Derivatives of Exponential and Logarithm
Functions
10/17/2011
The Derivative of y = e x
Recall!
e x is the unique exponential function whose slope at x = 0 is 1:
m=1
The Derivative of y = e x
Recall!
e x is the unique exponential function whose slope at x = 0 is 1:
m=1
e 0+h − e 0
eh − 1
= lim
=1
h→0
h→0
h
h
lim
The Derivative of y = e x ...
eh − 1
=1
h→0
h
lim
The Derivative of y = e x ...
eh − 1
=1
h→0
h
lim
d x
e x+h − e x
e = lim
h→0
dx
h
The Derivative of y = e x ...
eh − 1
=1
h→0
h
lim
d x
e x+h − e x
e = lim
h→0
dx
h
ex eh − 1
= lim
h→0
h
The Derivative of y = e x ...
eh − 1
=1
h→0
h
lim
d x
e x+h − e x
e = lim
h→0
dx
h
ex eh − 1
= lim
h→0
h
h
e −1
= e x lim
h→0
h
The Derivative of y = e x ...
eh − 1
=1
h→0
h
lim
d x
e x+h − e x
e = lim
h→0
dx
h
ex eh − 1
= lim
h→0
h
h
e −1
= e x lim
h→0
h
= ex ∗ 1
The Derivative of y = e x ...
eh − 1
=1
h→0
h
lim
d x
e x+h − e x
e = lim
h→0
dx
h
ex eh − 1
= lim
h→0
h
h
e −1
= e x lim
h→0
h
= ex ∗ 1
So
d x
e = ex
dx
The Chain Rule
Theorem
Let u be a function of x. Then
d u
du
e = eu .
dx
dx
Examples
Calculate...
1.
d 17x
dx e
2.
d sin x
dx e
3.
√
d
x 2 +x
e
dx
Examples
Calculate...
1.
d 17x
dx e
2.
d sin x
dx e
3.
√
d
x 2 +x
e
dx
= 17e 17x
= cos(x)e sin x
=
√
2x+1
x 2 +1
√
e
2 x 2 +x
Examples
Calculate...
1.
d 17x
dx e
2.
d sin x
dx e
3.
√
d
x 2 +x
e
dx
= 17e 17x
= cos(x)e sin x
=
√
2x+1
x 2 +1
√
e
2 x 2 +x
Notice, every time:
d f (x)
e
= f 0 (x)e f (x)
dx
The Derivative of y = ln x
To find the derivative of ln(x), use implicit differentiation!
The Derivative of y = ln x
To find the derivative of ln(x), use implicit differentiation!
Remember:
y = ln x
=⇒
ey = 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 e y = x to get
dy y
e =1
dx
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 e y = x to get
dy y
e =1
dx
So
1
dy
= y
dx
e
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 e y = x to get
dy y
e =1
dx
So
1
1
dy
= y = ln(x)
dx
e
e
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 e y = x to get
dy y
e =1
dx
So
1
1
1
dy
= y = ln(x) =
x
dx
e
e
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 e y = x to get
dy y
e =1
dx
So
1
1
1
dy
= y = ln(x) =
x
dx
e
e
d
dx
ln(x) =
1
x
Does it make sense?
d
dx
ln(x) =
1
x
1
f (x) = ln(x)
1
2
3
4
1
2
3
4
-1
3
2
f (x) =
1
x
1
Examples
Calculate
1.
d
dx
ln x 2
2.
d
dx
ln(sin(x 2 ))
Examples
Calculate
2x
2
=
2
x
x
1.
d
dx
ln x 2 =
2.
d
dx
ln(sin(x 2 )) =
2x cos(x 2 )
sin(x 2 )
Examples
Calculate
2x
2
=
2
x
x
1.
d
dx
ln x 2 =
2.
d
dx
ln(sin(x 2 )) =
2x cos(x 2 )
sin(x 2 )
Notice, every time:
d
f 0 (x)
ln(f (x)) =
dx
f (x)
The Calculus Standards: e x and ln x
To get the other derivatives:
ax = e x ln a
ln x
loga x =
ln a
The Calculus Standards: e x and ln x
To get the other derivatives:
ax = e x ln a
ln x
loga x =
ln a
For example:
d x
2
dx
The Calculus Standards: e x and ln x
To get the other derivatives:
ax = e x ln a
ln x
loga x =
ln a
For example:
d x ln(2)
d x
e
2 =
dx
dx
The Calculus Standards: e x and ln x
To get the other derivatives:
ax = e x ln a
ln x
loga x =
ln a
For example:
d x ln(2)
d x
e
= ln(2) ∗ e x ln(2)
2 =
dx
dx
(ln(2) is a constant!!!)
The Calculus Standards: e x and ln x
To get the other derivatives:
ax = e x ln a
ln x
loga x =
ln a
For example:
d x ln(2)
d x
e
= ln(2) ∗ e x ln(2) = ln(2) ∗ 2x
2 =
dx
dx
(ln(2) is a constant!!!)
The Calculus Standards: e x and ln x
To get the other derivatives:
ax = e x ln a
ln x
loga x =
ln a
For example:
d x ln(2)
d x
e
= ln(2) ∗ e x ln(2) = ln(2) ∗ 2x
2 =
dx
dx
(ln(2) is a constant!!!)
You try:
d
log2 (x)
dx
The Calculus Standards: e x and ln x
To get the other derivatives:
ax = e x ln a
ln x
loga x =
ln a
For example:
d x ln(2)
d x
e
= ln(2) ∗ e x ln(2) = ln(2) ∗ 2x
2 =
dx
dx
(ln(2) is a constant!!!)
You try:
d
1
log2 (x) =
ln(2) ∗ x
dx
Differential equations
Suppose y is some mystery function of x and satisfies the equation
y 0 = ky
Goal: What is y ??
Differential equations
Suppose y is some mystery function of x and satisfies the equation
y 0 = ky
Goal: What is y ??
1. If k = 1, then y = e x has this property and thus solves the
equation.
Differential equations
Suppose y is some mystery function of x and satisfies the equation
y 0 = ky
Goal: What is y ??
1. If k = 1, then y = e x has this property and thus solves the
equation.
2. For any k, y = e kx solves the equation too!
Differential equations
Suppose y is some mystery function of x and satisfies the equation
y 0 = ky
Goal: What is y ??
1. If k = 1, then y = e x has this property and thus solves the
equation.
2. For any k, y = e kx solves the equation too!
This equation,
d
dx y
= ky is an example of a differential equation.