HW 10
25 – 1
Verify the rule
d
[sec x] = sec x tan x.
dx
Solution:
d d
[sec x] =
(cos x)−1
dx
dx
d
[cos x]
dx
= −(cos x)−2 · (− sin x)
1
sin x
=
·
cos x cos x
= sec x tan x.
= (−1)(cos x)−2 ·
25 – 2
Find the derivative of each of the following functions:
√
(a) f (x) = 5 4x − sec 3x.
(b) f (t) = tan−1 e5t .
Solution:
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(a)
f 0 (x) =
d √
5
4x − sec 3x
dx
d
[4x − sec 3x]
dx
= 51 (4x − sec 3x)−4/5 · (4 − sec 3x tan 3x(3))
4 − 3 sec 3x tan 3x
√
=
.
5( 5 4x − sec 3x)4
= 15 (4x − sec 3x)−4/5 ·
(b) The expression reads “the inverse tangent of e5t ,” so the chain rule, with outside
function the inverse tangent function, is required:
d −1 5t tan e
dt
1
d 5t =
·
e
2
1 + (e5t ) dt
5e5t
=
.
1 + e10t
f 0 (t) =
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25 – 3
Find the derivative of each of the following functions:
(a) y = 2csc 7x tan(ln 4x).
3t e
.
(b) y = sec−1
1 + t2
Solution:
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(a)
d csc 7x
2
tan(ln 4x)
dx
d
d csc 7x 2
tan(ln 4x) + 2csc 7x
[tan(ln 4x)]
=
dx
dx
d
d
= 2csc 7x ln 2 [csc 7x] tan(ln 4x) + 2csc 7x sec2 (ln 4x) [ln 4x]
dx
dx
y0 =
= 2csc 7x ln 2 (− csc 7x cot 7x(7)) tan(ln 4x) + 2csc 7x sec2 (ln 4x)
1
4x
(4).
(b)
3t d
e
sec−1
dt
1 + t2
3t e
1
d
s
=
·
2
dt 1 + t2
e3t
e3t
−1
1 + t2
1 + t2
d 3t d (1 + t2 )
e − e3t
1 + t2
1
dt
dt
s
=
·
2
(1 + t2 )2
e3t
e3t
−1
1 + t2
1 + t2
y0 =
1
=
e3t
1 + t2
s
e3t
1 + t2
·
2
−1
(1 + t2 )e3t (3) − e3t (2t)
.
(1 + t2 )2
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25 – 4
x
Differentiate f (x) = tane x.
x
Solution: The expression on the right means (tan x)e . Neither base nor exponent is
constant, so logarithmic differentiation is required. For consistency with the notation used
when that method was introduced, we write y for f (x):
y = (tan x)e
x
ln y = ln(tan x)e
x
= ex ln tan x
Using implicit differentiation, we get
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d x
d
[ln y] =
[e ln tan x]
dx
dx
1 0
d x
d
·y =
[e ] ln tan x + ex
[ln tan x]
y
dx
dx
1 0
1
· y = ex ln tan x + ex
sec2 x
y
tan x
0
x
x 1
2
y = y e ln tan x + e
sec x .
tan x
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Therefore,
0
0
ex
f (x) = y = tan
1
2
x e ln tan x + e
sec x .
tan x
x
x
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25 – 5
Find the derivative of the following function without using intermediate steps:
esec 3x
6
f (x) = sin
.
4x3 − 2x + 5
Solution:
esec 3x
esec 3x
f (x) = 6 sin
· cos
4x3 − 2x + 5
4x3 − 2x + 5
3
sec 3x
(4x − 2x + 5)e
sec 3x tan 3x(3) − esec 3x (12x2 − 2)
×
.
(4x3 − 2x + 5)2
0
5
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26 – 1
Use a linearization to approximate
√
4
16.16.
√
Solution: With f defined by f (x) = 4 x the goal is to approximate f (16.16). We need
to choose a. We want it to be close to 16.16 so that the approximation will be good, but
we also want it to be a number at which both f and f 0 are easily evaluated. We choose
a = 16. We have f 0 (x) = 14 x−3/4 , so
L(x) = f (a) + f 0 (a)(x − a) = f (16) + f 0 (16)(x − 16) = 2 +
1
32 (x
− 16).
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Therefore,
√
4
1
16.16 = f (16.16) ≈ L(16.16) = 2 + 32
(16.16 − 16) = 2.005.
√
(The actual value is 4 16.1 = 2.00498 to five decimal places.)
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26 – 2
Let f (x) = 1 + 1/x. Find dy and ∆y using x = 1 and dx = 1/2 (= ∆x). Sketch the graph
of f and label the distances corresponding to the computed quantities.
Solution:
First, f 0 (x) =
and dx = 1/2, we have
d
d [1 + 1/x] =
1 + x−1 = −x−2 . Therefore, when x = 1
dx
dx
dy = f 0 (x)dx = f 0 (1)dx = (−1)( 12 ) = − 21 .
Next, when x = 1 and ∆x = 1/2, we have
∆y = f (x + ∆x) − f (x) = f (1 + 12 ) − f (1) =
Here is the sketch:
5
3
− 2 = − 31 .
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