Partial Solutions for Section 3 - University of Colorado Boulder

Partial Solutions for Section 3.5 Michael Roy

See also the answers in the back of the book.

1. Use sum rule and constant multiple rule: f (x) = 4(- sin x)+2(cos x) = -4 sin x+ 2 cos x.

3. Product rule: f (x) = (-4x2)(- sin x) + (cos x)(-8x).

5.

Quotient rule:

f (x) =

(5+sin

x)(sin x)-(5-cos (5+sin x)2

x)(cos

x)

.

If you want, you can simplify

a bit by noting that sin2 x + cos2 x = 1, but it's not necessary.

6.

f

(x)

=

. (x2+sin x)(cos x)-(sin x)(2x+cos x) (x2+sin x)2

7. f (x) = sec x tan x - 2 sec2 x.

8. f (x) = (x2 + 1)((sec x tan x) + (sec x)(2x).

11. f (x) = (sec x)(sec2 x) + (tan x)(sec x tan x).

13.

f

(x)

=

. (1+csc x)(- csc2 x)-(cot x)(- csc x cot x) (1+csc x)2

If

you

want,

you

can

simplify

further

by

noting

that

cot2 x - csc2 x

=

-1.

Then

f

(x)

=

csc x 1+csc

x

.

21.

dy dx

=

[(x)(cos x) + (sin x)(1)] - 3(- sin x)

=

x cos x + 4 sin x.

Then

d2y dx2

=

[(x)(- sin x) + (cos x)(1)] + 4 cos x = -x sin x + 5 cos x.

25. Let f (x) = tan x, so f (x) = sec2 x. Then:

(a) f (0) = 0 and f (0) = 1, so the tangent line is y - 0 = 1(x - 0) or y = x.

(b)

f

(

4

)

=

1

and

f

(

4

)

=

2,

so

the

tangent

line

is

y

-1

=

2(x -

pi 4

),

or

y

=

2x

-

pi 2

+

1.

(c)

f

(

- 4

)

=

-1

and

f

(-

4

)

=

2,

so

the

tangent

line

is

y

+

1

=

2(x

+

pi 4

),

or

y

=

2x

+

pi 2

-

1.

27. We can show this by plugging in the appropriate derivatives and seeing that both sides are equal. Since we'll need the second derivative, let's start by computing them: y = x sin x, so y = x cos x + sin x, y = 2 cos x - x sin x. (It may seem like we'd actually need to compute the fourth derivative for part (b), and while you can do it this way, we can avoid this by a little trick.)

(a) Plug in: y + y = (2 cos x - x sin x) + (x sin x) = 2 cos x as desired.

1

(b) Instead of actually computing more derivatives of y, just take the second derivative of both sides of our answer in (a): using the sum rule twice, the second derivative of y + y is y(4) + y . Similarly, the second derivative of 2 cos x is -2 cos x. (If this isn't clear to you, you can also do this part by calculating y(4) directly using the techniques of part (a) and then plugging in.)

28. Proceed just as in #27: find the second derivative y and add it to y to verify that they equal 0.

(a) For y = cos x, y = - sin x and y = - cos x, so y + y = - cos x + cos x = 0. For y = sin x, y = cos x and y = - sin x, so again y + y = 0.

(b) Either proceed as in (a) by finding the second derivative (it will be -(A sin x+ B cos x)), or derive it from part (a) using the constant multiple and sum rules.

31. Note that here our independent variable is and our dependent variable is x,

whereas we're normally used to having x be our independent variable and y be

our dependent variable. The idea here is to find an equation relating x and

using trig.

sin =

x 10

is probably the simplest.

Solving for x (because it's the

independent variable) gives x = 10 sin . Taking a derivative with respect to

(NOT with respect to x) gives x

= 10 cos , so x

=60

=

10(

1 2

)

feet-per-radian.

Note that we end up with feet-per-radian, not feet-per-degree, since our rule

for the derivative of sin only works in radians. To convert to feet-per-degree,

we'd

have

to

multiply

by

the

scaling

factor

180

,

giving

about

0.087

ft/degree.

Annoying, isn't it? That's why radians are better than degrees.

35. Let's look at the first few derivatives and see if a pattern emerges:

(a)

d dx

[sin

x]

=

cos

x.

d2 dx2

[sin

x]

=

-

sin

x.

d3 dx3

[sin

x]

=

-

cos

x.

d4 dx4

[sin

x]

=

sin

x.

Hey... wait a second... isn't that what we started with? So, taking four

derivatives of sin x is the same as doing nothing. Then taking 87 derivatives

is the same as taking 87 = 4(21) + 3 derivatives, so the 4(21) derivatives

will just cycle through the above pattern 21 times and end up back where

we

started,

so

d87 dx87

[sin x]

=

d3 dx3

[sin x]

=

- cos x.

(b)

d dx

[cos

x]

=

-

sin

x.

d2 dx2

[cos

x]

=

-

cos

x.

2

d3 dx3

[cos

x]

=

sin

x.

d4 dx4

[cos

x]

=

cos

x.

Once again, four derivatives take us back where we started. Since 100 =

4(25), taking 100 derivatives will just cycle through this pattern 25 times

and

end

up

back

where

it

started,

so

d100 dx100

[cos x]

=

cos x.

EOF

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