42. R 0(p) = p · f (p) + f (p)
=sin
9000
600. Increasing the price by a
2· 1 = f (p)
2 + pf (p), so R (120)
2+
2 + 120 · ( 80) =
(x3.
x)=
4. f 0(x) = 4 sin x cos x (x + sin x)
f 0 (x)
4x2psin
x 8x cos x
4.
f
(x)
=
4
sin
x
cos
x
small
amount
p
dollars
would
decrease
the
revenue
by
about
600
dollars.
Exercise Set 2.5
Exercise Set
p
10 2.5
0 = sin x(5
2x(5
2.7.f 0f(x)
+
cos
x
+
sin
x)
cos
cos
x)
1
+
5(sin
x cos
x)
= sec
x
tan
x
2
sec
x
0 (x)
3 x(5 + sin x)n cos x(5
4.x f 00 (x)
= 4 sin x cos x
5.f 0f(x)
= 1 + 5(sin
sin
cosn x)1)
cos =
x)
· (0)2 1 · (nx
n 1.
0(x)== 1x
n 1 4 sin x + 2 cos x
(x)
1.5.
cos+xxsin
x)
(5 + sin
x)f2 nx
(x)== n4 sin
=
43.
ff(x)
so fx0+
(x)2(5
=
=
=
.
2
2
2n
n+1
sin
x)
+sec
sinx x)
x 2 2 + 1) sec(5x +
x x)(2x) = (x2 +
x(5
0f 0 (x) = (x
8.
tan
x
+
(sec
1)
tan
x
+
2x
sin sec
x(5x+ sin x) cos x(5 cos x)
1 + 5(sin x cos x)
3. f (x) = 4x102sin x 8x cos x
5. 2x
f 00(x)
(x)
= 10 + cos x
=
(x odds
+
sin
x)
cos x sin x(2x + cos x)
x2 cos x2.
sin x
0 1=
f 0f(x)
+
cos
x
2
p2.6.
116/
–
25
f
=
2
2
(x)
=
=
(5
+
sin
x)
(5 + sin x)2
3
3
+2.5
sin
x)(xx
cos
xcsc
x(x2cos
xsin x)
2x
2sin
x(x4 csc
2+
2x(2x + cos x)
2 sin xx
Exercise
Set
00 (x) =
+
sin
x)
+
9.
f
x
cot
x
0
(x)==4 sin x cos x
=
4.6.ff(x)
(x2 + sin x)2
(x2 + sin x)2
00
2
p
(x22 sin
+ sin
cos
x x sin x(2x + cos x)
x2 cos x 2x sin x
3.
f
(x)
=
4x
sin
x
8x
cos
x
3.
ff 00(x)
=
4x
x x)8x
cos
1.7.ff(x)
=
4
+
2
cos
x
00
2
secx(5
x tan
sec
x
6.
(x)
=
=
10. f (x)
(x)==sin
sin
x+xsin
cscx)p
x2 +
x
csc
x
cot
x
cos x(5 cos x)
1 + 5(sin x cos x)
(x2 + sin x)2
(x2 + sin x)2
0
5.7.ff0 (x)
=
(x)== sec x tan x
2 sec22x
2
0
10
(5
+
sin
x)
(5
+
sin
x)
4.8.f 0f0(x)
=
4
sin
x
cos
x
00
2
f (x)
sinxx cos x p
2
2xx tan x + (sec x)(2x) = (x2 + 1)
3 sec 4.
(x)
++1)cos
sec
x tan
x +=
2x4sec
2.
11.f f(x)
(x)===(x
sec
3 x(sec x) + (tan x)(sec x tan x) = sec x + sec x0 tan x
x
0
2
2
7.
f
(x)
=
sec
x tan
2 sec2 x
8. f (x) = (x2 + 1) sec x tan x + (sec x)(2x) = (x +2 1) sec x tan x + 2x sec
xx
sin
x(5
+
sin
x)
cos
x(5
cos
x)
1
+
5(sin
x
cos
x)
0
2
(x
+
sin
x)
cos
x
sin
x(2x
+
cos
x)
x
cos
2x
sin
x
sin x(5
cos x(5 cos x)
1 + 5(sin x cos x)
9.ff0f00f(x)
= 42 cscx)(
x cotcsc
x+
2 + sin x)
0(x)=
2 csc x
30
5.
6.
12.
