Math 1132
HW 6.7
Fall 12
1. Differentiate f (x) = ln(ln(4x))
Solution:
Use the chain rule:
f 0 (x) =
Z
2. Evaluate
0
8
1
4
1
·
=
ln 4x 4x
x ln 4x
4x − 1
dx
x+1
Solution:
Let u = x + 1. Then x = u − 1 and du = dx. So we have
Z 9
Z 8
4(u − 1) − 1
4x − 1
dx =
du
x
+
1
u
1
0
Z 9
4u − 5
=
du
u
1
Z 9
Z 9
5
=
4du −
du
u
1
1
= 4(8) − [5 ln |u|]91 = 32 − 5 ln 9
1
Z
14x dx
3. Evaluate the integral
−1
Solution:
The anti-derivative for any bx , b > 0 is
Z
bx
bx dx =
+C
ln b
Since
bx = eln(b)x
So applying that to our equation, we get
Z
1
−1
September 10, 2012
1
14x
ln 14 −1
1
1
=
14 −
ln 14
14
14x dx =
Dave Miller
1
Math 1132
HW 6.7
Fall 12
4. Find the derivative of y = (6x)6x
Solution:
We are going to use implicit differentiation and properties of logs to evaluate this. First, take natural log of both sides:
ln y = 6x ln(6x)
Now take derivative of both sides, using implicit differentiation on the left,
and product rule on the right:
6x
y0
= 6 ln(6x) +
·6
y
6x
Now solve for y 0 :
y 0 = y(6 ln(6x) + 6)
Plug in y
y 0 = (6x)6x (6 ln(6x) + 6)
1
5. What is the average value of f (x) =
on the interval [1, p] for p > 1?
x
What is the average value of f as p → ∞?
Solution:
The average value of a function f (x) on a closed interval [a, b] is given by
1
b−a
Z
b
f (x)dx
a
Plugging our information into this equation gives
Z p
1
1
Av =
dx
p−1 1 x
1
=
ln(p)
p−1
Now, we want to find lim Av. We’re going to use L’Hospital’s Rule, since
p→∞
lim
p→∞
September 10, 2012
ln p
∞
=
p−1
∞
Dave Miller
2
Math 1132
HW 6.7
Fall 12
is in indeterminant form. L’Hospital’s Rule says if lim
x→a
minant form, then
f (x)
is in indeterg(x)
f 0 (x)
f (x)
= lim 0
x→a g (x)
x→a g(x)
lim
Using this on our function gives
lim
p→∞
September 10, 2012
ln p
1/p
= lim
=0
p − 1 p→∞ 1
Dave Miller
3