Steven J' Hillman

1
CALCULUS
Part Two
"Makes Derivatives Almost Fun!" -Mathemagazine
Calculus
I counted to Infinity. Twice.
Infinite Limit Edition
By Sig Ma and E. U. Clid
Steven J. Hillman
2
SLOPE FORMULA USING LIMIT DEFINITION
1) The Derivative Is Our Friend.
The derivative of a function f at value a is
denoted f(a)
lim
f(x) - f(a)
f(a)
X a
If limit exists
x - a
lim
f(xh) - f(x)
f'(a)
h 0
Slope Instantaneous Velocity Derivative
h
If tangent line to y f(x) at point (4,3) passes
through point (0,2) f(4) 3 f(4) slope of
tanget
Y
(4,3)
f(x)
(0,2)
1
lim
f(4h) - f(4)
f'(4)

h 0
4
h
X
3
2) The Derivative Is Very Useful. (We know this
because people use it all the time.)
BASIC DIFFERENTIATION RULES AND FORMULAS
Constant Rule
Power Rule
The derivative of a constant function is 0
If n is a positive integer, then
n-1
d
d
d
(x )
n
5-1
4
d
(c)
0
(6)
0
nx
(x )
5
5x



5x
dx
dx
dx


dx
Difference Rule
Sum Rule
If f and g are both differentiable, then
If f and g are both differentiable, then
d
d
x²-x²³
x²x²³
dx
d
dx
d
f(x)-g(x)
f(x)g(x)


dx
dx
d
d
d
d
x²
x²³
x²
x²³

-


dx
dx
dx
dx
d
d

d
d

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

dx
dx
dx
dx
2x - 23x²²
2x 23x²²
4
3) The Derivative Is Very Old. (This result can
be obtained by looking at how long the derivative
has been around, and then subtracting from Now.)
BASIC DIFFERENTIATION RULES AND FORMULAS
Product Rule
Quotient Rule
If f and g are both differentiable, then the
derivative of f(x)g(x) is f(x)g(x) g(x)f(x)
If f and g are both differentiable, then the
derivative of f(x)/g(x) is g(x)f(x) - f(x)g(x)
all divided by (g(x))²
g(x)f'(x) f(x)g'(x)
d


d
f(x)g(x)
f(x)g'(x) g(x)f'(x)
f(x)

dx

dx
g(x)
2
g(x)
d
x²(3x²2) - (x³2x)x²


x³(2x²)
x³(4x) 2x²(3x²)
d

x³2x
dx

dx
x²
2
4
4
4
(x²)
4x 6x 10x

3x 2x² - x - 2x³
4
5

4
d
x
x³(3)
x³(0) 3(3x²)

