5A_Derivative_of_Logarithmic_Functions.pdf

8/10/2019 5A_Derivative_of_Logarithmic_Functions.pdf

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Calculus and Vectors How to get an A+

5A Derivative of Logarithmic Function2010 Iulia & Teodoru Gugoiu - Page 1 of 3

5A Derivative of Logarithmic Function

A Review of Logarithmic Function

yxby bx log==

0,1,0,log)( >>== xbbxxfy b

xxyx

xxxy

bbb

bbb

logloglog

loglog)(log

=

+=

axx

xnx

b

ab

bn

b

logloglog

loglog

=

=

xx

xx e

10loglog

logln

=

=

1log

01log

=

=

bb

b

Ex 1. Use the graph of the logarithmic function to evaluateeach limit.

a) xx

lnlim0+

=+

xx

lnlim0

b) xx

5.00

loglim+

=+

xx

5.00

loglim

c) xx

loglim

=

xx

loglim

d) xx

1.0loglim

=

xx

1.0loglim

B Derivative of xln

xx

dx

d

xx

1ln

1)'(ln

=

=

(1)

Proof:

xx

xy

eyye

exexxy

y

y

yy

1)'(ln

1'

1''1

)'()'(ln

====

===

Ex 2. Differentiate and simplify.

a) xx ln2

)1ln2(ln2

1ln2)'(lnln)'()'ln( 2222

+=+=

+=+=

xxxxx

xxxxxxxxxx

b)x

xln

222

'

ln1

ln1

)')((ln)'(lnlnx

xx

xx

xx

xxxxxx =

==

c) xex ln

+=+=+=x

xex

exexexexe xxxxxx 1

ln1

ln)'(lnln)'()'ln(

C Derivative of )(ln xf

Using (1) and the chain rule:

)(

)('

)(ln

)(

)(')]'([ln

xf

xf

xfdx

d

xf

xfxf

=

=

(2)

Ex 3. Differentiate and simplify.

a) )ln( 23 xx +

xx

x

xx

xx

xx

xxxx

+

+=

+

+=

+

+=+

223

2

23

2323 2323)'()]'[ln(

b)1

1ln

+

x

x

1

2

)1(

)1)(1()1)(1(

1

1

1

1

1

1

1

1ln

22

''

=

+

+

+=

+

+=

+

xx

xx

x

x

x

x

x

x

x

x

c) )ln( xx ee +

xee

ee

ee

eeee

xx

xx

xx

xxxx tanh

)'()]'[ln( =

+

=

+

+=+


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D Derivative of xblog

xbx

dx

d

xbx

b

b

)(ln

1log

)(ln

1)'(log

=

=

(3)

Proof:

xbx

bbxxb

)(ln1)'(ln

ln1

lnln)'(log

'

==

=

Ex 4. Differentiate.a) xlog

xx

)10(ln

1)'(log =

b) xx 32 log

3lnlog2

)3(ln

1log2)'log( 3

233

2 xxx

xxxxxx +=+=

c)x

x

10

log

xx

xx

x

xx

x

xx

xx

xxx

10

)10)(ln(log)10(ln

1

)10(

)10)(ln10)((log10)10(ln

1

)10(

)'10)((log10)'(log

10

log

2

2

'

=

=

=

E Derivative of )(log xfb

Using (3) and the chain rule:

)()(ln

)(')(log

)()(ln

)(')]'([log

xfb

xfxf

dx

d

xfb

xfxf

b

b

=

=

(4)

Ex 5. Differentiate.

a) )1log( 2 +x

)1)(10(ln

2

)1)(10(ln

)'1()]'1[log(

22

22

+=

+

+=+

x

x

x

xx

b) )2(log 22x

x

2ln

2ln2

2)2(ln

)2)(ln2()2(2

2)2(ln

)'2()]'2([log

2

2

2

22

2

x

x

x

xx

x

xx

x

xx

x

xx

+=

+==

c) xlnlog

xxx

xx

ln)10(ln

1

ln)10(ln

)'(ln)'ln(log ==

Ex 6. Find the equation of the tangent line to the

curve xexfy x ln)( == at the point )0,1(P .

ex

eyx

ey

efm

x

exexf xx

11)1(

10

1)1('

1ln)1()('

==

==

+=

Ex 7. Find local extrema points forx

xxf ln)( = .

eefexatxf

x

x

x

xx

xxf

1)(,0)('

ln1)1)((ln

1

)(' 22

===

=

=

x e

)(xf e/1

)(' xf + 0 -

Local maximum point: )/1,( ee .


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Ex 8. Find the inflection points for xxxf ln)( 2= .

3

22/32/3

2/3

2

2

3

2

3)()(

12/3ln0)(''

3ln211

2ln2)(''

ln21

ln2)('

eeef

ee

exxwhenxf

xx

xxxf

xxxx

xxxxf

=

=

====

+=++=

+=+=

x )/(1 ee

)(xf )2/(3 3e

)('' xf - 0 +

The inflection point is:

32

3,

1

eee.

Ex 9. Find the global extrema forx

xxf log)( = over ]10,1[ .

1.0)10(

0)1(

1598.010

10ln/1)10(

718.21010ln/1log0)('

log10ln

1)1)((log)10(ln

1

)('

10ln/110ln/1

10ln/1

22

=

=

=

===

=

=

f

f

f

xxwhenxf

x

x

x

xxx

xf

The global minimum point is )0,1( .

The global maximum point is )1598.0,718.2( .

Ex 10. Differentiate.

a)

x

xy=

)1)(ln()'ln)(()'()'(' lnln +==== xxxxeexy xxxxxx

b) ||ln xy=

0,1

'

0,1

)'(1

0,1

'

0),ln(

0,ln

=

=

=

xx

y

xx

xx

xx

y

xx

xxy

Reading: Nelson Textbook, Pages 571-574Homework: Nelson Textbook: Page 575 #3ef, 4ace, 5ab, 6abc, 9a, 10, 11

Reading: Nelson Textbook, Pages 576-577Homework: Nelson Textbook: Page 578 #1ad, 2cf, 3a, 4bd, 5, 8, 10

5A_Derivative_of_Logarithmic_Functions.pdf

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