Solutions to homework from chapter 4.4
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Some theory
In this chapter, the following two formulas can be useful:
d x
a = ax ln a,
dx
d
1
1
loga x = ·
.
dx
x ln a
However, you don’t need to memorize them because they are easy to derive.
Note that a = eln a and ax = (eln a )x = ex ln a . Thus, using the chain rule,
d x
d x ln a
d
a =
e
= ex ln a ·
(x ln a) = ex ln a · ln a = ax ln a.
dx
dx
dx
To prove the second one, use the base change formula loga x =
ln x
ln a :
d
d ln x
1
d
1
1
loga x =
=
·
ln x =
· .
dx
dx ln a
ln a dx
ln a x
We can also write the versions of these formulas for the chain rule:
d f (x)
a
= af (x) · ln a · f � (x),
dx
d
1
1
loga f (x) =
·
· f � (x).
dx
f (x) ln a
These will be used in the last three exercises.
Solutions
2.
By the chain rule,
d −0.1x
d
e
= e−0.1x ·
(−0.1x).
dx
dx
Since
d
dx (−0.1x)
= −0.1,
d −0.1x
e
= e−0.1x · (−0.1) = −0.1 · e−0.1x .
dx
8.
By the chain rule,
√
√
√
2
d √
d √
d 1
1
3
3
3
3
e x=e x·
( 3 x) = e x ·
(x 3 ) = e x · · x− 3 .
dx
dx
dx
3
1
15.
By the product rule,
d 2 x2
(x e ) =
dx
We know that
chain rule:
d 2
dx x
d 2
x
dx
�
·e
x2
2
+x ·
�
d x2
e
dx
�
.
2
= 2x. To calculate the derivative of ex , we need to use the
2
2
d x2
d 2
e = ex ·
(x ) = ex · 2x.
dx
dx
Finally,
22.
�
2
2
d 2 x2
(x e ) = 2x · ex + x2 · ex · 2x.
dx
By the quotient rule,
d
d
( dx
x) · (1 + e−x ) − x · dx
(1 + e−x )
d
x
=
.
−x
−x
2
dx 1 + e
(1 + e )
We know that
d
dx x
= 1 and, using the chain rule,
d
d −x
d
(1 + e−x ) =
e = e−x ·
(−x) = e−x · (−1) = −e−x .
dx
dx
dx
Finally,
d
x
1 · (1 + e−x ) − x · (−e−x )
1 + e−x + x · e−x
=
=
.
dx 1 + e−x
(1 + e−x )2
(1 + e−x )2
32. By the rule for exponents, 3x · 5x = 15x . Since 15 = eln 15 , we can write
15x = (eln 15 )x = ex ln 15 . So,
d x x
d x ln 15
d
(3 · 5 ) =
(e
) = ex ln 15 ·
(x · ln 15) = ex ln 15 · ln 15 = 15x ln 15.
dx
dx
dx
d x
Of course, we could use formula dx
a = ax ln a with a = 15 to get the answer
x
instantly. If you didn’t note that 3 · 5x = 15x , the product rule could work for
you as well:
�
�
�
�
d x x
d x
d x
x
x
(3 · 5 ) =
3 ·5 +3 ·
5
dx
dx
dx
= (3x ln 3) · 5x + 3x (5x ln 5).
This answer is already fine, but after some simplification we can get
(3x ln 3) · 5x + 3x (5x ln 5)
=
=
=
=
2
3x · 5x · ln 3 + 3x · 5x · ln 5
3x · 5x · (ln 3 + ln 5)
15x · (ln 3 + ln 5)
15x · ln 15.
34.
Using the chain rule,
d
ln(x4 + x2 + 3)
dx
=
=
=
42.
1
d 4
·
(x + x2 + 3)
x4 + x2 + 3 dx
1
· (4x3 + 2x)
4
x + x2 + 3
4x3 + 2x
.
4
x + x2 + 3
First, we need the quotient rule:
d
( dx
ln(x2 )) · (ex + e−x ) − ln(x2 ) ·
d ln(x2 )
=
dx ex + e−x
(ex + e−x )2
d
x
dx (e
+ e−x )
.
To get the answer, we need derivatives of ln(x2 ) and (ex + e−x ). To find the
derivative of ln(x2 ), we can use two approaches. We could use the chain rule
right away:
1
d 2
1
1
2
d
ln(x2 ) = 2 ·
x = 2 · 2x = · 2 = .
dx
x dx
x
x
x
Or, alternatively, we can note that ln(x2 ) = 2 ln x, and therefore
d
d
d
1
2
ln(x2 ) =
2 ln x = 2 ln x = 2 · = .
dx
dx
dx
x
x
Now, after calculating
d
x
dx (e
d ln(x2 )
=
dx ex + e−x
50.
+ e−x ) = ex − e−x , we can get the final answer:
2
x
· (ex + e−x ) − ln(x2 ) · (ex − e−x )
.
(ex + e−x )2
As in #42, we can take two approaches.
1
d 7
1
1
7
d
ln(x7 ) = 7 ·
(x ) = 7 · 7x6 = 7 · = .
dx
x dx
x
x
x
Or,
52.
d
d
1
7
ln(x7 ) =
7 ln x = 7 · = .
dx
dx
x
x
Use the product rule:
dy
dx
=
=
=
To find where
dy
dx
�
�
�
�
d
d x
x · ex + x ·
e
dx
dx
ex + x · ex
ex · (1 + x).
= 0, we need to solve for x the following equation:
ex · (1 + x) = 0.
3
The product is zero when at least one of the factors is zero. Note that ex > 0
for any value of x, therefore
56.
1+x
=
0,
x
=
−1.
Using the product rule,
dy
= 4x3 ex + x4 ex .
dx
To find where
dy
dx
= 0, we need to solve for x the following equation:
4x3 ex + x4 ex = 0.
To do it, we need to factor:
x3 · ex · (4 + x) = 0.
Noting that ex is never zero, the last equation is equivalent to
x3 = 0 or (4 + x) = 0.
So, the derivative is zero at x = 0 and x = −4.
62.
Using the chain rule,
dy
d
=
(ln x)3
dx
dx
d
ln x
dx
1
3 · (ln x)2 · .
x
3 · (ln x)2 ·
=
=
To find where
dy
dx
= 0, solve for x:
3 · (ln x)2 ·
Note that
1
x
1
= 0.
x
is never zero, therefore the last equation is equivalent to
(ln x)2 = 0,
or
ln x
=
0,
loge x
=
0,
x
=
e0 ,
x
=
1.
4
71.
Using the chain rule,
f � (x)
=
=
Derivative of log3 (5x − 1).
1
1
d 3x
·
·
(e + x2 )
2
+ x ln 2 dx
1
1
·
· (3e3x + 2x).
3x
2
e + x ln 2
e3x
Using the chain rule,
d
1
1
d
1
1
log3 (5x − 1) =
·
·
(5x − 1) =
·
· 5.
dx
5x − 1 ln 3 dx
5x − 1 ln 3
Derivative of 5x · log5 x.
Using the product rule,
�
�
d x
d x
d
5 · log5 x =
5 · log5 x + 5x ·
log5 x
dx
dx
dx
1 1
= 5x · ln 5 · log5 x + 5x · ·
.
x ln 5
5