Chapter 3, PVA, and Related Rates
I. 3.1: Derivatives of Polynomials and Exponential Functions
a. Derivative of a Constant
i. Derivative of a Constant Function
1. Example:
Question: Differentiate
Answer:
b. Power Functions
i.
ii. The Power Rule
If
is a positive integer (or any real number), then
1. Example:
Question: Differentiate
How to:
Answer:
c. New Derivatives from Old
i. The Constant Multiple Rule
If
is a constant and
is a differentiable function, then
1. Example:
Question: Evaluate
How To:
Answer:
ii. The Sum Rule
If
and
are both differentiable, then
1. Example:
Question:
How To:
Answer:
iii. The Difference Rule
If
and
are both differentiable, then
1. Example
Question:
How To:
Answer:
d. Exponential Functions
i. Derivative of the Natural Exponential Function
1. Example
Question:
How To:
Answer:
II. 3.2: The Product and Quotient Rules
a. Product Rule
i. If g and h are both differentiable, the rule is:
then
1. Example:
Question: Calculate the derivative of the function f(x) = ex(x2 + 1).
How To: If we consider g(x) = ex and h(x) = x2 + 1, then f(x) =
g(x)h(x).
g '(x) = ex and h'(x) = 2x
Answer: f'(x) = ex(x2 + 1) + ex2x = ex(x2 + 2x + 1) = ex(x + 1)2.
b. Quotient Rule
i. If g and h are both differentiable
then
1. Example:
Question: Differentiate
How To: If we consider
.
and
then
.
and
Answer:
III. 3.4: Derivatives of Trigonometric Functions
1.
2.
3.
4.
5.
6.
Examples: Find
1. y=
Use the Product Rule & Derivative of Sine
2. y=
y’=
3. y=
y=
(
)
y’=
y’=
4. Find an equation of the tangent line to the curve
STEPS
at the point (
WORK
1. Find y’ with the product rule
y’=
y’=
2. Plug in x=
y’=
y’=
y’= -1
3. Form the equation of the tangent line
using the slope and point.
Special Trig Limits:
1.
2.
at (
.
.
a. Ex1
i.
ii.
iii.
x
=
=2x1=
2
b. EX2
i.
x
=
=0
IV. 3.5: The Chain Rule
If f and g are both differentiable and F  f  g is the composite function defined
by F ( x)  f ( g ( x)) then F ' is given by the product
If Leibniz notation, if y  f (u ) and u  g (x) are both differentiable functions,
then
dy dy du

dx du dx
a) The Power Rule Combined with the Chain Rule
If n is any real number and u  g (x) is differentiable, then
d n
du
(u )  nu n 1
dx
dx
b)
d x
(a )  a x ln a
dx
c) Tangents to Parametric Curves
dy
dt
dy
dx
 dx if
0
dx
dt
dt
V. 3.6: Implicit Differentiation
Consists of differentiating both sides of the equation with respect to x and then solving the
resulting equation for y ' .
a) Arcsin Rule
d
1
(sin 1 x) 
dx
1 x2
b) Arctan Rule
d
1
(tan 1 x) 
dx
1 x2
VI. 3.7: Derivatives of Logarithms
Deriving log a x  y :
This statement is equivalent to: a y  x
y
When we derive this, we get a * ln( a) * y'  x' (remember ay = x)
x'
Simplifying, we get y ' ln( a ) * x this is the derivative of a logarithm (x) in base a.
Derive y  log 22 ( x 2 sin( x))
Answer:
Derive y  log 4 ( x)  4
Answer:
2 x sin( x)  x 2 cos( x)
ln( 22) * ( x 2 sin( x))
1
ln( 4) * ( x)
2x  8
 3x 2
Answer:
2
ln( 9) * ( x  8 x  16)
Derive y  log 9 ( x 2  8 x  16)  x 3
Because ln (e) = 1, the derivative of any natural log function (ln(u)) is
The integral of
Integrate
u'
u
u'
 ln | u |  c
u
3 x  12
dx
2
 4x
 0 .5 x
Derive (ln(cos( x))) * ( x 2  4)
e can be defined as
1
lim
(1  x) x
x0
Answer: 3 ln( 0.5 x 2  4 x)  c
Answer: (
 sin( x)
) * ( x 2  4)  (ln(cos( x))) * (2 x)
cos( x)
VII. 3.8:Linear Approximations and Differentials
If we want to differentiate an equation using logarithms:
Example: y 
x 1
x2
1) Take Natural Log of both sides
ln( y )  ln( x  1)  ln( x  2)
2) Differentiate and isolate y’ (
Derivative of both sides:
dy
)
dx
1
1
1
dy  (

) dx
y
x 1 x  2
Move dy and dx together:
dy
1
1
 y(

)
dx
x 1 x  2
Replace y with the equivalent:
dy
x 1
1
1
(
)(

)
dx
x  2 x 1 x  2
Derive y  2  (
x 3 * (4 x  62)
)
65  x
Derive yx 2  x 3  4 x
Answer: (
x 3 * (4 x  62)
3x 2
4
1
)*( 3 

