Math 140
Lecture 16
Greg Maloney
with modifications by Todor Milev
University of Massachusetts Boston
October 21-25, 2013
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Outline
1
Derivatives of Logarithmic Functions
Logarithmic Differentiation
The Number e as a Limit
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
d
1
(loga x) =
.
dx
x ln a
Proof.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
d
1
(loga x) =
.
dx
x ln a
Proof.
Let y = loga x.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
d
1
(loga x) =
.
dx
x ln a
Proof.
Let y = loga x.
Then ay = x.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
d
1
(loga x) =
.
dx
x ln a
Proof.
Let y = loga x.
Then ay = x.
Differentiate implicitly:
FreeCalc Math 140
Lecture 16
=
October 21-25, 2013
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
d
1
(loga x) =
.
dx
x ln a
Proof.
Let y = loga x.
Then ay = x.
Differentiate implicitly: ay (ln a)y 0 =
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
d
1
(loga x) =
.
dx
x ln a
Proof.
Let y = loga x.
Then ay = x.
Differentiate implicitly: ay (ln a)y 0 =
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
d
1
(loga x) =
.
dx
x ln a
Proof.
Let y = loga x.
Then ay = x.
Differentiate implicitly: ay (ln a)y 0 = 1
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
d
1
(loga x) =
.
dx
x ln a
Proof.
Let y = loga x.
Then ay = x.
Differentiate implicitly: ay (ln a)y 0 = 1
y0 =
FreeCalc Math 140
Lecture 16
ay
1
ln a
October 21-25, 2013
Derivatives of Logarithmic Functions
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
d
1
(loga x) =
.
dx
x ln a
Proof.
Let y = loga x.
Then ay = x.
Differentiate implicitly: ay (ln a)y 0 = 1
1
ln a
1
=
.
x ln a
y0 =
FreeCalc Math 140
Lecture 16
ay
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) =
Let g(x) =
Then
FreeCalc Math 140
f (x) = g(h(x)).
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) = 5x + 1.
Let g(x) =
Then
FreeCalc Math 140
f (x) = g(h(x)).
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) = 5x + 1.
Let g(x) =
Then
FreeCalc Math 140
f (x) = g(h(x)).
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) = 5x + 1.
Let g(x) = log3 x.
Then
FreeCalc Math 140
f (x) = g(h(x)).
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) = 5x + 1.
Let g(x) = log3 x.
Then
f (x) = g(h(x)).
Chain Rule: f 0 (x) = g 0 (h(x))h0 (x)
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) = 5x + 1.
Let g(x) = log3 x.
Then
f (x) = g(h(x)).
Chain Rule: f 0 (x) = g 0 (h(x))h0 (x)
=
(
FreeCalc Math 140
Lecture 16
)
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) = 5x + 1.
Let g(x) = log3 x.
Then
f (x) = g(h(x)).
Chain Rule: f 0 (x) = g 0 (h(x))h0 (x)
1
=
(
h(x) ln 3
FreeCalc Math 140
Lecture 16
)
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) = 5x + 1.
Let g(x) = log3 x.
Then
f (x) = g(h(x)).
Chain Rule: f 0 (x) = g 0 (h(x))h0 (x)
1
=
(
h(x) ln 3
FreeCalc Math 140
Lecture 16
)
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) = 5x + 1.
Let g(x) = log3 x.
Then
f (x) = g(h(x)).
Chain Rule: f 0 (x) = g 0 (h(x))h0 (x)
1
=
(5x ln 5)
h(x) ln 3
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule)
Differentiate f (x) = log3 (5x + 1).
Let h(x) = 5x + 1.
Let g(x) = log3 x.
Then
f (x) = g(h(x)).
Chain Rule: f 0 (x) = g 0 (h(x))h0 (x)
1
=
(5x ln 5)
h(x) ln 3
5x ln 5
= x
.
(5 + 1) ln 3
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
1
d
(loga x) =
.
dx
x ln a
ln x = loge x, so plug in a = e to find the derivative of ln x.
1
d
(ln x) =
dx
x ln e
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
1
d
(loga x) =
.
dx
x ln a
ln x = loge x, so plug in a = e to find the derivative of ln x.
1
d
(ln x) =
dx
x ln e
1
=
x( )
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
1
d
(loga x) =
.
dx
x ln a
ln x = loge x, so plug in a = e to find the derivative of ln x.
1
d
(ln x) =
dx
x ln e
1
=
x(1)
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
1
d
(loga x) =
.
dx
x ln a
ln x = loge x, so plug in a = e to find the derivative of ln x.
