HW 7
17 – 1
Find the derivatives of the following functions:
(a) f (x) = 4x5 − 8x3 + x2 − x + 7
√
4
(b) f (t) = 3 t + √
3
(c) f (x) =
t
x4 − 2x +
x
√
x−3
(Hint: First rewrite.)
Solution:
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(a) f 0 (x) = 20x4 − 24x2 + 2x − 1.
(b) First rewrite to get f (t) = 3t1/2 + 4t−1/3 . Then
f 0 (t) = 32 t−1/2 − 43 t−4/3
3
4
= √ − √
.
3
2 t 3( t)4
2
f (x) = 3x −
1 −3/2
2x
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(c) First rewrite to get f (x) = x3 − 2 + x−1/2 − 3x−1 . Then
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−2
+ 3x
1
3
= 3x2 − √ 3 + 2 .
x
2( x)
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17 – 2
Verify the sum rule (see 17.2).
Hint: Let s(x) = f (x) + g(x). The sum rule says that s0 (x) = f 0 (x) + g 0 (x). Use the
definition of the derivative (as applied to the function s) to show that this equation holds.
(It might help to study the verification of the constant multiple rule.)
Solution:
Following the hint, we have
s(x + h) − s(x)
h
(f (x + h) + g(x + h)) − (f (x) + g(x))
= lim
h→0
h
f (x + h) − f (x) g(x + h) − g(x)
+
= lim
h→0
h
h
f (x + h) − f (x)
g(x + h) − g(x)
= lim
+ lim
h→0
h→0
h
h
= f 0 (x) + g 0 (x),
s0 (x) = lim
h→0
as claimed.
18 – 1
Find the derivatives of the following functions:
(a) f (x) = 4ex − 3x + 5x2
√
(b) f (t) = 7t − 6 3 t + 8et
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Solution:
(a) f 0 (x) = 4ex − 3x ln 3 + 10x.
(b) Mentally rewriting
√
3
t as t1/3 in order to use the power rule, we have
2
f 0 (t) = 7t ln 7 − √
+ 8et .
( 3 t)2
19 – 1
Find all points on the graph of f (x) = cos x at which the tangent line has slope
√
3/2.
0
Solution: The general slope function for this function is its derivative, which
√ is f (x) =
0
− sin x. We
√ points by solving f (x) = 3/2, that is,
√ get the x-coordinates of the desired
− sin x = 3/2, or in other words sin x = − 3/2. Picturing the unit circle and using
the 30-60-90
triangle we find that x = −π/3 (−60◦ ) in the fourth quadrant yields a sine
√
of − 3/2, as does x = −2π/3 in the third quadrant. A multiple of 2π added to these
angles produces the same sine. This gives x = −π/3 + 2πn, −2π/3 + 2πn. Finally, the
corresponding y-coordinates are obtained by evaluating f at these x values. The desired
points are (−π/3 + 2πn, 1/2), (−2π/3 + 2πn, −1/2) (n, integer).
20 – 1
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Let f (x) = (x2 − 3x + 1)(x4 + 9x2 ).
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(a) Find the derivative of f by first expanding the right-hand side so as to avoid using
the product rule.
(b) Find the derivative of f by using the product rule and verify that the result is the
same as that obtained in part (a).
Solution:
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(a) Expanding first, we have
f (x) = x6 + 9x4 − 3x5 − 27x3 + x4 + 9x2
= x6 − 3x5 + 10x4 − 27x3 + 9x2 ,
so
f 0 (x) = 6x5 − 15x4 + 40x3 − 81x2 + 18x.
(b) Using the product rule, we have
d 2
[(x − 3x + 1)(x4 + 9x2 ) ]
dx
d 4
d 2
x − 3x + 1 (x4 + 9x2 ) + (x2 − 3x + 1)
x + 9x2 ]
=
dx
dx
= (2x − 3)(x4 + 9x2 ) + (x2 − 3x + 1)(4x3 + 18x)
f 0 (x) =
5
3
4
2
5
3
4
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2
= 2x + 18x − 3x − 27x + 4x + 18x − 12x − 54x
+ 4x3 + 18x
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= 6x5 − 15x4 + 40x3 − 81x2 + 18x,
in agreement with part (a)
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20 – 2
Find the derivatives of the following functions:
(a) f (x) = (2x5 − 3x2 + 1) sin x
√
5
(b) f (t) = 3 tet −
t
Solution:
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(a) Using the product rule, we have
d 5
(2x − 3x2 + 1) sin x
dx
d 5
d
=
2x − 3x2 + 1 sin x + (2x5 − 3x2 + 1) [sin x]
dx
dx
= (10x4 − 6x) sin x + (2x5 − 3x2 + 1) cos x.
