Exercise Set 2.5

4. f (x)
=2.5
4 sin
cos x
Exercise
Set
Exercise
2 x
3. f ⇧⇧⇧(x)
= Set
4x
sin2.5
x+ sin8xx)cos cos
x2 x(5
sin10
x(5
91
1 + 5(sin x cos x)
cos x)
9.
f (x) =
4 csc x cot x + csc x
2.
=
5. f ⇧⇧(x)
(x) =
= x23 + cos
2
8. ff⇧ (x)
(x) =
= (x
+ 1)
sec
x+
tan
+2 (sec
+ 5(sin
1)sin
secx
x 2tan
+ 2x sec x
sin
+
x)
1.
4 sin
x+
+
2(5cos
x xx)
+
cosx x)
sin
x(5
sin
x)
cos
x(5 x)(2x)
cos x)= (x1 (5
⇧⇧ ⇧
4.
f
(x)
=
4
sin
x
cos
x
=
5.
f
(x)
=
10. f ⇧(x) = Set
sin
csc x + x csc x2cot x
Exercise
2 x2.5
2
(5 +
(5 + sin x)
3. f ⇧(x)
= 2.5
4x2 sin x 8x
cossin
x 2x)
Exercise
Set
91
+ sin
x) cos
9. f⇧⇧ (x) = (x10
4 csc
x cot
x +x cscsin
xx(2x + cos x) x2 cos x 2x sin x
6.
f
(x)
=
=
22 x)
2+
2.
(x) =
= sec
cos
cos
sin43x(sec
x(5
sin
x)
cos
x(5
cosx)x)= sec13+x 5(sin
2(tan
2 x tan
1. ff⇧⇧⇧(x)
sin+x+
2x(x
cos
11.
x)
+
x)(sec
+ 2sec
tan
x
+x sin
x)
(x
+xx
sin
x)
2
xsin
=
5.
=
+
sin x
x)
cos
x x)sin
x(52 +
cossin
x x)2x
4. ff ⇧(x)
= 4(x
x cos
2 x(2x + cos x)
2 sin x
⇧(x)
(5
+
sin
p
172/
1
–
25
odds
6. f (x) =Set
=
10.
sin x
csc x 2+ x csc x 2cot x
Exercise
2.5
sin x)
(x23 x
+ sin
x)x2cot2 x
2(x + 2
10
2xx)(
12.
x)( csc x cot x) = csc
csc
7. fff⇧⇧⇧⇧(x)
(x) =
sec
tan
xcsc
2+
secx(cot
x x(5
3.
(x)
= (csc
4x
sin
x+
8xx)
cos
2.
+
cos
x
1
+
5(sin
x
cos
x)
sin
x(5
sin
x)
cos
cos
x)
3
2
2
x4 sin
2 x
+ sin
xx sin
x(2xx +
cos
x)
x3 xcos
x x2x
sin
2x)
=
5. ff⇧⇧⇧⇧(x)
1.
=
x+
2 cos
cos
11.
= (x
sec
x(sec
x)
++(tan
x)(sec
tan
x)
=
sec
+
sec
tan
x
2
2
⇧(x) =
2
6.
f
(x)
=
(5
sin
x)
(5
+
sin
x)
2
2sec x 2
2
7.
f
x
tan
x
2
⇧ (x) = sec
2+
2 cot x) 2 csc x( 2csc x
2
(1
csc
x)(
csc
x)
cot
x(0
csc
x
⇧
(x
+
sin
x)
(x
+
sin
x)
8. ff⇧⇧ (x)
(x) =
= 4(xsin
+
1)
sec
x
tan
x
+
(sec
x)(2x)
=
(x
+
1)
sec
x
tan
x
+
2x
sec xcsc x + cot x)
4.
x
cos
x
2
13.
f
(x)
=
=
, but 1 + cot2 x = csc2 x
3. f ⇧(x) = 4x10sin x 8x2 cos
x+ csc x)2
2 x)2
3
(1+
(1x +
csc
⇧ (x) = (csc
2
2
12.
f
x)(
csc
x)
(cot
x)(
csc
x
cot
x)
=
csc
x
csc
cot
x
2.
+sin
cos
⇧(x) = (x32 +
x)xcos
x x22+
sin(sec
x(2x
+ cos=x)(x2csc
x 1)
cos
2x xsin
8. fffff⇧⇧⇧(x)
(x)=
=sec
(x
+
x+tan
x)(2x)
+
secx
x tan
+x2x seccsc
x x
x(
9.
csc
x1)
x4 x
6.
(x)
=
7.
tan
xsec
2csc
sec
⇧
2 x x)
2 xxx(5
1=+ 5(sin
xcsc xcos 1)
x)
sin
x(5
+cot
sin
cos
cos
x)
2
2
2
2
⇧
(identity),
thus
cot
x
csc
x
=
1,
so
f
(x)
=
=
.
