4.1 through 4.5 Differentiation of Logarithmic and Exponential

Functions• The derivative of lnx• Using product/quotient/chain rules with lnx involved too• The derivative of ex

• Using product/quotient/chain rules with ex involved• Careful not to confuse ex with xn

• Differentiating something with both lnx and ex

A. The derivative of ln x

• The derivative of ln x is

Memorize it. This is for differentiating just plain ln x.

x

1

B. Using product/quotient/chain rules with “ln x” in it

( )( )xf

xxxf

′= Find

ln3

( )x

xxf

ln ateDifferenti =

( )1ln 2 +xdx

d

( )34 1ln −xdx

d

( )xxdx

dln2:You try 2

23

ln:You try

x

x

dx

d

( )15ln:You try 3 +− xxdx

d

( )24 3ln :You try +xdx

d

C. The derivative of ex

• The derivative of just plain ex is

ex

Memorize it. It seems like you haven’t done anything to it! That’s OK. It is ITS OWN DERIVATIVE. Make peace with it. This is only for just plain ex

D. Using product/quotient/chain rules with ex involved

x

e

dx

d x

( )1

2

=x

xexdx

d

( ) ( ).1 find ,:You try fxexf x ′=

14 +xedx

d

22x

edx

d

313

:You tryx

edx

d +

Notice a small pattern here:( )

( )

( )

e.predictabl is rulechain the

on x, multipleconstant ajust is e when therSo

11

1717

44

171717

444

xxx

xxx

xxx

eeedx

d

eeedx

d

eeedx

d

−−− −=−⋅=

=⋅=

=⋅=

x

5xe :You try

E. Careful not to confuse ex with xn

• Word of caution not to confuse the power rule with the exponential form:

xxnn eedx

dxnx

dx

d =⋅= − BUT 1

Also, recall that e without an variable exponent on it is just a

number...

exdx

d

:TRICK a is one thisSo

( ) trick.]a s[It' :You try 22xedx

d

F. Differentiating something involving both lnx and ex

• These problems could use product, quotient, or chain rules.

• As usual, product rule is used when it starts off with a function times a function, or “something times something.” And of course, the quotient rule is used when it is a function divided by another function. BUT the typical CHAIN-type can look a few different ways:

Just to review types of problems we have used the chain rule with:

• (some fancy function of x)n, where n is some number. [on last exam] It becomes:

n(that f’n of x)n-1(the deriv. of that f’n)• e(some fancy function of x), which becomes

e(that f’n of x)(the deriv. of that f’n)• ln (some fancy function of x), which becomes

1 (the deriv. of that f’n)

(that f’n of x)

( ).1ln of derivative theFind xe+

xe4ln ateDifferenti

( )xx eedx

dTry 3ln: +−

++− 5

2

1ln

222 xexxxdx

d

( )( )xexdx

dTry x −−+ 2/4 41ln: