The Derivative
3.1
Calculus
Derivative –
instantaneous rate of
change of one variable wrt
another.
Differentiation
– process of finding
the derivative
Finding Average Rate of Change
 A piece of chocolate is pulled from a refrigerator (6° C) and
placed on a counter (22° C). The temperature of the chocolate is
given by:
Min
0
4
Temp
6.00
9.87
8
12
16
20
24
28
32
36
12.81 15.04 16.72 18.00 18.97 19.70 20.26 20.68
What is the average rate of change in the temperature
of the chocolate from 8 to 20 minutes?
T 18  12.81

 .4325 C
min
t
20  8
The rate of change was not constant thru out the
process. This only tell us what happened on
average over a period of time!
We take the
limit of the
average rate
of change as
we let the
intervals get
smaller and
smaller
∆x  0
y
(b, f(b))
(a, f(a))
a
Tangent
Line
b
x
Tangent Line
Definition of Tangent Line
The tangent line to the graph of y = f(x) at x = c is
the line through the point (c, f(c)) with slope
f ( x )  f (c )
lim
x c
xc
provided this limit exists.
If the instantaneous rate of change of f(x) with
respect to x exists at a point c, then it is the
slope of the tangent line at that point.
Find the slope of the tangent line at x = 3 if
f ( x)  2 x  3
2
f ( x )  f (c )
lim
x c
xc
2 x  3 x  3
 lim
x 3
x 3
2 x 2  3  21
lim
x 3
x 3
 lim 2 x  3
2 x 2  18
 lim
x 3
x 3
 12

2 x2  9
 lim
x 3
x 3

x 3
Derivative at any point
The derivative is
same as the slope
of the tangent line
To Find slope of tangent line at a given
point. Plug given point in f’(x)
f  x
“f prime x”
y
“y prime”
or
“the derivative of f with respect
to x”
dy
dx
“dee why dee ecks”
or
“the derivative of y with
respect to x”
df
dx
“dee eff dee ecks”
or
“the derivative of f with
respect to x”
d
f  x  “dee dee ecks uv eff uv ecks” or “the derivative
dx
of f of x”
( d dx of f of x )

4
3
2
y  f  x
1
0
3
The derivative
is the slope of
the original
function.
1
2
3
4
5
6
7
8
9
2
The derivative is defined at the end points
of a function on a closed interval.
1
0
-1
-2
1
2
3
4
5
6
7
8
9
y  f  x

6
5
y  x 3
2
4
3
2
1
-3
-2
-1 0
-1
1
x
2
3
y  lim
 x  h
2
h 0
-2

3 x 3
2

h
-3
6
5
4
3
2
1
-3 -2 -1 0
-1
-2
-3
-4
-5
-6
x  2 xh  h  x
y  lim
h 0
h
2
1 2 3
x
2
y  lim 2 x  h
2
0
h 0
y  2 x

Find the derivative of f(x) and use it to find the equation
of the tangent line at the point x = 4
f ( x)  5 x  3x 2
f ( x  h)  f ( x )
f ' ( x)  lim
h 0
h
5( x  h)  3 x  h   (5 x  3x 2 )
lim
h 0
h
f (4)  5(4)  3(4) 2  28
f ' (4)  5  6(4)  19
2
5 x  5h  3x 2  6 xh  3h 2  5 x  3x 2
 lim
h 0
h
 lim 5  6 x  3h
h 0
 5  6x
Slope: -19 and point (4, -28)
y  mx  b
 28  19(4)  b
48  b
y  19 x  48
Find dy/dx for
dy
f (t  h)  f (t )
 lim
dt h0
h
4
4

t (t  h)
dy
t

h
t

 lim
t (t  h)
dt h0
h
dy
4t  4(t  h)
 lim
dt h0 ht (t  h)
dy
4t  4t  4h
 lim
dt h0 ht (t  h)
4
y
t
dy
4
 lim
dt h0 t (t  h)
dy  4
 2
dt
t