Download General Log and Exponential Functions

Updated: January 16, 2016
Calculus II
6.4*
Math 230
Calculus II
Brian Veitch •
Fall 2015 •
Northern Illinois University
General Logarithmic and Exponential Functions
Up to this point, we’ve only dealt with the exponential function ex and the log function ln x.
These two functions have base e. So what about all those exponential and log functions with
a different base? What’s their derivative and integral?
Rule 1
ax = ex ln a
If x and y are real numbers and a, b > 0, then
1. ax+y = ax ay
2. ax−y =
ax
ay
3. (ax )y = axy
4. (ab)x = ax bx
Definition 1: Derivative for ax
d x
(a ) = ax ln a
dx
1
Updated: January 16, 2016
Calculus II
6.4*
We will use the definition from the beginning of this section.
d x
d x ln a
a =
e
dx
dx
d
= ex ln a (x ln a)
dx
x ln a
= e
ln a
= ax ln a
The biggest problem students face with this section is where does ln a go. In fact, I
sometimes find it easier to change ax to ex ln a .
Example 1
Differentiate the following
1. y = 5x (1 + 5 ln x)
2. y = 2x · 51/x
3. Use logarithmic differentiation to find y 0 when y = x
√
x
1. y = 5x (1 + 5 ln x)
d
d
(1 + 5 ln x) + (1 + 5 ln x) · 5x
dx
dx
5
x
x
= 5 (1 + ) + (1 + 5 ln x) · 5 · ln 5
x
y 0 = 5x ·
2. y = 2x · 51/x
d 1/x
d
(5 ) + 51/x · (2x)
dx
dx
1
= 2x · 51/x ln 5 · − 2 + 51/x · 2
x
1/x
2 ln 5 · 5
=
+ 2 · 51/x
x
y 0 = 2x ·
2
Updated: January 16, 2016
Calculus II
6.4*
3. Use logarithmic differentiation to find y 0 when y = x
√
x
(a) Take ln of both sides.
√
ln y = ln x
ln y =
√
x
x ln x
(b) Differentiate both sides with respect to x
√ 1
1 0
1
x · + ln(x) · √
·y =
y
x
2 x
1
ln x
y0 = y √ + √
x 2 x
√
2 + ln x
x
0
√
y = x
2 x
√
(c) OR rewrite function as y = e
x ln x
d √
x ln x
dx
√
√ 1
1
x ln x
√ ln x + x ·
= e
·
x
2 x
√
ln x
1
√ +√
= e x ln x ·
2 x
x
√
2 + ln x
√
= x x
2 x
√
y0 = e
y0
Z
Definition 2: Evaluating
and do the chain rule
x ln x
·
ax dx
Z
ax dx =
Example 2
Z 2
Find
5x dx
0
3
ax
+C
ln a
Updated: January 16, 2016
Calculus II
Z
0
2
6.4*
2
5x 5 dx =
ln 5 0
50
52
−
=
ln 5 ln 5
24
=
ln 5
x
Example 3
Z
Find 8tan x · sec2 (x) dx
1. Let u = tan(x)
2. du = sec2 (x) dx
3. Substitute
Z
tan(x)
8
Z
2
· sec (x) dx =
8u du
8u
+C
ln 8
8tan(x)
=
+C
ln 8
=
Definition 3: Derivative of loga x
loga x =
ln x
ln a
d
1
(loga x) =
dx
x ln a
d
1
du
(loga u) =
·
dx
u ln a dx
So if you were asked to find log8 5, you evaluate it as
4
ln 5
= 0.773976 on your calculator.
ln 8
Updated: January 16, 2016
Calculus II
6.4*
Example 4
Differentiate the following
1. y = log5 (xex )
1. y = log5 (xex )
There are two ways of doing this problem. Recall that when you’re taking the derivative
of a log, check to see if you can break it up into smaller logs.
(a) Straightforward Derivative
d
1
· (xex )
ln 5 dx
1
· (xex + ex )
xex ln 5
1
· ex (x + 1)
x
xe ln 5
ex (x + 1)
xex ln 5
x+1
x ln 5
y0 =
xex
=
=
=
=
(b) Or, rewrite log5 (xex ) as y =
ln x + ln ex
1
ln(xex )
=
=
(ln x + x)
ln 5
ln 5
ln 5
1
y =
ln 5
0
y0 =
Wasn’t that so much easier??
5
1
+1
x
1 x+1
·
ln 5
x
Updated: January 16, 2016
Calculus II
6.4*
Example 5
Let’s do some more integrals.
Z
log10 (x + 1)
1.
dx
x+1
Z
3x
2.
dx
3x + 1
Z
1.
log10 (x + 1)
dx
x+1
(a) Let u = log10 (x + 1)
1
dx
(b) du =
dx → ln(10) du =
(x + 1) ln 10
x+1
(c) Substitute
Z
log10 x + 1
dx =
x+1
Z
u · ln(10) du
1 2
u ln(10) + C
2
ln 10
=
· (log10 (x + 1))2 + C
2
=
Z
2.
3x
dx
3x + 1
(a) Let u = 3x + 1
(b) du = 3x · ln 3 dx →
1
du = 3x dx
ln 3
(c) Substitute
Z
3x
dx =
3x + 1
Z
1 1
·
du
u ln 3
1
· ln u + C
ln 3
1
=
· ln(3x + 1) + C
ln 3
=
6