(x)=
+x)(cot
x)( csc x=cot=x)(5=+2sin
cscx)
x
5.
f x02(x)2csc
= x cot
=
2
3.
f (x)
(x)
==4x(csc
sin x (5
8x
xsin
2
2 x)
64
Chapter
2(sec x)(2x) = (x2 +
2 x + 2x
+cos
sin
(x
+
(x
+
sin
x)
0
2 x)2
8.
f
(x)
=
(x
+
1)
sec
x
tan
x
+
1)+
sec
x2
tan
(5
+
sin
x)
(5
sin
x)
9. f 0 (x) = 4 csc x cot x + csc x
10. f0 (x) = sin x csc p
x + x csc
2 x cot x
csc
x) cot x(0 csc xxcot
csc
x(x csc
x csc2 x + cot2 x)
2(1x+cos
2 x)
64
2
4.
f00f(x)
4 sin
xx)(
0 = (x
0 sin
2 csc x cot x + csc2 x , but 1 + cot2 xChapter
2sin x(2x + cos x)
+
sin
x)
cos2xcsc
cos x9.=f2x
13.
=
(x)
=
4
0 (x)
tan
x
(x
+
sin x) cos
sin x(2x + cos x)
x2=coscsc
x 2 x2x sin x
7.
f
(x)
=
sec
x
tan
x
sec
x
0
2 x
10.
(x)== sin x 2 csc(because
x + x(1csc
xcsc
cot
x2 x = (sin
6.
f ff(x)
= x)(1/
+
x)
(1
+
csc
x)
17.
sin
x
sec
cos
x)
=
tan
x),
so
2
6.
f
(x)
=
=
0(x) =
3
2
2
2
2
+ sinx)(sec
x) x tan x) = sec x +
(xsec
+xsin
11. f (x) = sec
x(sec
x)(x+ (tan
tanx) x
2
2
1 +x(5
x tan
x
(x2 + sin x)2
csc x(
csc(xx + sin x)
sin
+
sin
cos2x(5 cos x) 0 12 + 5(sin
x csc
cos
x) 1)
0 x
2x)
0(identity),
2 tan
xsec
10.
f
(x)
sin
csc
x
+
x
csc
x
cot
x
2=2x 2sec
0
2 xp
31) sec
thus
cot
x
csc
x
=
1,
so
f
(x)
=
=
.
5.
=
8.
f
(x)
=
(x
+
1)
tan
x
+
(sec
x)(2x)
=
(x
+
x
tan
x
+
x
2
2
2
11.
(x) =
= sec
x(sec
+ sin
(tan
xx[x(sec
tan
x)(sin
=x)
sec
x +cos
sec
xcsc
tan
2x
2sintan
2
(1 +
x tanx)
x)(sec
x)x)(sec
+ (tan
x)(1)]
sec
1
17. ff00(x)
(because
x sec
x=
x)(1/
x)
=
tan
x), x
so tan xp x
2
3
sin
x)
2x)
(1
+
x)
2
64
Chapter
2
12.f 0ff(x)
(x)
(csc
x)(
x)
+ (cot
csc
x
csc
x cot2 x 1 + csc
7.
x tan
x csc
2 sec
x x)( csc x 2cot x) =(5 +
(x)===sec
2
1+
x tan
x(5 +
7. f 0 (x)== (1
sec+xxtan
xx)2 =
2 sec
x x tan x)2 (because sec x
(1
+
x
tan
x)
tan
(1
+
0
2
3
00
2
2 + (tan x)(sec x tan x) = sec x + sec x tan2 x
2
2 x(sec
11.
sec
2 tan
2
3f (x) =sec
9. fftan
(x)2 =
4(1
xx cot
xx)(sec
+x)
csc2+2 x)
x x)(
+x)(
tan
tan(sec
x[x(sec
x)
+2=(tan
x)(1)]
x
x sec23 x
1
2csc
2x)x
12.
=
(csc
csc
(cot
csc
x
cot
csc
x
csc
x
cot
x 22xx)
64
Chapter
2
x
=
(x
+
sin
x)
cos
x
sin
x(2x
+
cos
x)
cos
x
2x
sin
x
00f 00(x)
21).
2 cot
(1
+
tan
x)(sec
x
tan
x)
x)(sec
x)
sec
x
tan
x
+
sec
x
tan
x
csc
x)(
csc
x)
cot
x(0
csc
x
x)
csc
x(
csc
x
csc
+ cot
(x)
=
=
= x) =
(because
x2
8.
ffff(x)
=
(x +
(sec x)(2x) =2 (x=+ 1)=sec
x tan
2
0 (x)
2
0 x + 2x sec
2 x
2 cscsec
tan1)xsec x tan x +(1
2 tan x(1
2
6.
