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15 EDerivatives of Exponential
Functions
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
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.
x xde e
dx
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
xda
dx
ln xade
dx( and are inverse functions.)
xe ln x
lnx ade
dx
ln lnx a de x a
dx (chain rule)
( 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
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
Next class…..next class.
Homework
Page 375 (#1 – 6)
Chapter 15 Section F
So far 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
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
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
Homework 15 F
1 ) ln(7 ) ln(7) ln( )a y x or x
1 11 ) 0dy
adx x x
Homework 15 F
1 ) ln(2 1)b y x
1 ) ln( ) 2 1b u u x
1dy
du u 2
du
dx
2
2 1
dy du
du dx x
Homework 15 F
21 ) ln( )c y x x 21 ) ln( )c u u x x
1dy
du u 1 2
dux
dx
2
1 2dy du x
du dx x x