Chapter 3_2.pptx

Chapter 3.2

In Chapter 3 we will develop

derivative formulas for functions of interest in the analysis of business applications.

This includes derivatives for functions we have been using to model data.

In this Section we consider derivatives of some basic functions.

b

We start by considering a constant

function.

Suppose where b is a constant. What is ?

x f(x)

= b

x f(x + h) = b

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

� ′ ( � )=lim

h →0

� −�

h =limh →0 0=0

where

  and

If ,  

Consider functions of the form where n is a positive integer. What is ?

We start with Note

x f(x)

= x

x f(x + h)

� ′ ( � )=lim

h →0

( �+h) − �

h =limh →0

h

h =limh →0 1=1

If ,  

�

(

�

+

h

)

=

�

+

h

where

�

(

�

)

=

�

and

 definition of the derivative

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

Now consider   What is ?

Note

x f(x)

= x2

x f(x + h)

= (x + h)2

� ′( � )=lim h →0

2h�+h2

h =limh →0

h(2� +h) h

If ,  

  � ′

( � )=lim

h →0

(2 �+h)=2 �  

� ′( � )=lim h →0

( �2+2h�+h2) − �2

h

 definition of the derivative

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

where   and

Consider now . What is ?

Note, we can show, after a few algebra steps that:

x f(x)

= x3 x

+ f(x + h)

= (x + h)3

� ′(�)=lim

h →0

3h �2+3h2�+h3

h =limh →0

h(3 �2+3 h�+h2)

h

If ,  

� ′(�)=lim

h →0 (3� 2

+3 h�+h2)=3 �2

 definition of the derivative

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

� ′( �)=lim

h →0

(�3+3h �2+3h2 �+h3)−�3

h

where   and

Using the definition of the derivative we have seen that if:

In general, if , and n is a positive integer,  

Example: If ,

What if the input variable is raised to a negative

integer power?

Consider What is ?

� ( �+h) − � ( �)

h =

1

h ( � ( �+h) − � ( �))

 definition of the derivative

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

� ( �+h) − � ( �)

h =

1

h

(

1

( �+h )2 −

1

�2

)

¿ h

h

(

− 2 � − h

(� +h )2 �2

)

−2 � − h

( � +h )2 �2

� ′ ( � )=lim

h →0

− 2 � − h

( �+h)2 �2 =

−2 � �4 =

−2

�3  

If ,  

¿ 1

h

(

�2−(�2+2 h�+h2)

( �+h)2 �2

)

1

h

(

−2h� −h2

(� +h)2 �2

)

� ( �+h) − � ( � )

h =

1

h

(

1

( � +h )2 −

1

�2

)

1

h

(

�2− ( �+h)2 (� +h)2 �2

)

This is the same general form of the derivative that we developed for

when n is a positive integer.

In general, if , and n is any non-zero integer,

Now we consider exponents other than integers.

But first it will be helpful to remember the factorization for a difference of

two squares:

Similarly:

What is ?

� ( �+ h) − � ( � )

h =

1

h (( � +h )

1/2

− �1/2

)

¿ 1

h ( (� +h )

1/2

− �1/2

) (� + h)

1/2

+ �1/2

(� + h)1/2+ �1/2  

¿ 1

h

( � +h )− �

( �+h)1/2+�1/2 =

1 h

h

(� +h )1/2+�1/2  

 definition of the derivative

� ′( �)=lim

h →0

� (�+h )− � ( �)

h

� ( �+ h) − � ( � )

h =

1 h

h

( � +h )1/ 2+ �1/2  

� ( �+ h) − � ( � )

h =

1

( � +h )1/ 2+ �1/ 2  

 definition of the derivative

� ′( �)=lim

h →0

� (�+h )− � ( �)

h

� ′ ( � )= 1 2 �

−1/2

If ,  

This is the same general form of the derivative that we developed for

when n is an integer.

In general, if , and n is any non-zero real number,

Examples: A. ,

B. , C. , D. ,

As we have discussed previously, the derivative of a function evaluated at a value for the input variable, is the:

• slope of the tangent line to the

function at the point of tangency, and • the instantaneous rate of change of

Example:

What is the rate of change of the function when

A. x = 2

When ,  

A. when x = 2,

slope of tangent at x = 2 is 4

Example:

What is the rate of change of the function when

B. x = -1

When ,  

B. when x = -1,

slope of tangent at x = -1 is -2

If   What is ?