(x)
==
14.
=
13.
(x)
=
,
but
1
+
cot
x
=
x
8.
f
(x)
=
(x
+
1)
sec
x
+
(sec
x)(2x)
=
(x
+
1)
sec
x tan x + 2x
+
x
tan
x)
(1
+
x
tan
x)
+
x
tan
x)
2
2
2
2
2
2
2 x = (sin x)(1/
17. f (x) =
(because
sinx)
x
socscx)x)2 2
(x
+(1
sin
x)
(xcos
+x)
sin
x)tan
+
tan
(1x),
++
tan
0=
x)sec
(1
12.
fcsc
(x)
=
(csc
x)( csc
csc
x)( csc x cot x) = csc3 x csc x co
2
2+ csc
2 x) + (cot
1 sin
+
tan
xx)(x +(1
10. f00tan
(x)2 =
x
csc
x
csc
x
cot
x
2 x
x
=
1).
(1
+
csc
csc
x)
cot
x(0
csc
x
cot
x)
x(
csc
x
x
+
cot
x)
+x1)
cot
x csc22x2
2
0 = (x
2
csc 0x 1)
csc x
tan
xcot
0 2 x csc
9.
4 csc
2x2p
13.
ff(x)
(x)
=
sec=
x(tan
xx+
tan
x)tan
sec
1)x(cos
2 x2.+ csc2 x , but 1 + cot x = csc x
2secxsin
18.f (identity),
(x)
==
cos
x
csc
=x)
(cos
x)(1/
sin
x)
==sec
cot
thus
cot
x+x(because
csc
=
1,
f=x
(x)
=
=42x),
2so
9.
(x)
csc
xcsc
cot
17.
(x)
(because
x=x)
sec
x x(tan
(sin
x)(1/
x)f x)
=
x),
so1+so
(13x
+tan
tan
x)(sec
x[x(sec
+ (tan
x)(1)]
x
tan
x
1
2x)
2tan
(1
+
csc
(1
x)
0
2
7. fff00=
(x)
=
sec
x
2
sec
x
cot
x
(1
+
csc
+
csc
x
2
2
2 x x)2
32
1 +2x(sec
x tan
(x)==sec
=
(1 + csc csc
x)( =
csc x) cot x(0 (because
csc x cotsec
x) x csc x( csc
11. f0 (x)
x) +
tan(1
x)2+=tan
secx)
xcsc
+13.
sec x tan
0 x x (1
(x (1
++
1) tan
cot
x (tan x)(sec
1)
(1
x2xtan
x)
+ x tan x)2 2x (1 + x2 tan x)2 2
(x)
=cot
202
2 x( fcsc
2 2
2
2x
2 x+
10.
=
sin
x
csc
+
x
cot
x
2csc
2
6x
cot
x
2x
cot
x
3(x
+
1)
csc
x =
(3
cot
x)[2x
cot
x
(x
+
1)
csc
x]
(x
+
1)
cot
x
csc
x
18.f (identity),
f0(x)
(x)
=
(because
cos
x
csc
x
=
(cos
x)(1/
sin
x)
=
x),
so
thus
cot
x
csc
x
=
1,
so
f
(x)
=
=
.
0
(1
+ xcot
tanxx)(sec x) tan x[x(sec2 x)
x)(1)]
x=x2 1tan
xx3+
1 x)x
2
+x
csc
(
10.
fcsc
(x)
=
sin
csc
x(1csc
cot
2sec
23
2 + (tan
00 (x)
x
=
1).
f(x)
=
.2 x
(1
+
x)
+
csc
x
(1
+
tan
x)(sec
x
tan
x)
(sec
x)(sec
x)
sec
x
tan
x
+
sec
x
tan
x
sec
x
8. ftan
=
(x
+
1)
sec
x
tan
x
+
(sec
x)(2x)
=
(x
+
1)
sec
x
tan
x
+
2x
sec
x
f
(x)
=
=
=
(because
sec
22
0
2
(3
cot
(3
cot x)
2
csc
x(
csc
x
1)
14.
(x)==(csc
=
15.f0f0f(x)
sin2x)(
x +2 cos
(identity),
sox2x)
f202(x)
12.
csc22 xx)=+1(cot
x)(
csc
cot
x)3= 0.2 csc3 x csc
x
cot
x
64
Chapter
2
0
(1
+
x
tan
x)
(1
+
x
tan
x)
(1
+
x
tan
x)
2
2
2
2
2
2
2
2
2
cot
x =x 1,3(x
so f+(x)
=
+ tan
x)
(12 x
+ tan
x) 6x
cotxx csc
2x cot
1) =
csc x
(3x(sec
cotx)
x)[2x
cot
x
(x x +
1)x)
csc= x]
cot
xthus
11. f (x)
+ (1
(tan
x)(sec
tan
sec x(x+ +
sec1)x(identity),
tan
0 2= sec
0 x csc
+ xcsc
x)22x
2
3 x)(sec x tan x) = sec3 x +(1
2
=(x
1).
=tan22 x
. sec
0f (x)x=
22 x)
11.xftan
(x)x =
sec
x(sec
x) +sec
(tan
tan
(1
+
tan
x)(sec
x
tan
(sec
x)(sec
x)
sec
+
sec
x
x
+
1)
cot
x
2
9. f tan
(x)
=
4
csc
x
cot
x
+
csc
x
2
2
0 sec x(tan x + tan x
sec
x(tan
1)sin
(32sec
cotxxx)
2 d2x)
22 sin x 2(3
sin
x x sin
sin
14.
==(1tan
=
= cot x)
18.
fdy/dx
(x)
(because
cos
x
=x x(cos
x)(1/
x)xx=
so
19. 0=
x sin
xx)(
+2cos
x,
=csc
cos
x2cot
sin
= cot
xx),
cos
xcsc
+
csc
csc
x) 2y/dx
cot
x)
csc
xalso,
x +=cot
0(x)=
2 csc
=
2xx(0
2f (x)
2 x)
2 so
xcot
16. 0f(x)
(x)==
23sec
tan
x x)
sec
x+(cot
tan
sec
xtan
=x)
23 xcsc3x(
=
0;tan
sec2x)x, but
tan12 x+=cot
122(identity),
x(because
tan
x)
(1
+
2 (1
2=3 csc
12.
x)(
csc
+
x)(
csc
x
cot
x)
csc
x
cot
x
13.
=
x = csc
x x + sec
(1
+
tan
x)
(1
+
17.fff0(x)
(x) =
= (csc
sin
x
sec
x
=
(sin
x)(1/
cos
x)
=
tan
x),
so
0
2
2
2
2 x)x+
cos
x
cos
x
(1
+
tan
x)(sec
tan
x)
(sec
x)(sec
x)
sec
12.
f
(x)
=
(csc
x)(
csc
(cot
x)(
csc
x cot x)
==
cscx3 tan
x csc x co
(1
+
csc
x)
(1
+
csc
x)
(x
+
1)
cot
x
0
0 = 1+
x
tan
x
10.
f
(x)
sin
x
csc
x
+
x
csc
x
cot
x
2
2
2
2
2
2
2
2
2
2
2
14.
f
(x)
=
f
(x)
=
0.
2
6x
cot
x
2x
cot
x
3(x
+
1)
csc
x
(3
cot
x)[2x
cot
x
(x
+
1)
csc
x]
(x
+
1)
cot
x
csc
2
2
2
3
19.
dy/dx
=
x
sin
x
+
cos
x,
d
y/dx
=
x
cos
sin
x
sin
x
=
x
cos
x
2
sin
x
sec
x(tan
x
+
tan
x
sec
x)
sec
x(tan
x
1)
18.
f
(x)
=
(because
cos
x
csc
x
=
(cos
x)(1/
sin
x)
=
cot
x),
so
2
0
20. f=
dy/dx
=
csc
x
cot
x,
d
y/dx
=
[(csc
x)(
csc
x)
+
(cot
x)(
csc
x
cot
x)]
=
csc
x
+
csc
x
cot
x
csc
x(
csc
x
1)
csc
x
(1
+
tan
x)
(1 + t
2
2
2
2
2 x= csc
(x) = (1
.
0 x)x+
23
=1,
2
0csc
cot
x22x)(sec
+
x)(
x)
cot
x(0
cot
csc x(seccsc
x)
(1thus
+
x
tan
tan
(tan
x= tan2 x x + cot
1 cot x)2
2 =
15.(identity),
x csc
++
cos
=
12 x)
(identity),
so
(x)
cot
x csc
x=
so
(x)
=
. 2 (3
2 csc
2 x) x)(1)]
2
0f 0(x) = sin
2 csc2 x
(3x[x(sec
cot
(1
(1ff+x)
tan
x)30.
2
2
0(x)
2tan x)
13.
f
=
=
,
but
1
+
cot
x
=
2
2
f
(x)
=
=
=
(because
sec
x
(1
+
csc
x)(
csc
x)
cot
x(0
csc
x
cot
x)
csc
x(
csc
(1
+x
csc
x)
16x
+
csc
2x2tan 2x) =
2 sec x2 +
2(1 +
2
2
11. f (x) = sec
x(sec
x) + (tan
x)(sec
sec
tan
xx(tan
2 xx(1 2x
0 sec
x+
tan
seccot
x)2 xx)2sec
x(tan
x 1)
2
2 x
2(x
+
csc
x)
csc
x)
cot
cot
x)[2x
+ [(csc
1) cscx)(
x] csc
(x2x)
+
x
csc
xx tan
(1
+
x=
tan
x)
(1
x)
tan
13.
fcot
(x)
=+
20. fdy/dx
dy/dx
=(3
csc
x cot
x, cot
d(1
y/dx
+1)
(cot
x)(
csc
x cot
x)] x
= csc+3 xx
+ csc
x23(x
cot2 +
x 1) csc x . =
0
=
=
21.
=
x(cos
x)
+
(sin
x)(1)
3(
sin
x)
=
x
cos
x
+
4
sin
x,
(x)
=
=
(
2
2 (1 + csc x) (1
2
x x 1) (1 + csc
csc x( sincsc
tanxx)sec
tan
2
2
0 22 sin x
2 x)2 + tan x)
0
2
2sec2cot
0f
2
2
(3
x)
2 2=
16.(identity),
(x)x=
==
21).
sec
x+tan
xx
xcsc
22xx)
tan
x(sec
xx
2 x3tan
0;sec
also,
fx(x)
tancot
x = 1 (identity),
so
2 x== sin
15.
(x)
=
sin
xtan
cos
xx)x
=
1(cot
so
(x)
0.sec
19.
=
xx)(
sin
x
+
cos
x,
d(identity),
y/dx
cos
x
sin
x==
x=xcos
sin
2
02
2sec
thus
cot
=x)(
1,=csc
so
fx=
(x)
.x3 xx (3
csc
x(
csc
x
1)
(1
+
x)(sec
tan
x)(sec
x)
+
tan
x
sec
3=
3 x xcsc
2
12.
ffdy/dx
(csc
csc
+
cot
x)
csc
x
x
cot
x
2
0
2
2
cos
x
cos
0d(x)
y/dx = x( x)
sin+x)(sin
+ (cos
+ 4 cos x = x=sin
5 cos
(1x++csc
x) x
(identity),
thus1 +
cotcscxx csc =x = 1, so f (x) =
=
14.
(x)
0
21.ffdy/dx
x)(1)x)(1)
23( sin x) = x cos x + 4 sin x,(1 + tan
(x)===0.x(cos
2 x)2
2
0
2
(1 + csc x)2
(1 2+ tan
15.
f
(x)
=
sin
x
+
cos
x
=
1
(identity),
so
f
(x)
=
0.
(x
+
1)
cot
x
2 2x) 2
2
2
3
2
sin
x
sin
x
19. fdy/dx
dy/dx
=
x sin
xcot
+ x,
cos
=
cos
sin
= csc
x cos
x2cscx)]
x3cot2x2x)
2 d y/dx
22 sin
2
2(1 +
02 = =
2xxx)(
20.
dx,
y/dx
= xx+
[(csc
x)xsin
+2+
(cot
csc
x cot
+ csctan
x cot
18.
(x)
(because
cos
csc
x)(1/
x)
=x)(
x),
so
csc
x)(
x)
cot
x(0
csc
xxcsc
cot
csc
x(cot
x=
+=csc
2=
2csc
2x)
2x
y/dx
sin
x)
+
x)(1)
44x)(sec
cos
x
x
sin
52x
cos
x=sec
2x)(2x)
16.
(x)
=(1
2=
sec
xx
tan
x(cos
2x)
tan
sec
x=
=(cos
sec
x =x 1 (identity), so
0d sec
x)(sec
x(cos
tan
x)
(sec
xxtan
xxalso,
tan
xfx,(x)
sec
x x , but
22.f0fdy/dx
x3+
(tan
sin
+csc
+
cos
x
x
sinx=
+
cos
x0;+
4 cos
xx)
x(tan
xx(
+cot
tan
xsec
sec
sec
x(tan
x=x)
1)
3 xsec
3xx+
13.
(x) =
==
1 + cot22 x = csc2 x
cos
cos
2
2
14.
f=(x)
=
=
=
tan
x) x sec xsin
tanxx + sec
(1tan
+ csc
(1++x)
csc
x) 2 x tan2 x) (sec
2x)
2 x)(sec
2tan x)2 22
2 +(1
2 x)(sec
2cot
0 csc
fd20 (x)
=2 0.(1 +
2 21) csc22sin
(1
+
x)
(1=
2x)[2x
2 x(sin
2 2 + 1)(1
3 x2xtan
tan
x)(sin
+x)
cot
xxx=
2x
3(x
+
cotx)
(x
csc
x]= xsin
(x
+
1)+4cos
cot
x
xsin
14.
f x]
(x)
16.
2tan
sec
tan
sec
xcot
x x+
sec
xsin
= 0;
(cos
x)
+
x)(2x)]
+cos
2[x(
x)
+
4=1)
x6x
(2
)csc
cos
4(x
1)
20.
cot
x,
d(cos
y/dx
=
[(csc
x)(
csc
x)
(cot
x)(
csc
x4=x
cot
x)]
x +x)
csc
cot
x=x x3.= 2 3 (1
0 y/dx=(3
2csc
2x
21.
x(cos
+
x)(1)
3(
sin
cos
sin
x,
2
csc
x(
csc
x
csc
x
fdy/dx
(x) == =
=
(1
+
tan
22.(identity),
dy/dx
xthus
( [xx
sin
x)
+
x)(2x)
+
4
x
=
x
sin
x
+
2x
cos
x
+
cos
x,
0
cos x
cos x + t
cot22 x sec
csc22 x)
x =(3 sec
1,cot
so
f 2(x)
= 1)
=
. (3 cot x)2
sec
x(tan
x
+
tan
x(tan
x
x)
0
2
f cos
(x)
0.
2
22 2
(1x++csc
x) x=
1 tan
+ csc
2 2x
=d(x)
= + 4so
2sin2x)2+ (cos x)(1)
0x = 2 x sin
= xx(+[xcos
cos
sec
x(tan
secx2 x)4(x sec
x(tan
2 5 x]
15.
= sin
x(sin
=x)+
1x)(1)
=sin
0.
d2y/dx
y/dx
x)
(sin
+
2[x(
+ 4cos
4 sin xx=+(2
xx ) cos
+ 1)
sin xx 1)
(1 +x)(
tan
(1
+fx)
tan
x)
23.f dy/dx
(sin
sin
+(identity),
(cosx)(2x)]
x)(cos
x)
=(x)
cos
x x)
sin
x,
21.
==
x(cos
x)(cos
+x)
3(
sin
=
x
cos
x
+
sin
x,
=
=
2
2
2
+22tan
x)
(1 + tan x)2
19. dy/dx
=2(1 +
x sin
+ cos x
x,tan
d y/dx
= x)(sec
x cos x2 x)sin xsec x
sintan
x=
xsec
cosx(1
xtan
xsec3 x
2
tanxsin
x)(sec
x)
(sec
x+
xsin
2 2(cos
2sinx)(cos
0 2 y/dx=
x)(
sin
x)
+
(cos
x)(
sin
x)
[(sin
x)
+
(sin
x)(cos
x)]
=
4
sin
x
cos
x
2
d
=
x(
x)
+
(cos
x)(1)
+
4
cos
x
=
x
x
+
5
cos
x
2
2
2
0
2
sin
x
sin
x
22.
dy/dx
x
(
sin
x)
+
(cos
x)(2x)
sin
x
+
2x
cos
x
+
4
cos
x,
14.
(x) = = (sin
=
=
23.fff(x)
x) + 2(cos
x)(cos2 x) =
x sin
0dy/dx
15.
x+
cos xsin
xsec
=
(x)cos
= 0.
2
16.
(x)==sin
2 sec
xx)(
tan
2 x,3 (1
=+
0; tan
also,
= sec2 x tan22x = 10 (identity), so
(1x1+(identity),
tantan
x)2x secso xf =
2 x) f (x)
2 2 3
32 x 15.
22
22 csc x2cot x, d2 y/dx2 = [(csc x)( cos
cos
x
f
(x)
=
sin
x
+
cos
x
=
1
(identity),
sosin
f (x)
20. dy/dx
=
csc
x)
+
(cot
x)(
csc
x
cot
x)]
=
csc
x
+
x cot
x
d0d y/dx
=2(cos
[x x)(
(cos
(sin (cos
x)(2x)] + sin
2[x( sin[(sin
x)
+ cos x] 4 sin x = (2 x ) cos x csc4(x
+ 1)
x = 0.
