The Instantaneous Rate of Change
If the average rates of change over [a, a+h], for h>0, and [a+h,a], for h<0, have a
limit as h approaches 0, this limit is called the instantaneous rate of change at a.
It is also called the derivative of f at a, written f '(a) , also called f prime at a.
f (a  h )  f (a )
f ' ( a )  lim
h 0
h
If this limit exists, then we say f is differentiable at a.
Example 1
a) Find f '(3) for
f (x)  x .
2
First write the difference quotient.
f ( 3  h )  f ( 3)
(3  h )  3
2

h
2
h
If f is continuous at a then the numerator and denominator both go to 0. Substitution
of h=0 gives the indeterminate form 0/0.
Second step, simplify and cancel h.
(3  h )  3
2
2
9  6h  h  9
2

h

h (6  h )
h

6h  h
2
h
6h
h
6+h is the simplified difference quotient.
Third step, take the limit as h approaches 0.
lim ( 6  h )  6  f ' ( 3 )
h 0
b) Find f '(a) for a general a if f ( x )
Do the exact same replacing 3 with a.
(a  h )  a
2
2
h

2
a  2 ah  h  a
2

 x .
2
2

2 ah  h
h
h(2a  h )
2
h
 2a  h
h
lim ( 2 a  h )  2 a  f ' ( a )
h 0
We can do this for any a, so f is differentiable at every point.
What makes a function not differentiable?
When does the limit fail to exist? Some possibilities follow.
Case 1. f is not differentiable at a if f is not continuous at a.
Example
4 x  2
f (x)  
4 x  3
x0
x0
This function has a gap, a jump of 1, at x=0.
The difference quotient for h<0 is
4h  2  3
h

4h  1
h
The limit as h approaches 0 from the left is infinity. This is not a number so the derivative
does not exist.
Note that for h>0, the difference quotients are all equal to 4.
Other types of discontinuity have similar results.
The problem is that when f is not continuous at a, the numerator of the difference
quotient does not approach 0 as h approaches 0 but the denominator, h, always
approaches 0.
Case 2.
f is not differentiable at a if the graph has a corner at a. The function is
continuous at a but the slope from the left is different than the slope from the right.
Example
3x  2
f (x)  
4 x  2
x0
x0
f is continuous at 0 since the pieces meet at (0,2) and f(0)=2.
For h<0,
lim
f (0  h )  f (0)
h
f (0  h )  f (0)


3h
3
h
 lim 3  3

h 0
h
f (0  h )  f (0) 4 h
For h>0,

4
h
h
f (0  h )  f (0)
lim
 lim 4  4


h 0
h 0
h
h 0
Case 3 f is not differentiable at a if the graph has a cusp or a vertical tangent at a.
Example of a cusp
f (x)  x
Example of a vertical tangent
2 3
at x=0
f (x)  x
1 3
at x=0
The Tangent Line to the graph of f at (a, f(a)).
The tangent line at (a,f(a)) is
y  f ' ( a )( x  a )  f ( a )
The slope is f '(a) and a point is (a, f(a)).
Example Find the equation of the tangent line to
f ( x )  x at (3, f(3)).
2
From the previous examples we know f '(3)=6 and f(3)=9 so we have the tangent line is
y  f ' ( 3 )( x  3 )  f ( 3 )  6 ( x  3 )  9