Suppose we have a constant times a function. What is the derivative?

 definition of the derivative

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

First note the derivative of a function , is written as:

 definition of the derivative

�′ ( � )=lim

h →0

�( �+h) −�( � )

h

Similarly, the derivative of a function , is written as:

If   where c is a constant,

what is ?

� ′( �)=lim

h →0

� � (�+h)− � �( � )

h =limh →0 �

�( �+h) − �(�)

h

 definition of the derivative

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

� ′( � )=lim

h →0 �

� (� +h)− �( � )

h

� ′( � ) ¿�

[

h →0

�( �+h) −� (� )

h

]

But the quantity in square brackets is : Therefore:

� ′

( � ) ¿ � �′( � )

 definition of the derivative

�′ ( � )=lim

h →0

�( �+h) −�( � )

h

If ,  

where c is a constant

Examples: A. ,

B. ,

C. , D. ,

Problem 1:

A. Find . B. Find .

Problem 2: A. Find .

B. Find .

Problem 1 Solved:

Problem 2 Solved:

Consider a function which is a sum of functions, such as the following:

How might we find the derivative ?  

 definition of the derivative

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

First note the derivative of a function , is written as:

 definition of the derivative

� ₁′( � )=lim h →0

� ₁( �+h) − � ₁ (� ) h

Similarly, the derivative of a function , is written as:

 definition of the derivative

� ′( � )=lim � 2( � +h )− � 2( �)  

Consider a function which is a sum of functions, such as the following:

 definition of the derivative

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

� (� +h )= � ₁ ( �+h)+ � ₂ (� +h)

� ( �+h) − � ( �)

h =

� ₁ (�+h)+ � ₂(� +h)− ₍ � ₁ ( �)+ � ₂( �)₎

h  

� ( �+h) − � ( �)

h =

� ₁ (�+h)+ � ₂(� +h)− ₍ � ₁ ( �)+ � ₂( �)₎

h  

� ( �+h) − � ( � )

h =

� ₁ (� +h) − � ₁( �)

h +

� ₂ ( �+h) − � ₂ (� ) h

 definition of the derivative

� ′ ( � )=lim

h →0

� (� +h )− � ( � )

h

� ′ ( �)=

lim

h→0 � 1( �

+h)− � ₁ ( �)

h +

lim

h →0 � 2( �

+h) − � ₂( �)

h

+

� ′ ( �)=

lim

h→0 � 1( �+h)− � 1 ( �)

h +

lim

h →0 � 2( �+h) − � 2( �)

h

But:

 definition of the derivative

� ₁′( � )=lim h →0

� ₁( �+h) − � ₁ (� ) h

 definition of the derivative

� ₂′( � )=lim h →0

� ₂( � +h )− � ₂( �) h

In summary: if

then

Note the result is equivalent to taking the derivative of term by term.

That is, if a function consists of a sum of terms, the derivative of the function can be taken term by

term.

Or said differently, the derivative of a sum of

functions is equal to the sum of the derivatives.

The derivative of a sum of functions is the sum of the derivatives:

If  ,  

Example:

Summary of derivative expressions developed in this section:

,

      ,    

Problem 3:

Find the derivatives of the following, reporting them using both the prime, , and notations:

A.  

Problem 3 Solved:

A.  

B.   C.  

D.  

E.  

F.   G.   

H.   

I.          

Problem 4:

Cumulative iPod sales during 2003 – 2008 are shown below.

A. Find a quadratic model, S(t), using the aligned. Round model parameters to 4 decimal digits.

year aligned input sales (million iPods)

2003 3 1.320

2004 4 5.736

2005 5 28.233

2006 6 67.642

Problem 4 Solved:

3 4 5 6 7 8

0 100 200

f(x) = 6.58 x² − 36.86 x + 50.49 R² = 1

iPod Sales

year (aligned to 2000)

iP o d s S o ld ( m ill io n s )

Problem 4:

Cumulative iPod sales during 2003 – 2008 are shown below.

B. Determine the rate of change of iPod sales in 2008 using the model. Provide the units for the rate of change.

C. Calculate the % rate of change of iPod sales in 2008.

Problem 4 Solved:

Rate of change of sales in 2008 is .  

Problem 4 Solved:

% rate of change of sales in 2008 is .

From prior slide:  

Problem 4 Solved:

0 100 200

f(x) = 6.58 x² − 36.86 x + 50.49 R² = 1

iPod Sales iP o d s S o ld ( m ill io n s )