dx
¾
½
0 9x² 9x²
-x 3 - 2x 2x


5
4) The Derivative was invented by two guys, one
of whom was Newton, who also makes a mighty nice
cookie.
TANGENT LINE EQUATIONS
The derivative gives the slope
f(x) x³ 4x 9 f(x) 3x² 4 Slope at point
(1, -4) is f(1) 3 4 7
Y
f'(x)
y y1 m(x x1) y (-4) 7(x 1) y 7x 7
4 y 7x - 11
Substitute to find the slope at a given point
X
f(x)
Plug the slope into m Plug the x and y values
into this equation
f(x) is tangent to f(x) at this point
Result is the tangent line to point (1,-4)
6
5) a The limit of the sum is the sum of the
limits. b The limit of the product is the
product of the limits. c The needs of the many
outweigh the needs of the few. Or the one.
DERIVATIVE OF SINE COSINE
Formulas for Derivatives of Trig Functions
If an angle is not just an X, then the derivative
of the trig function is multiplied by the
derivative of what is inside the trig function
d
(SinX) CosX
d
dx
Sin u(x) Cos u(x)u'(x)
dx
d
(CosX) -SinX
dx
The derivative of the trig function
The derivative of whats inside the trig function
d
(TanX) Sec²X
dx
The angle stays the same
y Cos3x³
d
(CscX) -CscXCotX
y' (-Sin3x³)(9x²)
dx
y' -9x²Sin(3x³)
d
(SecX) SecXTanX
dx
g(?) ?Sin? 4Cos(4?) g(?) 1Sin? ?Cos?
4-4Sin(4?) g(?) Sin? ?Cos? - 16Sin(4?)
d
(TanX) -Csc²
dx
7
6) The Derivative is Not Our Foe. (This result is
easily proved by assuming that the derivative is
our foe, then using 1 above to obtain a
contradiction.)
CHAIN RULE
When f(x) is a function u(x) raised to a power n,
then the derivative of f(x) is n times u(x)
raised to one less than n all multiplied by the
derivative of u(x)
f(x) Cos(x)³ f(x) 3-Sin(x)²
f(x) u(x)n f(x) nu(x)n1 u(x)
f(x) n(un-1) u
f(x) 2Sin(2x)² f(x) 4Cos(2x)(2) f(x)
8Cos(2x)
n
f(x) (x³3x²)²º f(x) 20(x³3x²)9(3x²6x)
f(x) 60x(x2)(x³3x²)9
u(x)
n-1
f(x) (2x³)(x²-5) f(x) 6x²(x²-5) 2x(2x³)
f(x) 6x4 30x² 4x4 f(x) 10x4 30x²
n
u(x)
u'(x)
8
GENERAL POWER RULE
7) When in doubt, refer to 1.
If n is any real number, then the derivative of
xn is nxn-1
f(x) xn f(x) nxn-1
f(x) x³ f(x) 3x3-1 f(x) 3x²
Rewrite radicals as exponents
f(x) vx³-2x²3 (x³ -2x²3)½ f(x)
½(x³-2x²3)-½(3x²-2x)
1
(3x²-2x)
(3x²-2x)
f'(x)

2(x³-2x²3)²
2(x³-2x²3)²
9
Q Why do you rarely find mathematicians spending
time at the beach? A Because they have sine and
cosine to get a tan and don't need the sun!
RATES OF CHANGE
In physics Position and velocity of a particle
moving along a path s(t) position function
s(t) v(t) instantaneous velocity function
s(t) v(t) a(t) acceleration function
s(t) 3t³ - 27t 8 v(t) 9t² - 27 a(t)
18t Object is at rest at v3 seconds v(t) 0
9t² - 27 27 9t² 3 t² tv3 After 12
seconds, acceleration is 216 m/s² a(12) 18(12)
216 When velocity is 2m/s, the position is
-34.682 m 2 9t² - 27 3.222 t² t 1.795
s(1.795) 3(1.795)³ - 27(1.795) 8 s(1.795)
5.783 - 48.465 8 -34.682
If v(t) gt 0 and a(t) gt 0 then particle is
speeding up If v(t) lt 0 and a(t) lt 0 then
particle is speeding up If v(t) gt 0 and a(t) lt 0
then particle is slowing down If v(t) lt 0 and
a(t) gt 0 then particle is slowing down
s(b) - s(a) b - a
Average Velocity
t is time (usually in seconds) When a particle
is at rest or turning, then v(t) 0 When v(t) gt
0, then particle is moving forward (right) When
v(t) lt 0, then particle is moving backwards (left)
10
Mathematics is made of 40 percent formulas, 40
percent proofs and 40 percent imagination.
IMPLICIT DIFFERENTIATION
d
x³ 3x²
d
dx
y³ 3y² y'
y³ y ² - 5y x² 4 3y³(y) 2y(y) 5y
2x 0 y(3y² 2y 5) 2x
dx
Variables agree
Variables disagree
Multiply by the derivative of y
y'
2x

3y² 2y - 5
x²y 5xy² 15 2xy yx² - (5y² 2y(y)(5x)
0 2xy yx² - 5y² - 2y(y)(5x) 0 y(x² -
10xy) 5y² - 2xy