(x
+
sin
x)
4. f (x) = 4 sin x cos x (x + 2sin x)
2
=
5.
cscx)
x)2 x( csc x
1 + csc
2
(1
x)2 cot x(0 csc x cot
csc
cscx
x + cot2 x)
(5 +csc
sin x)
(5(1x)
++sin
⇧⇧⇧
2 + csc x)(
2
3.
(x)
=
4x2sin
sin
cos
xcsc
9. fff⇧ (x)
(x)=
=
4+
csc
cot
xx cotx)(2x)
13.
=(x
, but 1 + cot2 x = csc2 x
10.
(x)
=
x1)xsec
csc8x
+ xcsc
x2
8.
xxtan
x(1
+
= (x2 + 1) sec=
x tan x + 2x (1
sec+xcsc x)2
⇧
2+(sec
csc
x)
1
+
5(sin
x
cos
x)
sin
x(5
+
sin
x)
cos
x(5
cos
x)
7. f (x) = (1
sec2 xtan
tanx)(sec
x
sec x)x (sec x)(sec2 x) 2sec x tan x + sec x tan2 x sec3 x
x2tan
= =x csc
5. ff ⇧⇧⇧(x)
(x) =
= (x+
+ sin2 x)(5
xsin x)
sin
x(2x
+ cos x)
cos
x x)
2x
csc
1) 2
csc x =
14.
22 xsin x
2cos
2 2x
+
+x(sin
2cot
4.
f
=
4
sin
x
cos
x
⇧⇧⇧(x)
2
10.
(x)
=
sin
x
csc
x
+
xtan
csc
x so
11.
f
(x)
sec
x(sec
x)
+
(tan
x)(sec
x tan
x) f=⇧ (x)
sec=3=x (5
+ sec
x tan(1
x+
6.
(identity),
thus
cot
x
csc
xx)=
1,
=
.
9. f ⇧(x) = 42csc x cot x(x(1
+2 +
csc
x
2
2+
22 tan x)
2
+
sin
x)
(x
sin
x)
(1
+
csc
x)
1
+
csc x
8. f (x) = (x + 1) sec
x
tan
x
+
(sec
x)(2x)
=
(x
+
1)
sec
x
tan
x
+
2x
sec
x
2
sec x(tan2x + tan
x sec2 x)
sec x(tan x 1)3 2
2 cot
⇧(x) = (csc
(x
+
sin
x)
x+ cos
sin
x(2x
+x
cos
cos
xx x2x
1sec
+x5(sin
cos
x)
x(5
+2csc
sin
x)
x(5
cos
x)
=
12. f=
f⇧⇧⇧(x)
x)(
x)+(tan
(cot
x)(
cotx)
cscsin
x
11.
(x)=
= sin
sec
x(sec
x)2cos
+
x)(sec
xxcsc
tan
x)
=x)
+ 3sec
tan
xx 2 x
2
2= xcsc
10.
sin
x
csc
x
x
csc
x
cot
2
2=
3
6.
f
(x)
=
⇧
2 x)
=
5.
(1
+
tan
x)
(1
+
tan
x)
2
(1
+
tan
x)(sec
x
tan
(sec
x)(sec
x)
sec
x
tan
x
+
2
2
2
2
7.
tan
x (5
sec
2
9. f (x) = sec4 x
csc
x cot
x
+2+
csc
(x+
sin
(x
+ sin
sin
x)xx)
(5
+ sin
x)2x) sec x tan x sec x =
14. f ⇧ (x) =
=
2
2
22
3 x( 2(1
(1 2+
tan
+ tan2x)
9212. f⇧⇧⇧ (x) = (csc
2
(1
+x)(
csc2x)(
x)
cot
xsec
cot
x)
csc
3=
csc2=csc
x)
+
(cotx)
x)(xx(0
csc
x
cot
x)
csccsc
x x + cot x) , but 1 + cot2 x = Chapter
2 x(sec
2
⇧csc
11.
fff(x)
==sin
sec
(tan
x)(sec
tan
x)
=
x +=sec
xxtan csc
x xxcotcsc
13.
(x)
csc2 x
15.
=
1
(identity),
so
f
(x)
=
0.
⇧⇧(x)
22x + cosx)x+
2
2
2sin
2x
2
8.
