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