Download Section 2.2 - Basic Differentiation Rules and Rates of Change

Section 2.2 - Basic Differentiation
Rules and Rates of Change
The Constant Rule
A constant function f x  k has derivative f ' x  0, or:

d
dx
k   0

Note: The constant function is a horizontal line with a
constant slope of 0.
Examples
Differentiate both of the following functions.
a. f x  13
f ' x 
d
dx
13  0
d
dx
e   0
The function is a horizontal line at
y = 13. Thus the slope is always
0.
b. gx  ei
g' x  
i
The function is a horizontal line because e,
Pi, and i all represent numeric values.
Thus the slope is always 0.
The Power Rule
For any real number n, the power function f x  x n has the
derivative f ' x  nx n 1, or:

d
dx
x
n
  nx

n 1
Ex:
f x  x 3
f ' x  3x 2
Examples
Differentiate all of the following functions.
a. f x   x 8
f ' x  
d
dx
x  8 x
8
81
 8x 7
1. Bring down the exponent 2. Leave the base alone 3. Subtract one from the original exponent.
b. gu  um
  
g' u  dud u m   m u m 1
The procedure does not change with variables.
 x 5 3

c. h x   x 2 
 
h' x   dxd x 5 3    53 x5 31   53 x 8 3
3
x
x1 3
x2
Make sure the function is written as a power function to use the rule.


The Constant Multiple Rule
The derivative of a constant times a function, is the
constant times the derivative of the function. In other
words, if c is a constant and f is a differentiable function,
then
d
dx
cf x  c
d
dx
f x 
Objective: Isolate a power function in order to take the
derivative. For now, the cf(x) will look like c x n .
Examples
Differentiate all of the following functions.
a. f x  3x 4
f ' x  
d
dx
4 1
4
d
4
 12x

3
4

3
x
x
3x


  dx
“Pull out” the
coefficient
3
Take the
derivative
x1   
b. gx  x  1 

 d
1
g' x   du 1 x   1 dud x1  1 1 x 11  1
Make sure the function is written as a power function to use the rule.

c. h x  

 x 5
 
h' x   dxd x 5    dud x 5    5 x 51  5x 6  5x 6
5
x


Make sure the function is written as a power function to use the rule.

The Sum/Difference Rule
The derivative of a sum or a difference of functions is the
sum or difference of the derivatives. In other words, if f
and g are both differentiable, then
d
dx
f x   gx 
d
dx
d
dx
OR
d
dx
f x   gx 
f x  
d
dx
gx 
f x  
d
dx
gx 
Objective: Isolate an expressions in order to take the
derivative with the Power and Constant Multiple Rules.
Example 1
 
If k  x   3  f  x   2 find k '  5 if f  5  10, f ' 5  3, h 5  2,
and h '  5  16 .
Evaluate the derivative of
k at x = 5
Find the derivative of k
h x 
k ' 5  3 f ' 5  1 h' 5
d
h x

k '  x   dx 3  f  x  
k ' x 
d
dx
2

3  f  x    dxd  12 h  x  
k '  x   3  f  x   
d
dx
1 d
2 dx
k '  x   3 f '  x   12 h '  x 
 h  x  
 
  2  
 3 3  12 16
 13
Example 2
Evaluate:
d
dx
x 12x  4 x 10x  6x  5
12x  4 x  10x  6x  
d
dx
8
d
x

  dx
8
5
5
4
4
d
dx
3
3
d
dx
d
dx
d
dx
5
Sum and Difference Rules
d
dx
x 12 x  4 x 10 x  6
8
d
dx
5
4
d
dx
d
dx
3
d
dx
x   dxd 5
Constant Multiple Rule
8x 81 12 5x 51  4 4 x 4 1 10 3x 31  6 1x11  0
Power Rule
8x  60x 16x  30x  6
7
4
3
Simplify
2
Example 3
Find f '  x  if f  x   2x  3  4 .
First rewrite the absolute value function
as a piecewise function
  2 x  3  4 if x   32 2 x  7 if x   32
f  x  