2 x)
y/dx
sin+
x)
2sec
2 x)(cos x) + (sin x)(cos x)] = 4 sin x cos x
(x)
=
0. =
sec
x(tan
x(2+x,sin
tan
x+
x)
sec
x(tan
x
1)
24.f dy/dx
=
sec
d2x)
y/dx
= 2+
2x)(2x)
sec2 xx)(
tan
x xx)
22.
x
(cos
+
4
cos
=
x
sin
x
+
2x
cos
x
+
4
cos
x,
2
sin
sin
x
2
0
2
2
16. f=(x) = 2 sec
xx)
sec2 x 2 tan=
x sec(1 x+=
2 3 = 0; also, f (x) = sec x tan x = 1 (identity),
sosin x
2
(1x2+tan
tan
tanxcos
x)
2 +
2
2 x,
cos
2
2 x(cos
23 x
21. dy/dx
=
x)
(sin
x)(1)
3(
x)2[x(
=
cos
xx)
+ sin
4 sin
0x
2 sin x 2 sin x
2 sin
2 +(sin
2 sin+
d
y/dx
=
[x
(cos
x)
+
x)(2x)]
sin
+
cos
x]
4
sin
x
=
(2
x
)
cos
x
4(x
+
1)
23.
dy/dx
=
(sin
x)(
x)
(cos
x)(cos
x)
=
cos
x
x,
16.
f
(x)
=
2
sec
x
tan
x
sec
x
2
tan
x
sec
x
=
2
= 0; a
0
24.f Let
dy/dx
==
sectan
x,x,
d then
y/dx f 0=
2 sec
x2tan
(x) f=(x)
0.
25.
(x)
= sec
x. x
cos3 x
cos3 x
2
0
2 =2 x( sin 2
dd22y/dx
x)
+
(cos
x)(1)
+
4
cos
x
=
x
sin
x
+
5
cos
x
f (x)
0. x)(cos x)] = 4 sin x cos x
y/dx
x)( x sin
+ (cos x)(sosin
x) =[(sin
x) +=(sin
15. f (x)
= sin= x(cos
+ cos
= 1x)(identity),
f 0 (x)
0. x)(cos
0 (cos x)(cos
2 x) = cos2 x
23.
= (sin
x)(x, then
sin
sin2 x,
25. dy/dx
Let
f
(x)
=
tan
=
sec
x.
0 x)f+(x)
(a) f (0) 2= 0 and f (0) = 1, so y 0 = (1)(x 2 0), y = x.
22. dy/dx
=2 x ( 2 sin x)
+ (cos
x)(2x)2 + 4 cos x =2 sin
x xsin x +sin
2xxcos x + 4 cos x,
2
24.
dy/dx
=2 =
sec(cos
x,
d2xy/dx
=+
22 sec
tan2 x
d
y/dx
x)(
sinx2 x)
(cosxxx)(
sin=x) [(sin
x)(cos
x) =
+ (sin
x)(cos
x)] =
= sec42sin
16. f 0 (x)
=
sec
x
tan
sec
tan
sec
x
2
0; also,
f (x)
x x cos
tan2xx = 1 (identity), so
⇣
⌘
⇣
⌘
⇣
⌘
3x
2
2 ⇡ = 0 2and f 0 (0) =
cos
cos
⇡ 1,x)(2x)]
⇡ (2 x2 ) cos x 4(x + 1) sin x
f (0)
so y +0 2[x(
= (1)(x
y⇡ =
d0(a)
y/dx
= [x (cos x) +
sin3x)x0),
+ cos
x]x.
4 sin x =
0 (sin
f 0. = 1 and f 0
= 2, so
y 1=2 x
, y = 2x
+ 1.
f (b)
(x) =
2 x,2then 2f 4
2 2 x.
4⌘
2
25.
f (x)
= sec
24. Let
dy/dx
=
sec
x, d y/dx ⇣=(x)
2 sec
x tan x
⇣ 4⇡=
⌘tan
⌘
⇣
⇡ x)(cos x) = cos2 x sin2⇡x,
⇡
23. dy/dx
x)(1 and
sin x)
(b) f=
f 0 + (cos
+ 1.
⇣ (sin
⌘=
⇣ =⇡2,
⌘ so y 1 = 2 x ⇣ , y⇡ =
⌘ 2x
⇡
4
4
4
2 + ⇡ 1.
0
0
2
2 = 0=
(c)
f0 (x)
=2 x.
y +[(sin
1 =0),
2 y x=+x.
