Things to Remember for MTH 2003, 2205, and 2207
(Last Updated 04/27/2009)
Limit Definition of the Derivative
f ( x  x)  f ( x)
lim
x 0
x
OR
lim
h0
Example
f ( x)  x 2  7 x  6
f ( x  h)  f ( x )

h
( x  h) 2  7( x  h)  6  x 2  7 x  6
lim

h 0
h
x 2  2 xh  h 2  7 x  7 h  6  x 2  7 x  6
lim

h 0
h
x 2  2 xh  h 2  7 x  7 h  6  x 2  7 x  6
lim

h 0
h
2 xh  h 2  7 h
h( 2 x  h  7)
lim
 lim

h 0
h

0
h
h
lim 2 x  h  7  2 x  7
f ( x  h)  f ( x )
h
lim
h 0


 

 

h 0
Differentiation Rule
Simple/General Power Rule
Example
f ( x)  x N
f ( x)  x 7
f ' ( x)  N  x N 1
f ' ( x)  7 x 6
g ( x)  A  x N
g ( x)  5 x 7
g ' ( x)  N  A  x N 1
g ' ( x)  35 x 6
Product Rule
h( x )  f ( x )  g ( x )
h' ( x )  f ( x )  g ' ( x )  g ( x )  f ' ( x )
OR
h( x)  [ FIRST ]  [ SECOND]
h' ( x)  [ FIRST ]  [ D. of SECOND]  [ SECOND]  [ D. of FIRST ]
File Name: 99013986
h( x)  (3 x 2  7 x  6)( 2 x  5)
h' ( x)  (3 x 2  7 x  6)( 2)  (2 x  5)(6 x  7)
h' ( x)  (6 x 2  14 x  12)  (12 x 2  16 x  35)
h' ( x)  18 x 2  2 x  23
Page 1 of 8
Things to Remember for MTH 2003, 2205, and 2207
(Last Updated 04/27/2009)
Differentiation Rule
Quotient Rule
Example
f ( x)
g ( x)
g ( x)  f ' ( x)  f ( x)  g ' ( x)
h' ( x ) 
[ g ( x)] 2
h( x ) 
h( x ) 
OR
TOP
h( x ) 
BOTTOM
[ BOTTOM ]  [ D. of TOP ]  [TOP ]  [ D. of BOTTOM ]
h' ( x ) 
[ BOTTOM ] 2
2x  5
3x  7
(3 x  7)( 2)  (2 x  5)(3)
h' ( x ) 
(3 x  7) 2
(6 x  14)  (6 x  15)
h' ( x ) 
(3 x  7) 2
6 x  14  6 x  15
h' ( x ) 
(3 x  7) 2
29
h' ( x )  
(3 x  7) 2
Chain Rule
h( x)  [ f ( x)] N
h( x)  (3 x 2  7 x  6)11
h' ( x)  N [ f ( x)] N 1  f ' ( x)
h' ( x)  11(3 x 2  7 x  6)10 (6 x  7)
OR
h( x)  [ INSIDE ]
N
h' ( x)  11(6 x  7)(3 x 2  7 x  6)10
h' ( x)  N [ INSIDE ] N 1  [ D. of INSIDE ]
Exponential Differentiation
h( x )  e f ( x )
h( x )  e 2 x
h' ( x )  e f ( x )  f ' ( x )
h ' ( x )  (e 2 x
OR
h( x )  e
EXPONENT
3
14 x 2 10 x 3
3
14 x 2 10 x 3
)(6 x 2  28 x  10)
h' ( x)  (6 x 2  28 x  10)e 2 x
3
14 x 2 10 x 3
h' ( x)  e EXPONENT  [ D. of EXPONENT ]
Logarithmic Differentiation
h( x)  ln[ f ( x )]
1
h' ( x ) 
 f ' ( x)
f ( x)
f ' ( x)
h' ( x ) 
f ( x)
h( x)  ln( 3x 2  7 x  6)
1
h' ( x )  2
 (6 x  7 )
3x  7 x  6
6x  7
h' ( x )  2
3x  7 x  6
OR
h( x)  ln[ INSIDE ]
1
h' ( x ) 
 [ D. of INSIDE ]
INSIDE
[ D. of INSIDE ]
h' ( x ) 
INSIDE
File Name: 99013986
Page 2 of 8
Things to Remember for MTH 2003, 2205, and 2207
(Last Updated 04/27/2009)
Logarithmic Rewrites
1) ln ([ FIRST ]  [ SECOND])  ln [ FIRST ]  ln [ SECOND]
TOP
2) ln
 ln [TOP ]  ln [ BOTTOM ]
BOTTOM
3) ln [ INSIDE ] N  N ln [ INSIDE ]
1) ln 3 xy  ln 3 x  ln y
2x
2) ln
 ln 2 x  ln 3 y
3y
4) e ln N  N
4) e ln(3 x 7 )  3 x  7
5) ln e N  N
5) ln e ( 2 x 5)  2 x  5
Use the …
Function
Original Function
f(x)
3) ln( 2 x  7) 4  4 ln( 2 x  7)
Purpose
 Find any y-coordinate on the graph
 Find any additional points for graphing
 Average Rate of Change
f (b)  f (a )
on the interval [ a, b]
ba
 Actual Change
y  f (b)  f (a) on the interval [ a, b]
First Derivative
f'(x)
 Slope of the Tangent Line
{Substitute any x-value into the first derivative}
 Instantaneous Rate of Change
{Find the derivative and substitute the given value into it}
 Critical Numbers
{Set the first derivative equal to zero and solve for x}
 Intervals of Increasing/Decreasing
{Use the critical numbers}
 Relative Extrema
{Based on the direction of the Increasing and Decreasing}
 Marginal Equations
 Differential Equations
dy  f ' ( x)  dx
Second Derivative
f''(x)
 X-coordinates for the possible points of Inflection
{Use the original function to get the y-coordinates.}
 Points of Diminishing Return (a.k.a inflection points)
 Intervals of Concavity
{Use the x-coordinates of the possible Inflection Points.}
 Relative Extrema
{Substitute the critical numbers from the first derivative into the second derivative. The
concavity will tell you which extrema you have, if any.}
Epsilon (DO NOT GET THIS MIXED UP WITH ELASTICITY)

