exercise

Find derivative functions below

1.f (x) = (4x + 7)10 , 2.g (x) = cos (2x + 3)

p
3.h(x) = tan (1 − x), 4.w (x) = x 2 + 5

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Hight Order Derivative
Let f (x) = x 5 + 2x 4 − 8x 3 + 3x 2

f 0 (x) = 5x 4 + 8x 3 − 24x 2 + 6x
f 00 (x) = 20x 3 + 24x 2 − 48x + 6
f 000 (x) = 60x 2 + 48x − 48

Reference:Varberg, D. E., Purcell, E. J., and Rigdon, S. E. (2007). Ninth Edition, Calculus . Pearson/Prentice Hall.

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Exercise

1
1 If f (x) = x 6 − 2x 5 + x 2 and g (x) = 5
1+x , find f (x) and g 5 (x)
2 If f (x) = sin x and g (x) = cos x, find f 5 (x) and g 5 (x)

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Maximum and Minimum

Definition
Let S domain f contain c. We say that
(i) f (c) is maximum value of f in S, if f (c)≥f (x) for all x in S
(ii) f (c) is minimum value of f in S, if f (c)≤f (x) for all x in S
(iii) f (c) is extreme value of f on S, if it is either the maximum value or
the minimum value

Theorem (Critical Point Theorem)

Let f defined on interval I containing the point c. If c is extreme values,
then c must be a critical point, that is, either c is
(i) an end point of I
(ii) a stationary point of f , that is, f 0 (c) = 0
(iii) a singular point of f , that is, f 0 (c) does not exist

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Example
Find the maximum and minimum values
(a) f (x) = x 2 on [−3, 3]
(b) f (x) = −2x 3 + 3x 2 on [−1/2, 2]
Answer
(a) f (x) = x 2 on [−3, 3]. f 0 (x) = 2x. Stationary point x=0. Critical points -3,0,3.
f (−3) = 9, f (0) = 0, f (3) = 9. Thus, the maximum value of f is 9 and the
minimum is 0.
(b) f (x) = −2x 3 + 3x 2 on [−1/2, 2]. f 0 (x) = −6x 2 + 6x = 6x(1 − x). Stationary
point x=0,1. Critical points −1/2, 0, 1, 2.
f (−1/2) = 1, f (0) = 0, f (1) = 1, f (2) = −4. Thus the maximum value of x is
1 and the minimum is -4.

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Exercise

Reference:Varberg, D. E., Purcell, E. J., and Rigdon, S. E. (2007). Ninth Edition, Calculus . Pearson/Prentice Hall.

Find the maximum and minimum values

(a) f (x) = x 2 + x − 1 on [−1, 1]
(b) f (x) = −x 2 + x on [−2, 2]
1
(c) f (x) = 1+x 2 on [−1, 3]

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QUIS

Time: 25 minute
Given a function y = 1 00
2−x , find y .
1

2 Given a function f (x) = x 2 − 3x + 5, find f 0 (x) and draw curve f 0 (x).

3 Find the maximum and minimum values f (x) = −3x 2 + 12x on
[−2, 2].

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