Topic 7: Calculus
7.1

Definition of a Limit
Let f be a function defined on an open interval containing c (except possibly at c)
and let L be a real number. The statement Lim = L means that for each  >0, there
x c
exists a  >0 such that
if 0 < x  c < 
then f ( x)  L <  .

Lim
x 0

cos x  1
0
x
Lim
x 0
sin x
1
x
Definition of a Derivative
f ( x  x)  f ( x)
provided that
x
x 0
the limit exists. For all x for which the limit exists, f ’ is a function of x .
The derivative of f at x is given by f ' ( x)  Lim

Notations (for first derivatives)
o f ’(x)
o y’
dy
o
dx
d
o
dx
Ex: f (x) = 3x + 2
f ( x  x)  f ( x)
f ’ (x) = Lim
x
x 0
(3( x  x)  2)  (3x  2)
= Lim
x
x 0
3x  3x  2  3x  2
= Lim
x
x 0
3x
= Lim
x 0 x
= Lim 3 = 3
x  0

Derivatives of
(All can be found in IB formula packet)
o x n  nx n 1
d
7x 29
dx
= 7  29  x ( 291)
= 203x 28
o sin x  cos x
o csc x   csc x cot x
o cos x  sin x
o sec x  sec x tan x
o tan x  sec 2 x
o cot x   csc 2 x
o ex  ex
d 3x
e  3e 3 x
Ex:
dx
1
o ln x 
x
x
o a  a x (ln a)
1
o log a x 
x ln a
1
o arcsin x 
1 x2
1
o arccos x  
a
1 x2
1
o arctan x 
1 x2
Derivatives are interpreted as gradient functions and as rates of change and can be
used to find equations of tangents and normals:
ex:

Finding equation of tangents
The slope of the tangent line is found by calculating the derivative of the function
at the point of intersection. The equation is then found by plugging that slope,
along with the original point, into:
y  y1  m( x  x1 )
Ex: Find an equation for the tangent line to the curve at the indicated point:

y  1  csc x  4 sin x ;
x
6


y1  1  csc  4 sin
6
6
1
y1  1  2  4( )
2
y1  1
y '   csc x cot x  4 cos x
m   csc


6
3
m  2 3  4
2
m  2 3  2 3

6
cot
 4 cos

6
m  4 3
y  y1  m( x  x1 )

y  1  4 3 ( x  )
6
2 3
y  4 3 x 
1
3
Identifying increasing and decreasing functions
o If f ' ( x)  0 for all x in (a, b), then f is increasing on a, b
o If f ' ( x)  0 for all x in (a, b), then f is increasing on a, b
o If f ' ( x)  0 for all x in (a, b), then f is constant on a, b
7.2

Chain Rule
Let y  f ( g ( x)) , then
dy
 f ' ( g ( x))  g ' ( x)
dx
Ex: Let f ( x)  sin( x) and g ( x)  cos( x)
f ( g ( x))  sin(cos x)
dy
  sin x cos(cos x)
dx

Sample Related Rates of Change Problems
If a balloon is deflating at a rate of 2 inches a second, find the change is surface
area and volume over time when the radius is 4 inches.
dr
dSA
dV
 2 inc
when r=4?
when r=4?
sec
dt
dt
dt
2
dSA
dr
SA  4r 2
 8r
 8  4  2  64 in
sec
dt
dt
3
4
dV
dr
V  r 3
 4r 2
 4  16  2  128 in
sec
3
dt
dt

Product Rule
Let y  f ( x)  g ( x) , then
y '  f ( x)  g ' ( x)  g ( x) f ' ( x)
Ex: y  x 3 (6 x  5)
y'  x 3  (6)  (6 x  5)  3x
y'  6 x 3  18 x 2  15 x

Quotient Rule
Let y 
g ( x)  f l ( x)  f ( x) g l ( x)
( g ( x)) 2
yl 
Ex: Let y =
sin 4 ( x)
cos 3 ( x)
yl 
=
=
=
=

f ( x)
, then
g ( x)
cos 3 x  4 sin 3 x cos x  sin 4 x  3 cos 2 x( sin x)
cos 6 x
4 cos 4 x sin 3 x  3 cos 2 x sin 5 x
cos 6 x
cos 2 x(4 cos 2 x sin 3 x  3 sin 5 x)
cos 6 x
4 cos 2 x sin 3 x  3 sin 5 x
cos 4 x
2
4 tan x sin x  3 tan 4 x sin x
Other Notations of Derivatives
o Second: y ll
o Third:
y lll
o Fourth: y ( 4)
o nth:
y (n )
d2
dx 2
d3
f lll (x)
dx 3
d4
f ( 4 ) ( x)
dx 4
dn
( n)
f ( x)
dx n
f ll (x)
d2y
dx 2
d3y
dx 3
d4y
dx 4
dny
dx n
7.3

Testing for Maximum or minimum using change of sign of the first derivative and
using sign of second derivative

First Derivative Test:
Let c be a critical number of a function f that is continuous on the open interval I
containing c. If f is differentiable on the interval, except possibly at c, then
f (c ) can be classified as follows:
1) If f l (x) changes from negative to positive at c, then f (c ) is a relative
minimum of f
2) If f l (x) changes from positive to negative at c, then f (c ) is a relative
maximum of f
3) If f l (x) does not change signs at c, then f (c ) is neither a relative minimum
nor a relative maximum of f
Ex: h( x)  x 3  6 x 2  15
h( x)  3x 2  12 x
 3 x( x  4)
Critical Values = 0,4
Relative Max: (0,15)
Increasing: (, o)  (4, )
Decreasing: (0,4)
Relative Min: (4,-17)

Second Derivative Test:
Let f be a function such that f l (c)  0 and that the second derivative of f exists
on the open interval containing c.
1) If f ll (c)  0 , then f (c ) is a relative minimum
2) If f ll (c)  0 , then f (c ) is a relative maximum
3) If f ll (c)  0 , then the test fails. In such cases, use the First Derivative Test.

Use of First and second derivative in optimization problems:
Ex: A fast-food restaurant has determined that the monthly demand for its
60000  x
hamburgers is p 
. Find the increase in revenue per hamburger
20000
(marginal revenue) for monthly sales of 20,000 hamburgers.
(Total revenue is given by R=xp)
60000  x
1
)
(60000 x  x 2 )
Solution: R  xp  x(
20000
20000
dR
1

(60000  2 x)
Marginal Revenue is:
dx 20000
When x=20000 the marginal revenue is:
dR
1
20000

[60000  2(20000)] 
 $1 / unit
dx 20000
20000
Ex: The radius of a sphere is measured to be 50 cm with a measurement error of +
.02 cm. Estimate the relative error in the computer volume of the sphere.
dV
V  4r 3
Relative error 
r  50cm
dr  0.02cm
V
Solution:
dV 12r 2 dr

V
4r 3
3dr
r
3(.02)

50
.06

50

dV
 .0012cm 3
V
Ex: A liquid form of penicillin manufactured by a pharmaceutical firm is sold in
bulk at $200 per unit. If the total production cost for x units is
C ( x)  500000  80 x  0.003x 2 and if the production capacity of the firm is at
most 30,000 units in a specified time, how many units of penicillin must be
manufactured and sold to maximize profit in that specified time?
C ( x)  500000  80 x  0.003x 2
p  xp  c
p  200 x  (500000  80 x  0.003x 2 )
 .003x 2  120 x  500,000
p l  .006 x  120
p l  0 , when  .006x  120
x=20,000 units
Justification using second derivative test:
p ll  .006
-.006 < 0 Therefore, when x = 20,000, the curve is concave down and the value is
a maximum
7.4

Definition of Antiderivative:
A function f is an Antiderivative of F on an interval I if f l  F (x) for all x on I.
Indefinite integral interpreted as a family of curves