1
d
(ln x) =
dx
x ln e
1
=
x(1)
1
= .
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Theorem (The Derivative of loga x)
1
d
(loga x) =
.
dx
x ln a
ln x = loge x, so plug in a = e to find the derivative of ln x.
1
d
(ln x) =
dx
x ln e
1
=
x(1)
1
= .
x
Theorem (The Derivative of ln x)
d
1
(ln x) = .
dx
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
=
FreeCalc Math 140
+ ln sec x.
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
=
dx
FreeCalc Math 140
+
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
=1+
dx
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u =
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u = sec x.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u = sec x.
Then ln sec x = ln u.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u = sec x.
Then ln sec x = ln u.
dy
d
du
Chain Rule:
=1+
(ln u)
dx
du
dx
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u = sec x.
Then ln sec x = ln u.
dy
d
du
Chain Rule:
=1+
(ln u)
dx
du
dx
=1+
FreeCalc Math 140
Lecture 16
(
)
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u = sec x.
Then ln sec x = ln u.
dy
d
du
Chain Rule:
=1+
(ln u)
dx
du
dx
1
=1+
(
u
FreeCalc Math 140
Lecture 16
)
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u = sec x.
Then ln sec x = ln u.
dy
d
du
Chain Rule:
=1+
(ln u)
dx
du
dx
1
=1+
(
u
FreeCalc Math 140
Lecture 16
)
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u = sec x.
Then ln sec x = ln u.
dy
d
du
Chain Rule:
=1+
(ln u)
dx
du
dx
1
=1+
( sec x tan x)
u
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u = sec x.
Then ln sec x = ln u.
dy
d
du
Chain Rule:
=1+
(ln u)
dx
du
dx
1
=1+
( sec x tan x)
u
1
=1+
sec x tan x
sec x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example (Chain Rule, Natural Logarithm)
Differentiate y = ln(ex sec x).
y = ln ex + ln sec x
= x + ln sec x.
dy
d
=1+
(ln sec x)
dx
dx
Let u = sec x.
Then ln sec x = ln u.
dy
d
du
Chain Rule:
=1+
(ln u)
dx
du
dx
1
=1+
( sec x tan x)
u
1
=1+
sec x tan x
sec x
= 1 + tan x.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example
Find f 0 (x) if f (x) = ln |x|.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example
Find f 0 (x) if f (x) = ln |x|.
ln x
f (x) =
ln(−x)
FreeCalc Math 140
Lecture 16
if x > 0
.
if x < 0
October 21-25, 2013
Derivatives of Logarithmic Functions
Example
Find f 0 (x) if f (x) = ln |x|.
ln x
f (x) =
ln(−x)
f 0 (x) =
FreeCalc Math 140
Lecture 16
if x > 0
.
if x < 0
if x > 0
if x < 0
October 21-25, 2013
Derivatives of Logarithmic Functions
Example
Find f 0 (x) if f (x) = ln |x|.
ln x
f (x) =
ln(−x)
1
x
f 0 (x) =
FreeCalc Math 140
Lecture 16
if x > 0
.
if x < 0
if x > 0
if x < 0
October 21-25, 2013
Derivatives of Logarithmic Functions
Example
Find f 0 (x) if f (x) = ln |x|.
ln x
f (x) =
ln(−x)
1
x
f 0 (x) =
FreeCalc Math 140
Lecture 16
if x > 0
.
if x < 0
if x > 0
if x < 0
October 21-25, 2013
Derivatives of Logarithmic Functions
Example
Find f 0 (x) if f (x) = ln |x|.
ln x
if x > 0
f (x) =
.
ln(−x) if x < 0
1
if x > 0
x
f 0 (x) =
1
if x < 0
(−1)
−x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example
Find f 0 (x) if f (x) = ln |x|.
ln x
if x > 0
f (x) =
.
ln(−x) if x < 0
1
if x > 0
x
f 0 (x) =
1
if x < 0
(−1)
−x
1
if x > 0
x
=
1
if x < 0
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Example
Find f 0 (x) if f (x) = ln |x|.
ln x
if x > 0
f (x) =
.
ln(−x) if x < 0
1
if x > 0
x
f 0 (x) =
1
if x < 0
(−1)
−x
1
if x > 0
x
=
1
if x < 0
x
1
= if x 6= 0.