f ( x  x)  f ( x)
 f ' ( x)
x
File Name: 99013986
OR

f ( x  h)  f ( x )
 f ' ( x)
h
Page 3 of 8
Things to Remember for MTH 2003, 2205, and 2207
(Last Updated 04/27/2009)
Elasticity of Demand
Elasticity Conclusions
p dx
d 

x dp
1.
d  1 then Demand is Elastic
2.
d  1then Demand is Inelastic
3.
d  1 then Demand is Unitary
Compounding
Continuously
A  Pe rt
A = Amount at the end of the period of
time
=
Amount at the start
P
r = percentage rate (change to decimal)
t = Amount of time in years
Used to determine…
Amount of a lump sum investment compounded
continuously over a period of time.
Non-Continuously
A = Amount at the end of the period of
time
=
Amount at the start
P
nt
 r
=
percentage rate (change to decimal)
r
A  P1  
 n
n = number of times in 1 year that the
amount is compounded
t = Amount of time in years
Used to determine…
Amount of a lump sum investment compounded a specific
amount of times in 1 year, over a period of time.
E.g.
Annually
n=1
Semi-annually n = 2
Quarterly
n=4
Monthly
n = 12
Bi-Monthly
n = 24
Weekly
n = 52
Daily
n = 365
Continuous Money Flow (Net Present Value)
1  e  rt 
A  P

 r 
File Name: 99013986
A = Present Value
P = Amount that flows uniformly
r = percentage rate (change to decimal)
t = number of years
Page 4 of 8
Things to Remember for MTH 2003, 2205, and 2207
(Last Updated 04/27/2009)
Finding Asymptotes
Vertical Asymptotes
Vertical asymptotes are found in the denominator of a
rational function. Simplify the rational function by
factoring then cancelling. Set whatever is left in the
denominator equal to zero and solve.
Anything that remains in the denominator after
cancelling is a vertical asymptote and is also a nonremovable discontinuity.
Anything that cancels in the denominator is a hole in the
graph and is also a removable discontinuity.
Example 1
2x  7
x 2  16
2x  7
f ( x) 
( x  4)( x  4)
( x  4)( x  4)  0
x40
x40
f ( x) 
x  4
x4
Example 2
x3
x2  9
x3
f ( x) 
( x  3)( x  3)
1
f ( x) 
x3
x3 0
x  3
f ( x) 
The ( x  3) cancels in the equation and becomes a hole in
the graph at x  3 . It is still excluded from the functions
domain.
Finding Asymptotes (continued)
Horizontal Asymptote (Three Conditions)
1. When the numerator’s highest exponent is larger
than the denominator’s highest exponent, there is
no horizontal asymptote.
f ( x) 
3x 2  7 x  6
 No Horizontal Asymptote
2x  5
f ( x) 
2x  5
y 0
3x  7 x  6
f ( x) 
2 x 2  5 x  11
2
y 
2
3
3x  7 x  6
OR
TOP BIGGER => NONE
2. When the denominator’s highest exponent is larger
then the numerator’s highest exponent, the
horizontal asymptote is y=0.
2
OR
BOTTOM BIGGER => ZERO
3. When the highest exponents in the numerator and
the denominator are equal, the horizontal
asymptote is the ratio of the leading coefficients.
OR
SAME => FRACTION of Leading Coefficients
Note: The horizontal asymptote can also be found by finding either the
File Name: 99013986
lim OR lim of the function.
x  
x
Page 5 of 8
Things to Remember for MTH 2003, 2205, and 2207
(Last Updated 04/27/2009)
Approximate Area under a Curve using Rectangles
Left-End Point
 width of 

