solutions to derivatives using the limit definition

8/7/2019 Solutions to Derivatives Using the Limit Definition

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SOLUTIONS TO DERIVATIVES USING THE LIMIT DEFINITION

SOLUTION 1 :

(Algebraically and arithmetically simplify the expression in the numerator.)

(The term now divides out and the limit can be calculated.)

.

SOLUTION 2 :


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(Algebraically and arithmetically simplify the expression in the numerator.)

(Factor from the expression in the numerator.)


.

SOLUTION 3 :

(Eliminate the square root terms in the numerator of the expression by multiplying

by the conjugate of the numerator divided by itself.)

(Recall that )


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.

SOLUTION 4 :

(Get a common denominator for the expression in the numerator. Recall that

division by is the same as multiplication by . )

(Algebraically and arithmetically simplify the expression in the numerator. It is

important to note that the denominator of this expression should be left in factored

form so that the term can be easily eliminated later.)


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.

SOLUTION 5 :

(At this point it may appear that multiplying by the conjugate of the numerator

over

itself is a good next step. However, doing something else is a better idea.)

(Note that A - B can be written as the difference of cubes , so that

. This will help explain

the next step.)


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(Algebraically and arithmetically simplify the expression in the nu merator.)


.

SOLUTION 6 :


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(Recall a well-known trigonometry identity :

.)

(Recall the following two well-known trigonometry limits :

and .)

.

SOLUTION 7 :

(Get a common denominator for the expression in the numerator. Recall that

division by is the same as multiplication by . )


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(Algebraically and arithmetically simplify the expression in the numerator. The

terms x3, 2x

2, -3x, and will subtract out. I would show this step, but this web

page is not wide enough. It is important to note that the denominator of this

expression should be left in factored form so that the term can be easily

eliminated later.)

(Factor from the numerator.)


.

SOLUTION 8 :


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(Eliminate the square root terms in the numerator of the expression by multiplying

by the conjugate of the numerator divided by itself.)

(Algebraically and arithmetically simplify the expression in the numerator. It is

important to note that the denominator of this expression should be left in factored

form so that the term can be easily eliminated later.)

(Factor from the numerator.)


.

SOLUTION 9 : The derivative at x=1 is


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.

Note that can be either positive or negative, and when ,

and when . Thus,

Further work requires the use of one-sided limits. First, the right-hand limit is

=

=

=

=

=

=


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=

= .

The left-hand limit is

=

=

=

=

= .

Thus, both one-sided limits exist and are equal, so that fis differentiable at x=1

with

.

SOLUTION 10 : The derivative at x=0 is


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(Note that .)

.

But for

so that implies that

.

Since , it follows from the Squeeze Principle that

.

In a similar fashion (Assume that .) it can easily be shown that

.


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Thus, both one-sided limits exist and are equal, so that function fis differentiable

at x=0 with

.

REMARK : What follows is a common INCORRECT attempt to solve this

problem using another method. Since for , it follows, using

the product rule and chain rule, that

for . Then

.

Because the term oscillates between 1 and -1 as approaches zero ,

this limit does not exist. An INCORRECT conclusion would be that f'(0) does not

exist, i.e., fis not differentiable at x=0. Iff' were continuous at x=0 , this would be

a valid method to compute f'(0) .

SOLUTION 11 : First rewrite f(x). That is,

f(x) = | x2 - 3x | = | x(x-3) |


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Assume that x < 0. Then

= 2x - 3 .

Assume that x>3. Then it is also true that f'(x) = 2x - 3. Assume that 0 < x < 3. Then


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= 3 - 2x .

Now check for differentiability at x=0, i.e., compute f'(0). Then

.

If , then

= 3.

If , then


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= - 3.

Since the one-sided limits exist but are NOT EQUAL, does

not exist, and fis not differentiable at x = 0. Now check for differentiability at x=3,

i.e., compute f'(3). Then

.

If , then so that

= 3.


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If , then so that

= - 3.

Since the one-sided limits exist but are NOT EQUAL,does not exist, and fis not differentiable at x = 3. Summarizing, the derivative

offis

Function fis not differentiable at x=0 orx=3.

SOLUTION 12 : First, determine iffis continuous at x=2, i.e., determine

if . Forx > 2


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= 0.

Forx < 2

= -3.

Thus, the one-sided limits exist but are NOT EQUAL, so that does not

exist and function fis NOT CONTINUOUS AT x=0 . Since function fis NOT

CONTINUOUS AT x=0 , function fis NOT DIFFERENTIABLE at x=2 .

REMARK 1 : Use of the limit definition of the derivative of fat x=2 also leads to a

correct solution to this problem.

REMARK 2 : What follows is a common INCORRECT attempt to solve thisproblem using another method. For x>2

.

Forx


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= 1 ,

and

= 1 .

An INCORRECT conclusion would be that f'(2) = 1. Iff' were continuous at x=2 ,

this would be a valid method to compute f'(2) .

solutions to derivatives using the limit definition

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