p. 185 #70.) Finding where the tangent line is horizontal is the same as finding
where the derivative is zero. If
f (x) = 3x5 − 32 x4
then
f ! (x) = 15x4 − 6x3
Setting this equal to zero gives
15x4 − 6x3 = 0
x3 (15x − 6) = 0
6
15x3 (x − 15
)=0
Dividing both sides by 15 gives
6
x3 (x − 15
) = 0 So x = 0 or x =
24
(x, f (x)) = (0, 0) and ( 25 , − 3125
).
6
15
=
2
5.
Thus, the coordinates are
p. 185 #74.) Note that for the line y = −x, the slope is -1. So finding
when the tangent line is parallel to this line is akin to finding where the slope of
the tangent line is -1. This is the same as finding where the derivative is -1. So if
f (x) = 4 − x2
then
f ! (x) = −2x
In this case, we want
f ! (x) = −2x = −1
This occurs when x =
nate is ( 12 , 15
4 ).
1
2.
For this x-value, f (x) =
15
4 .
p. 192 #2.) f (x) = (3x + 2x2 )(5x3 − 2)
Here, we will use the product rule. So
d
d
f ! (x) = (3x + 2x2 ) dx
(5x3 − 2) + (5x3 − 2) dx
(3x + 2x2 )
2
2
3
= (3x + 2x )(15x ) + (5x − 2)(3 + 4x)
= 45x3 + 30x4 + 15x3 − 6 + 20x4 − 8x
1
Thus, the coordi-
= 50x4 + 60x3 − 8x − 6.
p. 193 #10.) f (x) = (4 − 2x2 )2
We can use the chain rule:
d
f ! (x) = 2(4 − 2x2 ) dx
(4 − 2x2 )
!
2
f (x) = 2(4 − 2x )(−4x)
f ! (x) = −32x + 16x3 .
p. 193 #34.) h(t) =
√
a(t − a) + a
First, let us multiply out the parenthetical expression:
h(t) =
√
√
at − a a + a.
Now, since this is a sum, we can take the derivative (with respect to t) of
each term individually. Note that the a’s are constants. So
√
h! (t) = √a − 0 + 0
h! (t) = a.
p. 193 #35.) To find (f g)! (2), note first that this notation means
d
dx (f (2)g(2)).
Written this way, it is clear that we use the product rule. Then
d
dx (f (2)g(2))
= f ! (2)g(2) + f (2)g ! (2).
Plugging in the values given in the problem, we see that
f ! (2)g(2) + f (2)g ! (2) = (1)(3) + (−4)(−2) = 3 + 8 = 11.
Thus
(f g)! (2) = 11.
p. 193 #44.) First, let us multiply out the bracketed expression. So
y = [−2f (x) − 3g(x)]g(x) +
2g(x)
3
2
y = −2f (x)g(x) − 3g(x)2 +
2g(x)
3 .
So
d
y ! = − dx
(2f (x)g(x)) −
d
2
dx (3g(x) )
+
d 2g(x)
dx ( 3 ).
Let us examine each term individually. For the first term, we can use the
product rule:
d
dx (2f (x)g(x))
d
= 2 dx
(f (x)g(x)) = 2(f (x)g ! (x) + f ! (x)g(x)).
The second term requires the chain rule:
d
2
dx (3g(x) )
d
= 3 dx
(g(x)2 ) = 3(2g(x))(g ! (x)).
The third term can be evaluated directly
d 2g(x)
dx ( 3 )
d
= ( 23 ) dx
(g(x)) = ( 23 )g ! (x).
Plugging all of this information in gives
y ! = −2(f (x)g ! (x) + f ! (x)g(x)) − 3(2g(x))(g ! (x)) + ( 23 )g ! (x)
or
y ! = −2f (x)g ! (x) − 2f ! (x)g(x) − 6g(x)g ! (x) +
2g ! (x)
3 .
p. 193 #52.) In this case, we can use the quotient rule. Since
3x3 +2x−1
5x2 −2x+1 ,
f (x) =
the rule tells us that
f ! (x) =
d
d
(5x2 −2x+1) dx
(3x3 +2x−1)+(3x3 +2x−1) dx
(5x2 −2x+1)
.
(5x2 −2x+1)2
Now, we evaluate each part of this derivative separately.
d
3
dx (3x
d
2
dx (5x
+ 2x − 1) = 9x2 + 2
− 2x + 1) = 10x − 2.
Plugging this in gives
f ! (x) =
(5x2 −2x+1)(9x2 +2)+(3x3 +2x−1)(10x−2)
.
(5x2 −2x+1)2
3
p. 193 #68.) Again, we use the product rule. Since
g(s) =
1
2
3
s7
4
+s 7
s 7 −s 7
,
the rule states that
g ! (s) =
3
4
1
2
1
2
3
4
d
d
(s 7 −s 7 )+(s 7 −s 7 ) ds
(s 7 +s 7 )
(s 7 +s 7 ) ds
3
(s 7
4
+s 7 )2
.
We evaluate each part individually.
1
2
5
d
1 − 67
7
7
− 27 s− 7
ds (s − s ) = 7 s
3
4
4
3
d
3 −7
7
7
+ 47 s− 7 .
ds (s + s ) = 7 s
So
g ! (s) =
3
4
6
5
1
2
4
3
(s 7 +s 7 )( 17 s− 7 − 27 s− 7 )+(s 7 −s 7 )( 37 s− 7 + 47 s− 7 )
3
(s 7
4
+s 7 )2
.
p. 193 #88.) Once more, we use the product rule. Since
f (x) =
2f (x)+1
3g(x)
it follows that
f ! (x) =
d
d
g(x) dx
(2f (x)+1)+(2f (x)+1) dx
(3g(x))
.
g(x)2
Note that
d
!
dx (2f (x) + 1) = 2f (x),
d
!
dx (3g(x)) = 3g (x).
So
f ! (x) =
g(x)(2f ! (x))+(2f (x)+1)(3g ! (x))
.
9g(x)2
4