Basic rules of differentiation
y  f ( x)  c  f ( x)  0, where c is any constant.
y  f ( x)  x n  f ( x)  n x n1 , where n is any real number.
y  f ( x)  c u ( x)  f ( x)  c u ( x)
y  f ( x)  u ( x)  v( x)  f ( x)  u ( x)  v( x)
Recall that
f ( x)
y
dy
dx
Dx f ( x )
all represent the derivative. All rules follow
from the definition of the derivative.
Derivative of a constant
y  f ( x)  c
d
(c )  0
dx
f ( x)  lim
h0
f ( x  h)  f ( x )
cc
 lim
 lim 0  0
h

0
h0
h
h
Examples
y  
2
dy

0
dx
f ( x)  3  f ( x)  0
Power rule
d n
( x )  n x n1
dx
n is any real number
3
Not easy to prove in general. For y  f ( x)  x
f ( x  h)  f ( x )
( x  h) 3  x 3
f ( x)  lim
 lim
h0
h0
h
h
( x 3  3 x 2 h  3 xh 2  h 3 )  x 3
 lim
h0
h
h(3 x 2  3 xh  h 2 )
 lim
 lim(3 x 2  3 xh  h 2 )  3 x 2
h0
h0
h
d n
( x )  n x n1
dx
n is any real number
Examples
f ( x)  x10  f ( x)  10 x101  10 x 9
d 25

x   25x 251  25x 26
dx
1
1
1
f ( x)  x  x1 / 2  f ( x)  x1 / 21  x 1 / 2 
2
2
2 x
d 5 / 8 5 5 / 81 5 3 / 8

x  x
 x
dx
8
8
1
1
1
2
y   x  y   x   2
x
x
Derivative of a constant multiple of a function
d
d
(c u ( x))  c u ( x)
dx
dx
f ( x)  c u ( x)
f ( x  h)  f ( x )
c u ( x  h)  c u ( x )
 lim
h0
h0
h
h
u ( x  h)  u ( x )
 c lim
 c u( x)
h0
h
f ( x)  lim
Examples
d  1  d  1 7  1 d 7 1
7 8
8

x    7 x    x
 7  x 
dx  4 x  dx  4
4
4
 4 dx
y  5x
3/ 2
3 1 / 2 15 1 / 2
 y  5  x  x
2
2
Sum or difference rule
d
d
d
[u ( x)  v( x)]  [u ( x)]  [v( x)]
dx
dx
dx
Also applies to the sum and difference of more
than two functions.
Examples
d 4
d
d

x  5 x 2    x 4   5 x 2   4 x 3  5(2 x)  4 x 3  10 x
dx
dx
dx
d
d
d
d

5 x 8  x 1 / 4  10 x 5 / 2   5 x 8   x 1 / 4   10 x 5 / 2 
dx
dx
dx
dx
1 5 / 4 
5 3/ 2 
1 5 / 4


7
 58 x     x   10 x   40 x  x
 25x 3 / 2
4
 4

2

7
Problem 1
The position of an object (in feet) at time t (in
seconds) is s(t )  t 3  9t 2  27t  9
1. Find the velocity.
v(t )  s(t )  3 t 2  18 t  27
2. Find the velocity at t = 1 and t = 2.
s(1)  3(1) 2  18(1)  27  12 ft/sec
s(2)  3(2) 2  18(2)  27  3 ft/sec
3. Find the time when the velocity is zero.
s(t )  3 t 2  18 t  27  3 t 2  6t  9  3  t  3 t  3  0
The velocity is zero at t = 3.
Sales of digital cameras
According to projections made in 2004, the
worldwide shipment of digital point-and-shoot
cameras are expected to grow in accordance
with the rule:
N (t )  16.3 t 0.8766
(1  t  6)
where N(t) is measured in millions and t is measured in
years, with t = 1 corresponding to 2001.
a. How many digital cameras were sold in 2001?
N (1)  16.3
16.3 million digital cameras were sold in 2001
N (t )  16.3 t 0.8766
(1  t  6)
b. How fast were sales increasing in 2001?
N (t )  16.3(0.8766) t 0.1234  14.2886 t 0.1234
N (1)  14.3
Increasing at the rate of 14.3 million cameras per year
c. What were the projected sales in 2005?
N (5)  16.3 50.8766   66.8
66.8 million cameras
d. How fast were sales increasing in 2005?
N (5)  14.2886 (50.1234 )  11.7
Increasing at the rate of 11.7 million cameras per year