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
Differentiate y =
FreeCalc Math 140
(x − 1)5/3 sin3 x
√
.
ex + 1
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
ln y = ln
FreeCalc Math 140
(x − 1)5/3 sin3 x
√
ex + 1
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
ln y = ln
=
FreeCalc Math 140
+
Lecture 16
−
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
1 dy
=
+
−
y dx
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
1 dy
=
+
−
y dx
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
1 dy
5
1
=
+
−
y dx
3 x −1
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
1 dy
5
1
=
+
−
y dx
3 x −1
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
1 dy
5
1
3 cos x
=
+
−
y dx
3 x −1
sin x
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
1 dy
5
1
3 cos x
=
+
−
y dx
3 x −1
sin x
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
1 dy
5
1
3 cos x
1
ex
=
+
−
y dx
3 x −1
sin x
2 ex + 1
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
1 dy
5
1
3 cos x
1
ex
=
+
−
y dx
3 x −1
sin x
2 ex + 1
x
dy
5
e
=
+ 3 cot x −
y
x
dx
3(x − 1)
2(e + 1)
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Logarithmic Differentiation)
(x − 1)5/3 sin3 x
√
.
ex + 1
Take the natural logarithm of both sides:
Differentiate y =
(x − 1)5/3 sin3 x
√
ex + 1
ln y = (5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1) .
d
d
(ln y ) =
[(5/3) ln(x − 1) + 3 ln (sin x) − (1/2) ln (ex + 1)]
dx
dx
1 dy
5
1
3 cos x
1
ex
=
+
−
y dx
3 x −1
sin x
2 ex + 1
x
dy
5
e
=
+ 3 cot x −
y
x
dx
3(x − 1)
2(e + 1)
5
ex
(x − 1)5/3 sin3 x
√
=
+ 3 cot x −
.
3(x − 1)
2(ex + 1)
ex + 1
ln y = ln
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Steps in Logarithmic Differentiation
1
Take natural logarithms of both sides of an equation y = f (x).
2
Use the properties of logarithms to simplify.
3
Differentiate implicitly with respect to x.
4
Solve the resulting equation for y 0 .
Note: If f (x) < 0, then we use ln |f (x)| instead as ln f (x) is not defined.
We computed the derivative of ln |f (x)| in the previous lecture.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
=
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
1 0
y =
y
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
1 0
y =
y
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
1 0
d
d
y = (ln x)
(ln(3x + 1)) + (ln(3x + 1))
(ln x)
y
dx
dx
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
1 0
d
d
y = (ln x)
(ln(3x + 1)) + (ln(3x + 1))
(ln x)
y
dx
dx
1 0
y = (ln x)
+ (ln(3x + 1))
y
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
1 0
d
d
y = (ln x)
(ln(3x + 1)) + (ln(3x + 1))
(ln x)
y
dx
dx
1 0
1
y = (ln x)
· 3 + (ln(3x + 1))
y
3x + 1
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
1 0
d
d
y = (ln x)
(ln(3x + 1)) + (ln(3x + 1))
(ln x)
y
dx
dx
1 0
1
y = (ln x)
· 3 + (ln(3x + 1))
y
3x + 1
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
1 0
d
d
y = (ln x)
(ln(3x + 1)) + (ln(3x + 1))
(ln x)
y
dx
dx
1 0
1
1
y = (ln x)
· 3 + (ln(3x + 1))
y
3x + 1
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
1 0
d
d
y = (ln x)
(ln(3x + 1)) + (ln(3x + 1))
(ln x)
y
dx
dx
1 0
1
1
y = (ln x)
· 3 + (ln(3x + 1))
y
3x + 1
x
3 ln x
ln(3x + 1)
0
y =y
+
3x + 1
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = (3x + 1)ln x .
Take logarithms of both sides:
ln y = ln(3x + 1) ln x
ln y = ln x ln(3x + 1).
Differentiate implicitly with respect to x:
1 0
d
d
y = (ln x)
(ln(3x + 1)) + (ln(3x + 1))
(ln x)
y
dx
dx
1 0
1
1
y = (ln x)
· 3 + (ln(3x + 1))
y
3x + 1
x
3 ln x
ln(3x + 1)
0
y =y
+
3x + 1
x
3 ln x
ln(3x + 1)
ln x
= (3x + 1)
+
.
