Aim: How do we find the derivative function using the limit process?
Objectives: to derive the limit definition of derivative, to recognize the derivative form of a limit
process, to connect differentiability and continuity.
Lesson Development: If we are asked to find the derivative at another point, we need to go through the
tedious process of setting up the limit expression. Can we find a derivative function, in which we can
just plug in x = 1 and it will tell us the derivative?
f '(a )  lim
x a
f ( x )  f (a )
xa
Let h  x  a
f ( x )  f (a )
f (a  h)  f (a )
. We prefer this notation because h always goes to 0 and
 lim
x a
h0
xa
h
h is always in the denominator whereas previously it was all different depending on x value. But h is a
placeholder. It can be x  0
f '(a )  lim
Using the same thought process: f '(3)  lim
h 0
Now we can claim: f '( x)  lim
h 0
f (3  h)  f (3)
f (2  h)  f (2)
and f '(2)  lim
h

0
h
h
f ( x  h)  f ( x )
h
We commonly use f '( x) to denote the derivative function. But other commonly accepted notations are
dy d
,
[ f ( x)], y ' . f '( x) tells us the “slope at a point” or the (instantaneous) rate of change of f ( x)
dx dx
with respect to x.
EX1: If f ( x)  3x  x 2 , what is f '( x) ?
f '( x )  lim
h0
f ( x  h)  f ( x)
(3x  ( x  h)2 )  (3x  x 2 )
3x  x 2  2 xh  h 2  3x  x 2
 lim
 lim
 3  2x
h0
h0
h
h
h
*Remember to cancel out h before applying the limit.
Therefore, f '(1)  1, f '(2)  1
EX2: Given that f ( x)  x3 . Find f '(4) and f '(1)
[( x  h)3 ]  ( x 3 )
f '( x)  lim
h 0
h
3
2
x  3x h  2 xh 2  h3  x3
 lim
h 0
h
h(3x 2  2 xh  h 2 )
 lim
 3x 2
h 0
h
EX3: a) Find g '(t ) for g (t ) 
f '(4)  48
f '(1)  3
2
t
2t  2(t  h)
2
2

2t  2(t  h) 1
2h
2
t (t  h)
g '(t )  lim t  h t  lim
 lim
 lim
 2
h 0
h

0
h

0
h

0
h
h
t (t  h) h
t (t  h)h t
1
b) Find g '(2) . ANS:
2
EX4: What is the derivative function for any linear function or constant function? Explain.
Fact: The derivative of a linear equation = its slope
Fact: The derivative of a constant function = its slope = 0
If given a graph, how do we know if a tangent can be drawn? In other words, how do we know if
function is differentiable at the point?
Algebraically, f’(a) = undefined.
EX5: What derivative functions are these expressions representing?
a) lim
h 0
x 1 h  x 1
h
sin( x  h)  sin x
h 0
h
b) lim
f '( x), f ( x)  x  1
f '( x), f ( x)  sin x
e xh  e x
h 0
h
( x  h)3/2  x3/2
h 0
h
c) lim
d) lim
f '( x), f ( x)  e x
f '( x), f ( x)  x3/2
HW#11: P187 – 189: 3b, 7, 15, 17, 19, 27a, 28a
HW#11 Solutions:
3b) 3
15) f '( x)  
7) y = 5x – 16
27a) f ( x)  x , a = 1
28a)
1
x2
17) f '( x)  2 x  1 19) f '( x)  
1
2 x3/2