3
3
2
x

1
if
x


2
x

3

4
if
x






2
2
d
dx
 2x  7 
 dxd  2x   dxd  7 
 2 dxd  x1   dxd  7 
 2 1 x11  0
 2
Find the Left
Hand
Derivative
Find the Right
Hand
Derivative
2 if x   32

f '  x    if x   32
2 if x   3
2

d
dx
 2 x 1
 dxd  2x   dxd 1
 2 dxd  x1   dxd 1
 2 1  x11  0
2
Since the one-sided limits are not equal, the
derivative does not exist at the vertex
Example 4
Find the constants a, b, c, and d such that the graph of
f  x   ax2  bx  c contains the point (3,10) and has a
horizontal tangent line at (0,1).
What do we know:
1. f(x) contains the points (3,10) and (0,1)
2. The derivative of f(x) at x=0 is 0
Use the points to help find a,b,c
1  a  0   b  0   c 10  a  3  b  3  c
2
1 c
10  a  3  b  3  1
10  9a  3b 1
9  9a  3b
3  3a  b
2
2
We need another
equation to find a and b
Find the Derivative
 ax  bx  1
f '  x    ax    bx   1
f '  x   a  x   b  x   1
f ' x 
d
dx
2
d
dx
2
d
dx
2
d
dx
d
dx
d
dx
1
d
dx
f '  x   a  2 x21  b 1x11  0
f '  x   2ax  b
We know the derivative of f(x) at x=0
and x=1 is 0
Example 4 (Continued)
Find the constants a, b, c, and d such that the graph of
f  x   ax2  bx  c contains the point (3,10) and has a
horizontal tangent line at (0,1).
1 c
1. f(x) contains the points (3,10) and (0,1)
AND
What do we know:
3  3a  b
2. The derivative of f(x) at x=0 is 0
Use the Derivative
Find a
f '  x   2ax  b
3  3a  b
3  3a  0
1 a
0  2a  0  b
0b
a = 1, b = 0, and c = 1
Derivatives of Sine and Cosine
We will assume the following to be true:
d
dx
sin x  cosx
AND
d
dx

cosx  sin x
Example 1
Differentiate the function: f x 
Rewrite the ½
to pull it out
easier
d
dx
d
dx



1
2

 5 sin x  3 x 7
Rewrite the
73
1
radical to use
2 cos x  5 sin x  x

cosx
2

 cos x   5 sin x x 

1
2
d
dx
d
dx
Sum and Difference Rules
d
dx
the power rule
73
cos x   5 dxd sin x   dxd x 7 3 
Constant Multiple Rule
 sin x  5 cos x  x
1
7 7 31
2
3
Power Rule AND Derivative of Cosine/Sine
7 43
 sinx

5
cos
x

2
3x
Simplify
Example 2
Find the point(s) on the curve
tangent line is horizontal.
y  x 4  8x 2  4 where the
Horizontal
Lines have a
slope of Zero.
First find the derivative.
4
2
d
dx
y' 
y' 
d
dx
y' 
d
dx
x  8x  4

x

8x
   
x  8 x 
4
d
dx
4
d
dx
2
d
dx
4
2
d
dx
4
y'  4 x 4 1  8 2x 21  0
y'  4 x 16x
3
0,4 2,12 2,12
Find the x values where the
derivative (slope) is zero