3x + 1
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
=
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
1 0
y =
y
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
1 0
y =
y
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
1 0
d
d
y = (tan x)
(ln x) + (ln x)
(tan x)
y
dx
dx
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
1 0
d
d
y = (tan x)
(ln x) + (ln x)
(tan x)
y
dx
dx
1 0
y = (tan x)
+ (ln x)
y
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
1 0
d
d
y = (tan x)
(ln x) + (ln x)
(tan x)
y
dx
dx
1 0
1
y = (tan x)
+ (ln x)
y
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
1 0
d
d
y = (tan x)
(ln x) + (ln x)
(tan x)
y
dx
dx
1 0
1
y = (tan x)
+ (ln x)
y
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
1 0
d
d
y = (tan x)
(ln x) + (ln x)
(tan x)
y
dx
dx
1 0
1
y = (tan x)
+ (ln x) sec2 x
y
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
1 0
d
d
y = (tan x)
(ln x) + (ln x)
(tan x)
y
dx
dx
1 0
1
y = (tan x)
+ (ln x) sec2 x
y
x
tan x
0
2
y =y
+ (ln x) sec x
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
Logarithmic Differentiation
Example (Variable base and exponent)
Differentiate y = x tan x .
Take logarithms of both sides:
ln y = ln x tan x
ln y = tan x ln x.
Differentiate implicitly with respect to x:
1 0
d
d
y = (tan x)
(ln x) + (ln x)
(tan x)
y
dx
dx
1 0
1
y = (tan x)
+ (ln x) sec2 x
y
x
tan x
0
2
y =y
+ (ln x) sec x
x
tan x
tan x
2
=x
+ (ln x) sec x .
x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
f (1 + h) − f (1)
1 = f 0 (1) = lim
h
h→0
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
f (1 + h) − f (1)
f (1 + x) − f (1)
1 = f 0 (1) = lim
= lim
h
x
h→0
x→0
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
f (1 + h) − f (1)
f (1 + x) − f (1)
1 = f 0 (1) = lim
= lim
h
x
h→0
x→0
ln(1 + x) − ln(1)
= lim
x
x→0
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
f (1 + h) − f (1)
f (1 + x) − f (1)
1 = f 0 (1) = lim
= lim
h
x
h→0
x→0
ln(1 + x) − ln(1)
1
= lim
= lim ln(1 + x)
x
x→0
x→0 x
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
f (1 + h) − f (1)
f (1 + x) − f (1)
1 = f 0 (1) = lim
= lim
h
x
h→0
x→0
ln(1 + x) − ln(1)
1
= lim
= lim ln(1 + x)
x
x→0
x→0 x
1/x
= lim ln(1 + x) .
x→0
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
f (1 + h) − f (1)
f (1 + x) − f (1)
1 = f 0 (1) = lim
= lim
h
x
h→0
x→0
ln(1 + x) − ln(1)
1
= lim
= lim ln(1 + x)
x
x→0
x→0 x
1/x
= lim ln(1 + x) .
x→0
Then use the fact that the exponential function is continuous:
e = e1 =
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
f (1 + h) − f (1)
f (1 + x) − f (1)
1 = f 0 (1) = lim
= lim
h
x
h→0
x→0
ln(1 + x) − ln(1)
1
= lim
= lim ln(1 + x)
x
x→0
x→0 x
1/x
= lim ln(1 + x) .
x→0
Then use the fact that the exponential function is continuous:
1/x
e = e1 = e limx→0 ln(1+x) =
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
f (1 + h) − f (1)
f (1 + x) − f (1)
1 = f 0 (1) = lim
= lim
h
x
h→0
x→0
ln(1 + x) − ln(1)
1
= lim
= lim ln(1 + x)
x
x→0
x→0 x
1/x
= lim ln(1 + x) .
x→0
Then use the fact that the exponential function is continuous:
1/x
1/x
e = e1 = e limx→0 ln(1+x) = lim eln(1+x) =
x→0
FreeCalc Math 140
Lecture 16
October 21-25, 2013
Derivatives of Logarithmic Functions
The Number e as a Limit
The Number e as a Limit
Theorem (The Number e as a Limit)
e = lim (1 + x)1/x .
x→0
Proof.
Let f (x) = ln x. Then f 0 (x) = 1/x, so f 0 (1) = 1.
f (1 + h) − f (1)
f (1 + x) − f (1)
1 = f 0 (1) = lim
= lim
h
x
h→0
x→0
ln(1 + x) − ln(1)
1
= lim
= lim ln(1 + x)
x
x→0
x→0 x
1/x
= lim ln(1 + x) .
x→0
Then use the fact that the exponential function is continuous:
1/x
1/x
e = e1 = e limx→0 ln(1+x) = lim eln(1+x) = lim (1 + x)1/x .
x→0
FreeCalc Math 140
Lecture 16
x→0
October 21-25, 2013