,(sin
y =x)(cos
2x
(0)
f and
(0)x)
so sec
yx)(
02,sin
=so(1)(x
y/dx
==4(cos
x)(
sin
+1,(cos
x)
x)(cos
x)
+
x)]
25. d(a)
Let
ff(x)
tanand
x, 1then
f=
=
4⌘
2 = 4 sin x cos x
⇣ ⇡⌘
⇣ 4⇡ ⌘
⇣
⇡
⇡
⇣ ⇡ ⌘2 = 2 1 and
⇣f⇡0 ⌘ 2 = 2, so y + 1⇣ = 2 ⇡x⌘+
(c) f =
, y = 2x
1.
⇡ +
0 20 =
0 2 sec 4x tan x
24.
dy/dx
sec
x,
d
y/dx
4
4
(a)
(0) =
= sin
0 and
f (0)
1, =so2,
y sox.0y = (1)(x
26. (b)
Let ff(x)
then
cos
=
1x,and
f f =(x)
1 = 2 x0), y =, x.
y = 2x
+ 1.2
4
4
4
2
⇣ ⇡==⌘tan
⇣0 (x)
⇣
⌘
0 ⌘
2 x.
26. Let
Let
f
(x)
sin
x,
then
f
(x)
=
cos
0 f
⇡
⇡
⇡
25.
f
(x)
x,
then
=
sec
x.
⇣
⌘
⇣
⌘
⇣
⌘
(a) ff (0) ⇡
= 0 and f (0)
so
y = x.⇡
⇡2,y so 0y = (1)(x
⇡
(b)
f 0 f=0 1, =
x0),
+
(c)
f 4 ==1 and
1 and
= 2, so1y=+21 =
2 4x +, y =, 2x
y = 2x
+ 1. 1.
4
2
4
4
4
2
0 0
(a)
0),0),
y⇣
=
x. x.
(a) fff(0)
(0)
=
and
0=
(1)(x
y⇡),
=
0(0)==1,1,soso
(b)
(⇡)
=0⌘00and
andfff(0)
(⇡) =⇣ 1,⇡yso
0(1)(x
=
( 1)(x
y⇡=⌘ x + ⇡. ⇡
⇣ =⇡
⌘y y0 =
0
(c) ff(x)
1 and⇣f 0f(x)
, y = 2x +
1.
26. Let
x.2, so y +⇣1 = 2⇡ ⌘x +
⇣ ⇡=
⌘ sin=x, then
⌘ = cos=
⇡
0 0 ⇡ ⇣ ⌘4
4
⇣
⌘
(b) ff (⇡)
⇡),
y4=
⇡ ==01and
1andff (⇡)0= ⇡= 1,
⇡
1
(b)
2, so
so 1yy 01 =
= 2( 1)(x
x1
,1y ⇣
=
2x ⇡x⌘++⇡.1. 21
p +p .
2, y = p x
(c) f 4 = p and
f4
= p , so y p =4 p x
0
4⌘
0 ⇣1,
(0)4 =
0 and
f (0)f =
socos
y x.20 = (1)(x 20), y =2 x.
2
4 2
2
⌘ sin
⌘
⇣
26. (a)
Let ff(x)
x,2then
(x)4 =
⇣⇣ ⇡=
⇡
1
⇡ ⌘= p1 and f00 ⇣ ⇡ ⇡ ⌘= p1 , so y p1 ⇣= p1 ⇡ ⌘x ⇡ , y =
⇡ p1 x
p
p
(c)
f
+
.
(c) f 4 = 21 and f 0 4
= 2, so y + 1 = 2200 x + 2 , y =4 2x +
1.
0
27. (b)
(a) If
y=
x then
x +y 2x cos
x (and
y =⇡),
2 cos
x xx +
sin⇡.x 2so y200 + y 4= 2 cos x.2
4=x0sin
41, so
4y =
f
(⇡)
and
f
0
=
1)(x
0 (⇡)y== sin
(a) f (0) = 0 and f (0) = 1, so y 0 = (1)(x 0), y = x.
00
⇣y⇡==
⌘sin
⇣ xy (4)⇡x+
⌘sin
26.
x,1x
then
f 0 (x)
cos
27. Let
(a) f (x)
If
x sin
then
y 0⇣=⇡=⌘
sin
xx.+
costwice
x and1more
y 00 =to
+ yx.=
(b)
Di↵erentiate
the
of
part
get
y 00x=so1y2 cos
1 x(a)
12 cos
⇡ 2 cos 1x.
0
0 result