3.1 Derivative
of a Function
Calculus
You know this from last chapter, I want to find the slope between two points
on a secant line.
What are the
coordinates of
this other point?
secant line
( x + Dx, f (x + Dx))
What is the Δy?
f (x + Dx)- f (x)
(x, f(x))
Dx
How do we find the
slope (rate of
m=
change) on ANY
linear line?
y2 - y1
x2 - x1
f (x + Dx) - f (x) f (c + Dx) - f (c)
=
=
Dx
(x + Dx) - x
BUT, what if I wanted to find a slope of a
specific point???
Then I would want that second point of
the secant line closer to the first point.
In other words I would want to find the
limit of that slope as the Δx approaches 0
Definition of Tangent Line with Slope m
f (x + Dx) - f (x)
lim
=m
Dx®0
Dx
Find the slope of the graph of f(x) = x2 +1 at
the point (-1,2). Then, find the equation of the
tangent line.
(-1,2)
f ( x + Dx) - f ( x)
lim
Dx ®0
Dx
2
2
( x + Dx) + 1 - (x + 1)
= lim
Dx ®0
Dx
f(x) = x2 + 1
Therefore, the slope
at any point (x, f(x))
is given by m = 2x
This is the derivative!!
x + 2 xDx + (Dx ) + 1 - x - 1
= lim
Dx ®0
Dx
2
2
Dx(2 x + Dx)
= lim
=
2
x
Dx ®0
Dx
The equation of the tangent line is
y = -2x
2
What is the slope
at the point (-1,2)?
m = -2
The limit used to define the slope of a tangent
line is also used to define one of the two
fundamental operations of calculus --- differentiation
Definition of the Derivative of a Function
f ( x + Dx) - f ( x)
f ' ( x) = lim
Dx ®0
Dx
f  x
“f prime x”
y
“y prime”
dy
dx
or
“the derivative of f with respect to x”
“the derivative of y with respect to x”
dx does not mean d times x !
dy does not mean d times y !
dy
does not mean dy  dx !
dx
Find f’(x) for f(x) = x, and use the result to find
the slope of the graph of f at the points (1,1) & (4,2). What happens at the
point (0,0)?
f ( x + Dx) - f ( x)
f ' ( x) = lim
Dx ®0
Dx
f ' ( x) = lim
Dx ®0
x + Dx - x æ x + Dx + x ö
÷
× çç
÷
Dx
x
+
D
x
+
x
è
ø
x + Dx - x
= lim
Dx ®0 Dx
x + Dx + x
1
(
= lim
Dx ®0
(
Rationalize the numerator
1
x + Dx + x
)
)
=
1
2 x
f ' ( x) = m =
Therefore, at the point (1,1), the
slope is ½, and at the point (4,2),
the slope is ¼.
1
2 x
What happens at the point (0,0)?
The slope is undefined, since it produces division
by zero.
1
m=
4
1
m=
2
1
2
3
4
Find the derivative with respect to t for the
2
function y =
.
t
dy
f (t + Dt ) - f (t )
= lim
dx Dt ®0
Dt
2t - 2(t + Dt )
2
2
t (t + Dt )
t
+
D
t
t
= lim
= lim
Dt ®0
Dt ®0
Dt
Dt
2t - 2t - 2Dt 1
2 1
= lim
× =- 2
Dt ®0
t
t (t + Dt ) Dt
Alternate Form of the Derivative
(can only be used when finding the derivative at
a specific point!!)
The derivative of f at x = c is given by
f ( x ) - f (c )
f ' (c) = lim
x ®c
x-c
(x, f(x))
f ( x ) - f (c )
(c, f(c))
Dx = x - c
c
x
Derivative from the left and from the right.
f ( x ) - f (c )
limx ®c
x-c
f ( x ) - f (c )
lim+
x ®c
x-c
Example of a point that is not differentiable.
f (x) = x
is continuous at x = 0 but let’s
look at it’s one sided limits.
f (x) - f (0)
lim=
x®0
x-2
f (x) - f (0)
lim+
=
x®0
x-0
lim-
x -0
lim= x®0
x
= -1
x
x
lim+
= lim+ = 1
x®0 x - 0
x®0 x
x®0
x-0
x -0
The 1-sided limits are not equal.