 
 rectangle 


# of

 1
 rectangles

k 0
  lower 
 width of  
  k  
 
f  
 rectangle  
  bound 
Right-End Point


# of




# of

 1
 width of   rectangles

  
k 1
 rectangle 
  lower 
 width of  
  k  
 
f  
bound
rectangle





 width of  [Upper Bound]  [Lower Bound]

 
# of rectangles
 rectangle
Mid-Point
 width of   rectangles   lower  
1   width of  

 
   k    
 
f  

bound
rectangle
2


k 0
 rectangle 





Riemann Sum Formulas
Formula
n
Example
7
C  C  n
 3  3  7  21
5(5  1) 30

 15
2
2
k 1
3
3(3  1)( 2(3)  1) 84
k2 

 14

6
6
k 1
k 1
k 1
k 
n(n  1)
2
k 1
n
n(n  1)( 2n  1)
k2 

6
k 1
k 
n 2 (n  1) 2
k 

4
k 1
4 2 (4  1) 2 16(5) 2 400
k 


 100

4
4
4
k 1
Integration
Formula
Let K be a constant.
Example
n
n
3
 Kdu  K  u  C
n be an exponent  1 .
u n 1
n
u
du

C

n 1
5
4
3
 3dx  3x  C
Let
1
u  n 1
n
du

u
du

C
 un

 n 1
1
 u du  ln u  C
u
u
 e du  e  C
5
 x dx 
x6
 C OR
6
1
6
x6  C
1
x 4
1
5
dx

x
dx

 C OR  4  C
 x5

4
4x
1
 x dx  ln x  C
x
x
 e dx  e  C
Note: If the power is something other than x, usubstitution will be used.
File Name: 99013986
Page 6 of 8
Things to Remember for MTH 2003, 2205, and 2207
(Last Updated 04/27/2009)
u-Substitution for Integration
Use u-substitution when the function you are trying to integrate does not look like any of the forms listed on
page 6.
Another way to determine if you should use u-substitution is to look at the function as if you were taking the derivative.
If you have to use the product rule, quotient rule, chain rule, or an exponential e then the integral is a good candidate
for u-substitution.
* Remember that whatever you select as your u values, the derivative of that u must also exist in the integral *
Example 1
Example 2
 3x  7 dx
{This look like the chain rule}
3
let u  3 x  7
du  3dx
1
3 du  dx
u
1
3
 13 du
3
u
3
du
Example 3
Example 4
3 x 2 7 x 3
dx
let u  3x 2  7 x  3
du  6 x  7dx
e
3 x 2  7 x 3
e
u
 6 x  7dx
eu  C
e3x
2
 7 x 3
{This look like the
exponential rule}
 12 x  76 x
 6 x
5
2
File Name: 99013986
 7 x  6 dx
{This looks like a
product rule}
5
 7 x  6  12 x  7 dx
6x
5
du
u6
C
6
C
2
let u  6 x 2  7 x  6
du  12x  7dx
u
 du
{This looks like the quotient
rule}
2
ln 3x 2  5 x  3  C
1 u4

C
3 4
u4
C
12
3x  7 4  C
12
 6x  7e
6x  5
dx
 5x  3
let u  3x 2  5 x  3
du  (6 x  5)dx
1
 3x 2  5x  3  (6 x  5)dx
1
 u du
ln u  C
 3x
2

6
 7x  6
C 
6
1
6
6x
2

6
 7x  6  C
Page 7 of 8
Things to Remember for MTH 2003, 2205, and 2207
(Last Updated 04/27/2009)
Average Value
b
1
 f ( x)dx
ba
a

Consumer and Producer Surplus
( xe , pe ) is the point of equilibrium. Set the Demand and Supply equations equal to each other and solve for x e . Plug
this value into either the Demand or Supply equation to solve for p e .
Consumer Surplus =
xe
0
Producer Surplus =
( Demand )dx  ( xe )( pe )
OR

xe
0
( xe )( pe ) 
xe
0
(Supply)dx
OR

( Demand  pe )dx
xe
0
( pe  Supply )dx
Area Between 2 Curves
Find the area bound between f ( x), g ( x), x  a, x  b
If f ( x)  g ( x) between x  a and x  b then
If g ( x)  f ( x) between x  a and x  b then
b
  f ( x)  g ( x)dx
 g ( x)  f ( x)dx
a
a
File Name: 99013986
b
Page 8 of 8