Derivatives of Exponential and LogarithmFunctions

10/17/2011

The Derivative of y = ex

Recall!ex is the unique exponential function whose slope at x = 0 is 1:

m=1

limh→0

e0+h − e0

h= lim

h→0

eh − 1

h= 1

The Derivative of y = ex

Recall!ex is the unique exponential function whose slope at x = 0 is 1:

m=1

limh→0

e0+h − e0

h= lim

h→0

eh − 1

h= 1

The Derivative of y = ex ...

limh→0

eh − 1

h= 1

d

dxex = lim

h→0

ex+h − ex

h

= limh→0

ex(eh − 1

)h

= ex limh→0

eh − 1

h

= ex ∗ 1

Sod

dxex = ex

The Derivative of y = ex ...

limh→0

eh − 1

h= 1

d

dxex = lim

h→0

ex+h − ex

h

= limh→0

ex(eh − 1

)h

= ex limh→0

eh − 1

h

= ex ∗ 1

Sod

dxex = ex

The Derivative of y = ex ...

limh→0

eh − 1

h= 1

d

dxex = lim

h→0

ex+h − ex

h

= limh→0

ex(eh − 1

)h

= ex limh→0

eh − 1

h

= ex ∗ 1

Sod

dxex = ex

The Derivative of y = ex ...

limh→0

eh − 1

h= 1

d

dxex = lim

h→0

ex+h − ex

h

= limh→0

ex(eh − 1

)h

= ex limh→0

eh − 1

h

= ex ∗ 1

Sod

dxex = ex

The Derivative of y = ex ...

limh→0

eh − 1

h= 1

d

dxex = lim

h→0

ex+h − ex

h

= limh→0

ex(eh − 1

)h

= ex limh→0

eh − 1

h

= ex ∗ 1

Sod

dxex = ex

The Derivative of y = ex ...

limh→0

eh − 1

h= 1

d

dxex = lim

h→0

ex+h − ex

h

= limh→0

ex(eh − 1

)h

= ex limh→0

eh − 1

h

= ex ∗ 1

Sod

dxex = ex

The Chain Rule

TheoremLet u be a function of x. Then

d

dxeu = eu

du

dx.

Examples

Calculate...

1. ddx e17x

= 17e17x

2. ddx esin x

= cos(x)esin x

3. ddx e

√x2+x

= 2x+12√x2+x

e√x2+1

Notice, every time:

d

dxef (x) = f ′(x)ef (x)

Examples

Calculate...

1. ddx e17x = 17e17x

2. ddx esin x = cos(x)esin x

3. ddx e

√x2+x = 2x+1

2√x2+x

e√x2+1

Notice, every time:

d

dxef (x) = f ′(x)ef (x)

Examples

Calculate...

1. ddx e17x = 17e17x

2. ddx esin x = cos(x)esin x

3. ddx e

√x2+x = 2x+1

2√x2+x

e√x2+1

Notice, every time:

d

dxef (x) = f ′(x)ef (x)

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation!

Remember:y = ln x =⇒ ey = x

Take a derivative of both sides of ey = x to get

dy

dxey = 1

Sody

dx=

1

ey

=1

e ln(x)=

1

x

ddx ln(x) = 1

x

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation!Remember:

y = ln x =⇒ ey = x

Take a derivative of both sides of ey = x to get

dy

dxey = 1

Sody

dx=

1

ey

=1

e ln(x)=

1

x

ddx ln(x) = 1

x

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation!Remember:

y = ln x =⇒ ey = x

Take a derivative of both sides of ey = x to get

dy

dxey = 1

Sody

dx=

1

ey

=1

e ln(x)=

1

x

ddx ln(x) = 1

x

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation!Remember:

y = ln x =⇒ ey = x

Take a derivative of both sides of ey = x to get

dy

dxey = 1

Sody

dx=

1

ey

=1

e ln(x)=

1

x

ddx ln(x) = 1

x

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation!Remember:

y = ln x =⇒ ey = x

Take a derivative of both sides of ey = x to get

dy

dxey = 1

Sody

dx=

1

ey=

1

e ln(x)

=1

x

ddx ln(x) = 1

x

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation!Remember:

y = ln x =⇒ ey = x

Take a derivative of both sides of ey = x to get

dy

dxey = 1

Sody

dx=

1

ey=

1

e ln(x)=

1

x

ddx ln(x) = 1

x

The Derivative of y = ln x

To find the derivative of ln(x), use implicit differentiation!Remember:

y = ln x =⇒ ey = x

Take a derivative of both sides of ey = x to get

dy

dxey = 1

Sody

dx=

1

ey=

1

e ln(x)=

1

x

ddx ln(x) = 1

x

Does it make sense?