• Steps for Implicit Differentiation
• Differentiate both sides with respect to x
• Collect all terms that have y or dy/dx on one
  side of the equation
• Factor out all y or dy/dx
• Divide by the factor attached to y or dy/dx

y'
5y² - 2xy

x² - 10xy
11
The math faculty decided they got too few first
year students. So, they made a television
commercial to show how exciting mathematics can
be. Too get the biggest audience it was scheduled
at prime time 2 o'clock, 3 o'clock, 5 o'clock, 7
o'clock and 11 o'clock.
HIGHER ORDER DERIVATIVES
f(x) is the derivative of f(x) f(x) is the
derivative of f(x) f(x) is the derivative of
f(x) f4(x) is the derivative of f(x) fn(x)
continues the series
First order Second order Third order Fourth
order Nth order
X² y² 25 1st derivative 2x 2y(y) 0 y
-x/y 2nd derivative y (-y yx)/y² y
-y x²/y)/y² y -25/y³
Substitute
y x³ 2x² - x 2 y 3x² 4x 1 y 6x
4 y 6 y4 0 y5 0
Becomes a constant
4th order and higher all 0 for this equation
Velocity is a first order derivative of
position Acceleration is a second order
derivative of position and first order derivative
of velocity
y 5x² y 10x y 10 y 0
12
I accidentally divided by zero and my homework
burst into flames.
RELATED RATES
Compute the rate of change of one quantity in
terms of the rate of change of another quantity
(of which is more easily measured) by using an
equation that relates the two quantities. Then
use the Chain Rule to differentiate both sides
with respect to time.
The radius of a circle is increasing at a
constant 1 foot per second When r4ft, find the
rate of change for the circles area A ?r²
dA
dr
2 r
2
(4)(1)
8 ft/sec
dt
P
dt


P

P
13
DERIVATIVE TESTS
1st Derivative Test
2nd Derivative Test

• Suppose c is a critical number of a continuous
  function f
• If f changes from to at c, then f has a
  local maximum at c
• If f changes from to at c, then f has a
  local minimum at c
• If f does not change direction at c, then f does
  not have a local extreme at c

• Suppose c is a critical number of a continuous
  function f
• If f gt 0 for all x in some interval, the graph
  is concave up on that interval
• If f lt 0 for all x in some interval, the graph
  is concave down on that interval
• The Inflection Point is where f changes from
  concave up to concave down or vice versa

f(x) x 2SinX on 0 lt x lt 2? f(x) 1
2CosX 1 2CosX 0 2CosX -1 Critical Numbers
2?/3 and 4?/3
MAX
0,2?/3)
(2?/3,4?/3)
(4?/3,2?
1 2CosX


-
X 2SinX
2?/3
4?/3
2?
Local Maximum
Local Minimum
MIN
14
RELATIONSHIPS OF THE DERIVATIVES
f(x) x³-2x²5
f'(x) 3x²-4x
f''(x) 6x-4
f(x) crosses the x axis at the same x value that
f(x) has a minimum or maximum. Likewise, f(x)
crosses the x axis where f(x) has a minimum or
maximum.
15
THEOREMS
Rolle's Theorem
Mean Value Theorem

• Let f be a function that satisfies the following
  two hypotheses
• f is continuous on closed interval a,b
• f is differentiable on open interval (a,b)
• Then there is a number c in (a,b) such that
• or

• Let f be a function that satisfies the following
  three hypotheses
• f is continuous on closed interval a,b
• f is continuous on open interval (a,b)
• f(a) f(b)
• Then there is a number c in (a,b) such that f(c)
  0

f'(c)
f(b) - f(a)
b - a
f is always 0 when f is a minimum, maximum or
horizontal line
f(b) - f(a) f'(c)(b - a)
P1
Lines tangent to c1 and c2 are parallel to the
slope on interval (a,b)
P2
a
b
c1
c2
c3
a
b
c1
c2