0  4 x 3 16x
0  4 x x 2  4 
0  4xx  2x  2
x  0, 2,  2
Find the corresponding y
values
y  0  80  4  4
 y  24  822  4  12
4
2
y  2  82  4  12
4
2
Calculus Synonyms
The following expressions are all the same:
• Instantaneous Rate of Change
• Slope of a Tangent Line
• Derivative
DO NOT CONFUSE AVERAGE RATE OF
CHANGE WITH INSTANTANEOUS RATE
OF CHANGE.
Position Function
The function s that gives the position (relative to the origin)
of an object as a function of time t. Our functions will
describe the motion of an object moving in a horizontal
or vertical line.
s(t)
2
2
1
2
3
4
5
6
t
TIME:
0
3
1
2
5
7
6
4
Origin
-2
-2
-4
Description of
Movement:
No
Downward
Upward
Movement
-4
Displacement
Displacement is how far and in what direction something is from
where it started after it has traveled. To calculate it in one
dimension, simply subtract the final position from the initial
position. In symbols, if s is a position function with respect to
time t, the displacement on the time interval [a,b] is:
s  s  b   s  a 
s(t)
EX: Find the displacement
between time 1 and 6.
2
1
-2
-4
2
3
4
5
6
t
s  6   s 1  4  2
 6
Average Velocity
The position function s can be used to find average velocity
(speed) between two positions. Average velocity is the
displacement divided by the total time. To calculate it
between time a and time b:
It is the average rate of change
or slope.
s(t)
s
t

1
-4
Two Points
Needed
EX: Find the average velocity
between time 2 and 5.
2
-2
s a  sb 
a b
2
3
4
5
6
t
s  5  s  2 
5 2

4  2
5 2
 2
Instantaneous Velocity
The position function s can be used to find instantaneous
velocity (often just referred to as velocity) at a position if
it exists. Velocity is the instantaneous rate of change or
the derivative of s at time t:
v  t   lim
t 0
s  t t   s  t 
t
 s ' t 
One Point
Needed
EX: Graph the object’s velocity where it exists.
s(t)
v(t)
Corner Slope = 0
2
Slope = -4
1
2
3
4
5
6
2
t
-2 Slope = 2
-4
1
-2
Slope = 4
Slope = -2
Slope = 0
-4
2
3
4
5
6
t
Position, Velocity, …
Position, Velocity, and Acceleration are related in the
following manner:
Units = Measure of length (ft, m, km, etc)
Position:
s (t )
Velocity: s '(t )
The object is…
Moving right/up when v(t) > 0
Moving left/down when v(t) < 0
Still or changing directions when v(t) = 0
 v(t )
Units = Distance/Time (mph, m/s, ft/hr, etc)
Speed = absolute value of v(t)
Example 1
The position of a particle moving left and right with respect
to an origin is graphed below. Complete the following:
1. Find the average velocity between time 1 and time 4.
2. Graph the particle’s velocity where it exists.
3. Describe the particles motion.
s(t)
2
1
-2
-4
2
3
4
5
6
t
Example 2
Sketch a graph of the function that describes
the motion of a particle moving up and down
with the following characteristics:
The particle’s position is defined on [0,10]
The particle’s velocity is only positive on (4,7)
The average velocity between 0 and 10 is 0.
Example 3
The position of a particle is given by the equation
st  t 3  6t 2  9t
where t is measured in seconds and s in meters.
(a) Find the velocity at time t.

d
dt
v t  
d
dt
3
2
d
d
t

6t

  dt   dt 9t
v t  
d
dt
3
2
1
d
d
t

6
t

9
t
  dt   dt  
The derivative of the
position function is the
velocity function.


3
2
t

6t
 9t 

s' t   v t  
vt  3t 31  6 2t 21  9 t11
vt  3t 12t  9
2

Example 3 (continued)
The position of a particle is given by the equation
st  t 3  6t 2  9t
where t is measured in seconds and s in meters.
(b) What is the velocity after 2 seconds?

v 2  32 122  9  3 m/s
2
(c) What is the speed after 2 seconds?
 v 2  3  3 
m/s
Example 3 (continued)
The position of a particle is given by the equation
st  t 3  6t 2  9t
where t is measured in seconds and s in meters.
(d) When is the particle at rest?

0  3t 12t  9
2
The particle is at rest
when the velocity is 0.
0  3t 2  4t  3

0  3t 1t  3
t  1, 3

After 1 second and 3 seconds