f 0 (x) =
(b) Using the sum rule and constant multiple rule first and then the product rule, we
have
d √ t 5
f 0 (t) =
3 te −
dt
t
h
i
√
d 5
d
tet −
=3
dt
dt t
h i
√
√ d t
d
d −1 =3
t et + t
e
−
5t
dt
dt
dt
√
= 3( 12 t−1/2 et + tet ) − (−5t−2 )
√
5
3et
= √ + 3 tet + 2 .
t
2 t
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20 – 3
Verify the formula
d
[f (x)g(x)h(x)] =
dx
d
d
d
[f (x)]g(x)h(x) + f (x) [g(x)] h(x) + f (x)g(x) [h(x)] .
dx
dx
dx
Hint: Apply the product rule viewing f (x)g(x)h(x) as a product with the two factors f (x)
and g(x)h(x).
Solution: Following the hint we have
d
d
d
[f (x)g(x)h(x)] =
[f (x)] g(x)h(x) + f (x) [g(x)h(x)] .
dx
dx
dx
(1)
Working with the second term on the right, we use the product rule on the product g(x)h(x):
d
d
d
f (x) [g(x)h(x)] = f (x)
[g(x)] h(x) + g(x) [h(x)]
dx
dx
dx
d
d
= f (x) [g(x)] h(x) + f (x)g(x) [h(x)] .
dx
dx
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The second term on the right in (1) can now be replaced by this last expression:
d
[f (x)g(x)h(x)] =
dx
d
d
d
[f (x)]g(x)h(x) + f (x) [g(x)] h(x) + f (x)g(x) [h(x)] .
dx
dx
dx
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20 – 4
Find the derivative of f (x) = (2x − 5 +
√
3
x)8x cos x.
Hint: Use the formula in Exercise 20 – 3.
Solution:
Following the hint, we have
√
d (2x − 5 + 3 x)8x cos x
dx
√ √ d
d =
2x − 5 + 3 x 8x cos x + (2x − 5 + 3 x) [8x ] cos x
dx
dx
√
d
+ (2x − 5 + 3 x)8x
[cos x]
dx
√
= (2 + 13 x−2/3 )8x cos x + (2x − 5 + 3 x)(8x ln 8) cos x
√
+ (2x − 5 + 3 x)8x (− sin x).
f 0 (x) =
20 – 5
√
x4 + 3 x − 5
Let f (x) =
.
x
(a) Find the derivative of f by first rewriting the right-hand side so as to avoid using the
quotient rule.
(b) Find the derivative of f by using the quotient rule and verify that the result is the
same as that obtained in part (a).
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Solution:
(a) Rewriting first, we have
f (x) = x3 + 3x−1/2 − 5x−1 ,
so
f 0 (x) = 3x2 − 32 x−3/2 + 5x−2 .
(b) Using the quotient rule, we have
√
d x4 + 3 x − 5
0
f (x) =
dx
x
√
√
d
d 4
x
x + 3 x − 5 − (x4 + 3 x − 5) [x]
dx
dx
=
x2
√
3 −1/2
3
) − (x4 + 3 x − 5)(1)
x(4x + 2 x
=
x2
3 1/2
4
3x − 2 x + 5
=
x2
2
3 −3/2
+ 5x−2 ,
= 3x − 2 x
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in agreement with part (a).
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20 – 6
Find the derivatives of the following functions:
(a) f (x) =
3x2
5x − 7
(b) f (t) =
4 cos t − 1
2 + 3et
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Solution:
(a) Using the quotient rule, we have
d
3x2
0
f (x) =
dx 5x − 7
d
d 2
3x − 3x2
[5x − 7]
(5x − 7)
dx
dx
=
(5x − 7)2
(5x − 7)(6x) − 3x2 (5)
=
.
(5x − 7)2
(b) Using the quotient rule, we have
d 4 cos t − 1
0
f (t) =
dt
2 + 3et
d
d
(2 + 3et ) [4 cos t − 1] − (4 cos t − 1) [2 + 3et ]
dt
dt
=
(2 + 3et )2
(2 + 3et )(−4 sin t) − (4 cos t − 1)(3et )
=
.
(2 + 3et )2
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