CK-12-Calculus

(Marvins-Underground-K-12) #1

http://www.ck12.org Chapter 6. Transcendental Functions


d
dx[logbx] =wlim→x

logbw−logbx
w−x
=wlim→xlogwb−(wx/x)

=wlim→x

[ 1


w−xlogb

(w
x

)]


=wlim→x

[ 1


w−xlogb

(x+(w−x)
x

)]


=wlim→x

[ 1


w−xlogb

(


1 +w−xx

)]


=wlim→x

[ 1


x(w−x)logb

(


1 +w−xx

)]


=wlim→x

[ x
x(w−x)logb

(


1 +w−xx

)]


.


At this stage, leta= (w−x)/(x),the limit ofw→xthen becomesa→ 0 .Substituting, we get


=alim→ 0

[ 1


x

1


alogb(^1 +a)

]


=^1 xalim→ 0

[ 1


alogb(^1 +a)

]


=^1 xalim→ 0

[


logb( 1 +a)^1 /a

]


.


Inserting the limit,


=^1 xlogb

[


alim→ 0 (^1 +a)^1 /a

]


.


But by the definitione=lima→ 0 ( 1 +a)^1 /a,


d
dx[logbx] =

1


xlogbe.

From the box above, we can express logbein terms of natural logarithm by the using the formula logbw=lnw/lnb.
Then


logbe=lnlneb=ln^1 b.

Thus we conclude


d
dx[logbx] =

1


xlnb>^0 ,
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