MATH 3208

THE DERIVATIVE AS A FUNCTION

LIMIT DEFINITION OF THE DERIVATIVE OF A FUNCTION

We have considered the derivative of a function f at a fixed number a.

If we replace a with the variable x we get the derivative expressed as a function.

Thus, we have the following limit definition of the derivative as a function.

EXAMPLE 1

Given xxxf 3)( determine the derivative, )(xf . State the domain of )(xf and )(xf .

h

xxhxhxhhxx

h

xxhxhx

h

xfhxfxf

hhh

33223

0

33

00

33lim

][])()([lim

)()(lim)(

131)0()0(33133lim33

lim 22222

0

322

0

xxxhxhx

h

hhxhhx

hh

Domain of )(xf : x( – ∞ , ∞ ) Domain of )(xf : x( – ∞ , ∞ )

xxxf 3)( 13)( 2 xxf

EXAMPLE 2

(a) Determine )(xf if x

xxf

1)( .

h

xhx

hxxxhx

h

x

x

hx

hx

h

xfhxfxf

hhh

))((

))(1())(1(

lim

1)(1

lim)()(

lim)(000

))()((lim

))()((lim

1

))((

)()(

lim0

22

0

22

0 hxhx

h

hxhx

xhxhxxhxx

h

xhx

xhxhxxhxx

hhh

20

1

))(0(

1

))((

1lim

xxxxhxh

(b) Determine the equation of the tangent line and the normal line that touches the

curve of )(xf at ( 1 , 2 ).

EXAMPLE 3

(a) Determine the derivative )(xf if xxf )( .

(b) Find the slope of the graph of )(xf at the points x = 1 and x = 4.

xxf )( x

xf2

1)(

Domain of )(xf : x[ 0 , ∞ ) Domain of )(xf : x( 0 , ∞ )

SKETCHING THE GRAPH OF A DERIVATIVE

• The degree of the derivative is always one less than the degree of )(xf .

Graphically, this means there is one less turning point on the graph of )(xf .

• The x–coordinates of the local maximum/minimum of )(xf correspond to the

x–intercepts on the graph of )(xf .

• When the graph of )(xf is increasing the graph of )(xf lies above the x–axis

and when the graph of )(xf is decreasing the graph of )(xf lies below the x–axis.

EXAMPLE 4

Given the graph of )(xf sketch the graph of )(xf .

(a)

(b)

EXAMPLE 5

Match the graph of each function, )(xf , in (a) to (c) with its derivative graph, )(xf ,

in (i) to (iii).

(a) (i)

(b) (ii)

(c) (iii)

OTHER NOTATIONS OF A DERIVATIVE

If we use the traditional notation y = f(x) to indicate that the independent variable

is x and the dependent variable is y, then some common alternative notations for the

derivative are as follows:

The symbols D and dx

d are called differentiation operators because they indicate the

operation of differentiation, which is the process of calculating a derivative.

EXAMPLE 6

(a) Given y = 2x2 + 4 determine

dx

dy.

(b) Determine

xdx

d 1.

DIFFERENTIABLE FUNCTIONS

Theorem: If f(x) is differentiable at x = a then f(x) is continuous at x = a.

If f(x) is continuous at x = a then it is NOT necessarily that f(x) is differentiable at x = a.

So, continuity does not imply differentiability.

Remember: For a function to be continuous at x = a: )(lim xfax

= )(lim xfax

= f(a)

For a function to be differentiable at x = a then:

At x = a h

xfhxf

h

)()(lim

0

= h

xfhxf

h

)()(lim

0

= )(af ( )(af must exist )

HOW CAN A FUNCTION FAIL TO BE DIFFERENTIABLE?

A function is NOT differentiable at a point x = a if there is:

(i) A corner at x = a.

(ii) A cusp at x = a.

(iii) A vertical tangent at x = a.

(iv) A point of discontinuity at x = a.

1. A corner at x = a. Consider the function f(x) = | x | at x = 0.

f(x) = | x |

For x > 0:

11limlim)()(

lim||||

lim)()(

lim)(00000

hhhhh h

h

h

xhx

h

xhx

h

xfhxfxf

For x < 0:

11limlim)()(

lim||||

lim)()(

lim)(00000

hhhhh h

h

h

xhx

h

xhx

h

xfhxfxf

)(xf

01

01)(

xif

xifxf

f(x) is continuous at x = 0

f(x) has two tangent lines at x = 0

f(x) is not differentiable at x = 0

)0(f does not exist

2. A cusp at x = a. Consider the function 3

2

)( xxf at x = 0.

3

2

)( xxf

f(x) is continuous at x = 0

)(xf

)(lim

0xf

x

slope of tangent goes to ∞

)(lim

0xf

x

slope of tangent goes to – ∞

)0(f does not exist

3. A vertical tangent at x = a. Consider the function 3

1

)( xxf at x = 0.

3

1

)( xxf

f(x) is continuous at x = 0

vertical tangent at x = 0

(undefined slope)

)0(f does not exist

not differentiable at x = 0

4. A point of discontinuity at x = a. Consider

112

11)(

2

xifx

xifxxf at x = 1.

not continuous at x = 1 not differentiable at x = 1 )1(f does not exist

EXAMPLE 7

Let f(x) be given as

115

122)(

2

2

xifxx

xifxxxf

(a) Determine whether f(x) is continuous at x = 1.

52)1()1(2)22(lim)(lim 22

11

xxxf

xx

51)1(5)1()15(lim)(lim 22

11

xxxf

xx

f(1) = 2(1)2 + (1) + 2 = 5 f(x) is continuous at x = 1 since )(lim

1xf

x = )(lim

1xf

x = f(1) = 5

(b) Determine whether f(x) is differentiable at x = 1.

h

xxhxhx

h

xfhxf

hh

)22(2)()(2lim

)()(lim

22

00

h

xxhxhxhx

h

222242lim

222

0

14124lim24

lim0

2

0

xhx

h

hhxh

hh

the slope of the tangent line approaching 1 from the right = 4(1) + 1 = 5

h

xxhxhx

h

xfhxf

hh

)15(1)(5)(lim

)()(lim

22

00

h

xxhxhxhx

h

151552lim

222

0

5252lim52

lim0

2

0

xhx

h

hhxh

hh

the slope of the tangent line approaching 1 from the left = 2(1) + 5 = 7

Since h

xfhxf

h

xfhxf

hh

)()(lim

)()(lim

00

then f(x) is not differentiable at x = 1.

HIGHER DERIVATIVES

If )(xf is a differentiable function then its derivative )(xf is also a function, so,

)(xf may have a derivative of its own, denoted by )(xf .

The new function )(xf is called the second derivative of )(xf . We write: 2

2

dx

yd

dx

dy

dx

d

We interpret a second derivative as a rate of change of a rate of change. The most

familiar example of this is acceleration. The instantaneous rate of change of velocity

with respect to time is called the acceleration of the object.

The third derivative, )(xf , is the derivative of the second derivative, )(xf .

We can write: 3

3

2

2

dx

yd

dx

yd

dx

d

This process can be continued and the fourth derivative denoted as )()4( xf .

In general, if )(xfy we write: n

nnn

dx

ydxfy )()()(

EXAMPLE 8

Given 3)( xxf determine )(xf , )2(f , )(xf , and )4(f .

QUESTIONS Pages 92 – 94, # 3, 4, 5, 6, 9, 13, 17a, 19 – 27, 29a, 33 – 36, 42 – 43