calc03 9
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3.9: Derivatives of Exponential and Logarithmic Functions Mt. Rushmore, South Dakota
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-1
0
1
2
3
-3 -2 -1 1 2 3x
Look at the graph of xy e=
The slope at x=0 appears to be 1.
If we assume this to be true, then:
0 0
0lim 1
h
h
e e
h
+
→
−=
definition of derivative
→
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Now we attempt to find a general formula for the derivative of using the definition.
xy e=
( )0
limx h x
x
h
d e ee
dx h
+
→
−=
0lim
x h x
h
e e e
h→
⋅ −=
0
1lim
hx
h
ee
h→
⎛ ⎞−= ⋅⎜ ⎟
⎝ ⎠
0
1lim
hx
h
ee
h→
⎛ ⎞−= ⋅ ⎜ ⎟
⎝ ⎠
1xe= ⋅xe=
This is the slope at x=0, which we have assumed to be 1.
→
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( )x xde e
dx=
→
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xe is its own derivative!
If we incorporate the chain rule:
u ud due e
dx dx=
We can now use this formula to find the derivative ofxa
→
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( )xda
dx
( )ln xade
dx( and are inverse functions.)
xe ln x
( )lnx ade
dx
( )ln lnx a de x a
dx⋅ (chain rule)
→
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( is a constant.)ln a
( )xda
dx
( )ln xade
dx
( )lnx ade
dx
( )ln lnx a de x a
dx⋅
ln lnx ae a⋅
lnxa a⋅
Incorporating the chain rule:
( ) lnu ud dua a a
dx dx=
→
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So far today we have:
u ud due e
dx dx= ( ) lnu ud du
a a adx dx
=
Now it is relatively easy to find the derivative of .ln x
→
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lny x=
ye x=
( ) ( )yd de x
dx dx=
1y dyedx
=
1y
dy
dx e=
1ln
dx
dx x=
1ln
d duu
dx u dx=
→
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To find the derivative of a common log function, you could just use the change of base rule for logs:
logd
xdx
ln
ln10
d x
dx=
1ln
ln10
dx
dx=
1 1
ln10 x= ⋅
The formula for the derivative of a log of any base other than e is:
1log
lna
d duu
dx u a dx=
→
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u ud due e
dx dx= ( ) lnu ud du
a a adx dx
=
1log
lna
d duu
dx u a dx=
1ln
d duu
dx u dx=