Indefinite integrals of:
1 n 1
 x n dx 
x C
n 1
 sin xdx   cos x  C
 sixdx   cos x  C
 e x dx  e x  C

With Linear function ax + b:
Ex: f l x  cos(2 x  3)
f ( x)   f l ( x)dx
  cos( 2 x  3)dx
1
 sin( 2 x  3)  C
2
7.5
Anti-differentiation with a boundary condition to determine the constant term:
If dy/dx = 3x^2 + x and y =10 when x =0, find y.
y   (3x 2  x)dx
1 2
x C
2
1
10  (0) 3  (0) 2  C
2
c  10
1
y  x 3  x 2  10
2
 x3 
Definite Integrals:
5
 (  3v  4)dv
2
5
3 2
v  4v) |
2
2
3
3
 ( (5) 2  4(5))  ( (2) 2  4(2))
2
2
 39

2
(
Area Between a curve and the x-axis or y-axis in a given interval, areas between curves:
y  1  3 x ..........bounds : x  0, x  8, y  0
8
3
 (1  x )dx  ( x 
0
3 4/3 8
x )|
0
4
 20
Volumes of revolution:
Revolution about the x-axis and y-axis
b
Disk Method:   f ( x) 2 dx
a
b
Shell Method:  2 xf ( x)dx
a
7.6 Kinematic Problems: acceleration, velocity, and displacement
V=ds/dt
a = dV/dt = d2s/dt2 = V(dV/dt)
Area under velocity v. time graph represents distance traveled
V  3T 2  2......given : S (3)  7, find : S (2)
 (3t
2
 2)dt  t 3  2t  C  S (t )
S (3)  27  6  C  7
C  14
S (t )  t 3  2t  14
S (2)  10
7.7
Asymptotes:
Horizontal:
Horizontals asymptotes exist if the function approaches a value as x
approaches +/- infinite. They can be determined by taking the limits of the functions as
the approach +/- infinite.
Vertical:
Vertical asymptotes occur when the denominator of f(x) is zero and f(x) is,
therefore undefined. Ex. 1/x would have a vertical asymptote at x=0.
Oblique:
The graph of a rational function has an oblique asymptote if the degree of
the numerator is greater than that of the denominator. To calculate the equation of the
asymptote, perform long division and then drop the remainder.
Definition of Concavity:
Let f be differentiable on an open interval I. The graph of f is concave upward on
I if f’ is increasing on the interval and concave down on I if f’ is decreasing on the
interval.
Test for Concavity:
Let f be a function whose second derivative exists on an open interval I
1. If f ``(x)>0 for all x in I, then the graph of f is concave upward in I.
2. If f ``(x)<0 for all x in I, then the graph of f is concave donward in I.
Point of Inflection:
If (c, f(c)) is a point of inflection then either f ``(x) = 0 or f ``(x) does not exist at x
= c.
7.8
Implicit Differentiation:
The procedure of differentiating an implicit equation with respect to the
desired variable x while treating the other variables as unspecified
functions of x:
xy = 1
d
d
[ xy]  [1]
dx
dx
dy
dx
x y
0
dx
dx
dy
x y0
dx
dy  y

dx
x
7.9
Further Intergration
Substitution:
sec 2 x
2
 tan x dx..........let : u  tan x....du  sec ( x)dx
 sec
2
1
2
1
2
x(tan x) dx   u du
1
2u 2  C
2 tan x  C
Partial Fractions:
dx
dx
 x 2  5x  6  ( x  2)( x  3)
1
1
 ( x  2  x  3 )dx
 ln | x  2 |  ln | x  3 | C
ln |
x3
| C
x2
By Parts:
 udv  uv   vdu
 x cos xdx.......u  x....du  dx.....dv  cos xdx.....v  sin x
 x sin x   sin xdx
 x sin x  cos x  C
7.10 Separation of Variables:
x
dy 1
 y
dx y
1 y2
dx
y
y
dx
 1  y 2 dy   x
xdy 
1
ln | y 2  1 | C  ln x
2
C y2 1  x