International Business

Topics: Derivative, Integral, Calculus Pages: 11 (2080 words) Published: December 18, 2012
University of Business Technology

FALL – 2012
Instructor : Abdulraheem Zabadi

Table of Contents
Differential Calculus
Integral Calculus

Chapter One : Limits

Properties of Limits
If b and c are real numbers, n is a positive integer, and the functions ƒ and g have limits as x → c , then the following properties are true.

Scalar Multiple : limx→c (b f(x))=b limx→c fx

Sum or difference : limx→c ( fx± g(x)) = limx→c fx ± limx→c ( gx)

Product : limx→c (fx. g(x)) = limx→c fx . limx→c gx

Quotient: limx→c fxgx = limx→cf(x)limx→cg(x) , gx≠0

One-Sided Limits
limx → a+fx x approaches c from the right
limx → a-fx x approaches c from the left

Limits at Infinity
limx →∞f(x) = L or limx → -∞f(x) = L

The value of ƒ(x) approaches L as x increases/decreases without bound. y = L is the horizontal asymptote of the graph of ƒ.

Some Nonexistent Limits
limx→01x2 limx→0 xx limx→0sin1x
Some Infinite Limits

limx→01x2= ∞ limx→ 0+ ln x= - ∞

Exercise: What limx→0sinxx ?

a. 1 b. 0 c. ∞ d. DNE The answer is a ( you should memorize this limit)
Definition : A function ƒ is continuous at c if:
1. ƒ(c) is defined 2. limx→cfxexists 3. limx→cfx= fc

Graphically, the function is continuous at c if a pencil can be moved along the graph of ƒ(x) through (c, ƒ(c)) without lifting it off the graph.

Exercise: If f(x) = 3 x2+x2x , for x x ≠0
f(0) = k
and f is continuous at x = 0 , find the value of k ?
a. -32 b. 32 c. -1 d. 0 e.1 , answer is b

Intermediate Value Theorem
If ƒ is continuous on [a, b] and k is any number between ƒ(a) and ƒ(b), then there is at least one number c between a and b such that ƒ(c) = k.

Chapter Two Differential Calculus
f-x= limh →0fx+h – fxh , and if this limit exists f-c= limx →cf x – fcx-c Theorem : if f is differentiable at x = c , then f is continuous at x = c .

General and logarithmic Differentiation Rules

1. ddx cu =c u' , c is constant .
2. ddx u ±v =u' ± v' - sum rule .
3. ddx uv =u v'+ v u' - product rule
4. ddx c =0 , c is constant .
5. ddx uv = v u'- u v'v2 , v ≠0. – quotient rule
6. ddx x =1
7. ddx cu =c u'
8. ddx un =n un-1
9. ddx lnu = u'u
10. ddx eu = eu u'
11. ddx f(g(x) = f'gx.g'(x) - chain rule

Derivatives of the Trigonometric functions

1.ddx sinu =cosu . u'
2. ddx cosu =-sinu . u'
3. ddx tanu = sec2u . u'
4. ddx secu =secu tanu . u'
5. ddx cot u =- csc2u . u'
6. ddx cscu =- cscu cotu . u'
7. ddx sinu =cosu . u'
Derivatives of the Inverse Trigonometric functions
1.ddx arcsinu= u'1- u2
2. ddx arccosu= -u'1- u2
3. ddx arctanu= u' u2+1
4. ddx arccotu= -u' u2+1
5. ddx arccscu= -u'uu2-1
6. ddx arcsecu= u'uu2-1
Implicit differentiation : Implicit differentiation is useful in cases in which you cannot easily solve for y as a function of x. Example : Find dy/dx for : y3+ xy-2y- x2=0
Solution : 3y2 y’ + xy’ + y – 2 y’ - 2x = 0
3y2 y’ + xy’ = -y + 2x-2
y’ ( 3y2 + x ) = -y + 2x
y’ = -y+2x 3y2 +2 x-2
Higher Order Derivatives: These are successive derivatives of ƒ(x). Using prime notation, the second derivative of ƒ(x), ƒ''(x), is the derivative of ƒ'(x). The numerical notation for higher order derivatives is represented by : fn(x)=yn: The second derivative is also indicated by= d2y dx2

Derivatives of Inverse Functions: If y = ƒ(x) and x = ƒ-1(y) are differentiable inverse functions, then their derivatives are...
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