ACOW RATES OF CHANGE AND DERIVATIVES MODULE
Updated 5/28/2016
Page 1 of 2
Activity B3 – The Derivative
If we replace c with x in the formula used in Activity B2 we obtain the formula for the
derivative of a function(RS1). The derivative can be used to find slopes of tangent lines,
instantaneous rates of change, and the velocity of an object. There are several different
notations(RS2) used to represent the derivative.
In Activity B2 we used the formula
f (c  h )  f (c )
h 0
h
to find the instantaneous rate of change of f ( x) at x = c. When we find the derivative of
a function f ( x) we are finding the instantaneous rate of change for any value of x. For
example, if f ( x)  x 2 we can find f '( x) (RS3) for any value of x. Then if we want to
lim
find the slope of the line tangent to f ( x) at x = 2, x = –1, and x = 5, all we have to do is
substitute these values(RS4) into the answer we obtained for f '( x) .
1. Given g ( x)  4 x which of the following is the first step in finding g '( x ) ?
4h4
a) lim
h 0
h
4( x  h)  4 x
b)
h
4( x  h)  4 x
c) lim
h 0
h
f (4 x)  4 x
d) lim
h 0
h
2. Suppose f (t )  t 2 0  t  4 represents the position, in feet, of an object after t seconds.
Find f '(t ) by filling in the blanks with the appropriate number/letter.

lim
h 0

2
h 
h
2
x 2  2 xh  h 2 
h 0
h
2
2 xh  h
 lim
h 0
h
 2x  h
 lim
h 0
h
 lim  2 x  h 
 lim
h 0

3. Use the answer obtained in problem 2 above and find the instantaneous rate of change
of the object after
ACOW RATES OF CHANGE AND DERIVATIVES MODULE
Updated 5/28/2016
Page 2 of 2
a) 1 second.
b) 1.75 seconds.
c) 2.25 seconds.
4. Given h( x)  x find h '( x) by filling in the blanks with the correct numbers/letters.
lim
h 0
h 
 lim

h 
h 0
 lim
h 0
 lim
h 0
1

2
h

hx
xh  x
1
xh  x



h 
h 

