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 ,