Therefore, x is not differentiable at x = 2. Also, the
graph of f does not have a tangent line at the
point (2, 0).
A function is not differentiable at a point at
which its graph has a sharp turn or a vertical
tangent line(y = x1/3 or y = absolute value of x).
Differentiability can also be destroyed by
a discontinuity ( y = the greatest integer of x).
4
3
2
y  f  x
1
The derivative is
the slope of the
original function.
0
3
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

A function is differentiable if it has a
derivative everywhere in its domain. It
must be continuous and smooth.
Functions on closed intervals must have
one-sided derivatives defined at the end
points.
For example:
y = x -3
is not differentiable at x = 3
Basically anywhere there is a sharp turn (cusp), it is not differentiable.
p
• A function will not have a derivative at point
P(a , f(a)) where the slopes of the secant lines,
fail to approach a limit as x approaches a. The
following four examples are instances where this
occurs. For example, a function whose graph is
otherwise smooth will fail to have a derivative at a
point where the graph has:
1. A corner, where the one-sided derivatives differ,
example f(x) = | x |.
2. A cusp, where the slopes of the secant lines
approach infinity from one side and negative
infinity from the other (an extreme case of a
corner).
3. A vertical tangent, where the slopes of the secant
lines approach either infinity or negative infinity
from both sides.
4. A discontinuity (which will cause one or both of
the one-sided derivatives to be nonexistent).
Example, the Unit Step Function.
Finding Where a Function is Not Differentiable
• Find all points in the domain of f(x) = |x – 2| + 3 where
f is not differentiable.
• Think graphically! The graph of this function is the
same as that of y = | x |, translated 2 units to the right
and the 3 units up.
• This puts the corner at the point (2 , 3), so this function
is not differentiable at x = 2.
• At every other point, the graph is (locally) a straight line
and f has derivative of +1 or -1 (again, just like
y = | x |).
• Most of the functions we encounter in calculus are
differentiable wherever they are defined.
– They will not have corners, cusps, vertical tangent lines,
or points of discontinuity within their domains.
• The graphs will be unbroken and smooth, with a
well-defined slope at each point.
• Polynomials are differentiable, as are rational
functions, trigonometric functions, exponential
functions, and logarithmic functions.
• Composites of differentiable functions are
differentiable, and so are sums, products, integer
powers, and quotients of differentiable functions,
where defined.
Use the definition of the derivative to find the
equation of the derivative for the function
below:
f (x) = x - 6x - 4
2
f '(x) = lim
Dx®0
f (x + Dx) - f (x)
Dx
(x + Dx)2 - 6(x + Dx) - 4 - (x 2 - 6x - 4)
f '(x) = lim
Dx®0
Dx
x 2 + 2x(Dx) + (Dx)2 - 6x - 6Dx - 4 - x 2 + 6x + 4
f '(x) = lim
Dx®0
Dx
2x(Dx) + (Dx)2 - 6Dx
f '(x) = lim
Dx®0
Dx
= lim 2x + Dx - 6
Dx®0
= 2x - 6
Use the definition of the derivative to find the
equation of the derivative for the function
below:
f (x) = (x + 2) +8
2
Use the definition of the derivative to find the
equation of the derivative for the function
below:
f (x) = 2x - 5
Use the definition of the derivative to find the
equation of the derivative for the function
below:
1
f (x) =
3x -1
Find an equation for the line perpendicular
to the tangent to the curve y = x3 – 4x + 1 at
(2, 1)
Find equations for the tangents to the curve
at the points where the slope of the
curve is 8.