)
65  x
4 x  62 65  x
x
Answer: (
x2  4
2x
1
)*( 2
 )
x
x 4 x
VIII. Position, Velocity and Acceleration
S(t)= particle’s displacement (position) at time (t)
V(t)= particle’s instantaneous velocity at time (t)
Velocity= speed
and direction
A(t)= particles instantaneous acceleration at time (t)
Ex: A particle moves along a horizontal line such that its position at any time
where s is measured in meters and t in seconds.
is given by
1) Find any time when the particle is at rest.
a)
=
=0
=0
b) Find any time when the particle changes direction
V(t)= 0 at t= 1 seconds and t= 3 seconds
+
+
1
3
v(t)
Change of direction at t=1 and t=3
c) Find the intervals when the particle is moving left.
(1,3)
d) Find the intervals when the particle is moving right.
[0,1) u (3,
e) Find the total distance that the particle travels in the first 2 seconds
Solution without integral
Solution using integral
+ 1 =5
=3
4m
5-1=4m
5-3=2m
Total distance = 6 meters
2m
Total distance = 6 meters
f)
Find the velocity of the particle when the acceleration is zero.
a(t)=
=0
REMEMBER: displacement is final
6(
T=2
position -initial position.
Total distance uses the absolute
value, not displacement
g) Find the displacement of the particle after 6 seconds.
Solution without integral
Solution using integral
+ 1 =5
55-1
Total displacement = 54 meters
54
Total distance = 54 meters
Average Velocity within a given interval:
Slope of the segment whose endpoints are the initial and terminal points on the position function for the
given interval.
OR
Ex:
interval (0,7).
Solution without using integral
, what is the average velocity on the
Solution using integral
= 90
ft
=
Formula for the position function of a projectile or free fall object:
g= acceleration due to gravity (-32 ft/sec or -9.8 m/sec)
=initial velocity
initial position
time
Ex: A rock is thrown down from a 200 foot tall building at a rate of 22 ft/sec. Height is in feet and time is
in seconds. Find the following:
A) Height, velocity, and acceleration functions
B) Find the velocity when the rock hits the ground.
9143 sec
C) Find the position and velocity at t=3
But, the rock hits the ground at t= 2.9143, so the position is actually 0ft
But, the rock hits the ground at t=2.9143, so the velocity is actually 0ft/s
D) Find the velocity when the rock is at 92 ft.
t= 2 sec & 3 sec
E) Find the average velocity from t=0 to t=2
EX: A rocket is projected upward at 49 m/s from a 75m clift.
a) Find s(t), v(t), and a(t)
b) Find when the rocket hits the ground
c) Find the velocity when the rocket hits the ground
d) Find the maximum height of the rocket.
e) Find the velocity when the position is 150 meters.
f)
Find the total distance traveled by the rocket.
t=3 is not in domain.
g) Find the total displacement of the rocket.
h) Is the speed of the rocket increasing or decreasing at t=7
Both acceleration and velocity are the same sign, so the speed is increasing.
IX. Related Rates
Related Rates:
When given a rate of one quantity, it is possible to find the rate of change of a related quantity.
Circle example: A  r 2
When the area A is increasing by 10 square cm per second, how fast is r increasing when r = 5cm?
dA
 10,
dt
A   (5) 2  25 , r  5
We want to find the derivative (rate) of r.
If we take the derivative of both sides: A'  2r ( r ' )
Substituting our known values: 10  2 5(r ' ) 
1

 r'
The radius is increasing at a rate of 1/π cm per second.
Sphere example: V  4 r 3
3
The volume of the balloon is increasing at a rate of 100 cubic cm per second. When the diameter
is 50cm, how fast is the radius increasing?
dv
 100cm 3 / sec, d  50cm The radius is 25 (Half of the diameter).
dt
dV
dr
 4r 2
Derive both sides:
dt
dt
dr
1
dr

Substitute: 100  4 (25) 2

dt
25 dt
The radius is increasing at a rate of 1/25π centimeters per second.
What happens if we have more than 2 related quantities? We should isolate the quantity we
want.
Example (Rectangle): A = LW.
Find the rate of change of the width of the rectangle when the width = 10 feet, length = 20
feet, and rate of change of area = 50 square feet per minute.
A'  50, W  10,
L  20
If we derive A = LW right now, we end up with the quantity L’ in addition to W’. We don’t
know either of these. It is difficult (if not impossible) to solve equations with two variables, so
we should get rid of L entirely since we only want to find W’.
Because W = 10 and L = 20, we can say that W = ½L or L = 2W. In this case we want L = 2W.
So A  2W 2
Deriving this, we get: A'  4W (W ' )
Substitute: 50  4(10)
dW
dW
 1 .2 
dt
dt
The width is increasing by 1.2 feet per minute.
Now that we know
dW
dL
, we can find
if we so choose.
dt
dt
Remember, only substitute after you derive.