ddx ln(x) = 1

x

f (x) = ln(x) 1 2 3 4

-1

1

f (x) = 1x

1 2 3 4

1

2

3

Examples

Calculate

1. ddx ln x2

=2x

x2=

2

x

2. ddx ln(sin(x2))

=2x cos(x2)

sin(x2)

Notice, every time:

d

dxln(f (x)) =

f ′(x)

f (x)

Examples

Calculate

1. ddx ln x2 =

2x

x2=

2

x

2. ddx ln(sin(x2)) =

2x cos(x2)

sin(x2)

Notice, every time:

d

dxln(f (x)) =

f ′(x)

f (x)

Examples

Calculate

1. ddx ln x2 =

2x

x2=

2

x

2. ddx ln(sin(x2)) =

2x cos(x2)

sin(x2)

Notice, every time:

d

dxln(f (x)) =

f ′(x)

f (x)

The Calculus Standards: ex and ln x

To get the other derivatives:

ax = ex ln a

loga x =ln x

ln a

For example:

d

dx2x

=d

dxex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x

(ln(2) is a constant!!!)

You try:d

dxlog2(x) =

1

ln(2) ∗ x

The Calculus Standards: ex and ln x

To get the other derivatives:

ax = ex ln a

loga x =ln x

ln a

For example:

d

dx2x

=d

dxex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x

(ln(2) is a constant!!!)

You try:d

dxlog2(x) =

1

ln(2) ∗ x

The Calculus Standards: ex and ln x

To get the other derivatives:

ax = ex ln a

loga x =ln x

ln a

For example:

d

dx2x =

d

dxex ln(2)

= ln(2) ∗ ex ln(2) = ln(2) ∗ 2x

(ln(2) is a constant!!!)

You try:d

dxlog2(x) =

1

ln(2) ∗ x

The Calculus Standards: ex and ln x

To get the other derivatives:

ax = ex ln a

loga x =ln x

ln a

For example:

d

dx2x =

d

dxex ln(2) = ln(2) ∗ ex ln(2)

= ln(2) ∗ 2x

(ln(2) is a constant!!!)

You try:d

dxlog2(x) =

1

ln(2) ∗ x

The Calculus Standards: ex and ln x

To get the other derivatives:

ax = ex ln a

loga x =ln x

ln a

For example:

d

dx2x =

d

dxex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x

(ln(2) is a constant!!!)

You try:d

dxlog2(x) =

1

ln(2) ∗ x

The Calculus Standards: ex and ln x

To get the other derivatives:

ax = ex ln a

loga x =ln x

ln a

For example:

d

dx2x =

d

dxex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x

(ln(2) is a constant!!!)

You try:d

dxlog2(x)

=1

ln(2) ∗ x

The Calculus Standards: ex and ln x

To get the other derivatives:

ax = ex ln a

loga x =ln x

ln a

For example:

d

dx2x =

d

dxex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x

(ln(2) is a constant!!!)

You try:d

dxlog2(x) =

1

ln(2) ∗ x

Differential equations

Suppose y is some mystery function of x and satisfies the equation

y ′ = ky

Goal: What is y??

1. If k = 1, then y = ex has this property and thus solves theequation.

2. For any k , y = ekx solves the equation too!

This equation, ddx y = ky is an example of a differential equation.

Differential equations

Suppose y is some mystery function of x and satisfies the equation

y ′ = ky

Goal: What is y??

1. If k = 1, then y = ex has this property and thus solves theequation.

2. For any k , y = ekx solves the equation too!

This equation, ddx y = ky is an example of a differential equation.

Differential equations

Suppose y is some mystery function of x and satisfies the equation

y ′ = ky

Goal: What is y??

1. If k = 1, then y = ex has this property and thus solves theequation.

2. For any k , y = ekx solves the equation too!

This equation, ddx y = ky is an example of a differential equation.

Differential equations

Suppose y is some mystery function of x and satisfies the equation

y ′ = ky

Goal: What is y??

1. If k = 1, then y = ex has this property and thus solves theequation.

2. For any k , y = ekx solves the equation too!

This equation, ddx y = ky is an example of a differential equation.