=
(x
+
1)
sec
tan
+
(sec
x)(2x)
=
(x
+
1)
sec
x
tan
x
+
2x
sec
x
10.
sin
x
csc
x
+
x
csc
x
cot
2x
2x
(1
+
csc
x)
(1
+
csc
x)
(x
+
sin
x)
cos
x
x(2x
+
cos
x)
x
cos
x
2x
sin
x
7. ff ⇧ (x)
(x)
=
sec
x
tan
x
2
sec
sec x(tan x + tan x sec x)
sec x(tan x 1)
6. f=(x) =
= x(0 ⇧ csc=
2
csc
csc
x x)x(
1)
x 2 x + cot2 x)
2 cot
2+
2
2 csccsc
⇧
2 2
3x
2 csc
2 x(
(1
+
csc
x)(
x
x)
csc
2 (x
2
+
sin
x)
(x
sin
2x)
21, csc
2 sin
sin
(1
+
tan
x)
+cos
tan
x)xcot
12.
f
(x)
=
(csc
x)(
csc
x)
+
(cot
x)(
cot
x)
=
csc
cot
x 2 sin
⇧
92
(identity),
thus
cot
x
csc
x
=
so(1x
fx
(x)
=
. x2
19.
dy/dx
=2sec
sin
xcot
x=
xcsc
sin
x=
xcsc
cosxfx
2 x
2+ cos x, d2 y/dx =2
13.
f⇧⇧⇧(x)
(x)=
=
cot2 xChapter
=socsc2 2x
2=
2xcsc
2 3sin
11.
(x)
=
x(sec
x)
+
(tan
x)(sec
x
tan
x)
=
sec
x
+
sec
x
tan
x
16.
f
sec
x
tan
x
sec
x
2
tan
x
sec
x
=
2
0;
also,
(x)
x tan,2 but
x = 11 +
(identity),
9.
4
x
x
+
csc
x
(1
+
csc
x)
1
+
csc
x sec
2
2
8. f (x) = (x + 1) sec x tan x(1++(sec
=cos
(x3 x+ 1) sec
x3 tan
x + 2x(1
sec+xcsc=
csc x)(2x)
x)
x)
cos
x
⇧
2
2
2 x
(x)
=
⇧⇧ x cotcsc
x( 3 csc
csc x(
x csc
1) x 2 csccsc
7. fff⇧⇧⇧(x)
(x) =
= 0.
sec
x+
xx)(22 x
22sec
(12thus
csccos
csc
cot
15.
sin
xtan
+
=
12 x)
(identity),
so
ffcsc
=
2x2 cot x(0
20.x)
2x d
2x x)] x
2(x)
3+
(identity),
cot
csc
=
1,csc
soxx
(x)
==
=
. 3 x x)
2x)x
2 x
20.
dy/dx
=
csc
x
x,
y/dx
[(csc
x)(
csc
x)
+
(cot
x)(
csc
x
cot
=
csc
+ csc
x cot
⇧(x)
2
12.
f
(x)
(csc
x)(
x)
+
(cot
x)(
cot
x)
csc
x
csc
x
cot
13.
f
=
=
, but
1 2+xcot2 x = csc2 x
(1
+
tan
x)(sec
x
tan
(sec
x)(sec
x)
sec
x
tan
x
+
sec
x
tan
x
sec
10.
=
sin
x
csc
x
+
x
csc
x
cot
19.
dy/dx
=
x
sin
x
+
cos
x,
d
y/dx
=
cos
x
sin
x
sin
x
=
cos
x
2
sin
x
2
9. ff ⇧(x)
xcsc x)2
2 x =
(1 + csc x)
1 +x)csc
14.
(x) =
= 4 csc x cot x + csc
=
(1
+
(1
+
csc
2
2
tan x
x) x)(2x) = (x22sin
(1
+
tan
x)
⇧
x sec
x
8. f (x)
(x)== (x2 + 1) sec
x (1
tan+x2tan
+
+3csc
1)
tan
+1)
2x
x(
csc
xxx),
17.
(because
sin(sec
x sec x =2 (sin
x)(1/
cos
x) x=sin
tan
so sec x csc2 x
⇧csc
2x sec
2
92
Chapter so
2
2= csc
⇧⇧⇧ (x) =1 2
2 x)(
(1
+
csc
csc
x)
cot
x(0
x
cot
x)
csc
x(
x csc
+. cot
x) tan2 x2= 1 (identity),
16.
f
sec
x
tan
x
2
tan
x
sec
x
=
2
0;
also,
f (x)x =
sec2 x
(identity),
thus
cot
x
csc
x
=
1,
so
f
(x)
=
=
2
2
+
x
tan
x
2
2
2
2
2
11.
f
(x)
=
sec
x(sec
x)
+
(tan
x)(sec
x
tan
x)
=
sec
x
+
sec
x
tan
x
21.
dy/dx
=
x(cos
x)
+
(sin
x)(1)
3(
sin
x)
=
x
cos
x
+
4
sin
x,
⇧
10.
xtan
cscx,
xtan
csc
x=
cotsec
x
3x)(
sec x(tan
x+
xx d+xsec
x)
x(tan
1)3 xx)(1
20.
= (1sin
csc
xtan
cot
y/dx
[(csc
x)(x cos
csc
+=
(cot
x)]
=
csc33xx + csc
x cot
x cot2 x = csc2 x
+
x)(sec
x)
(sec
x)(sec
x)
sec
xcos
tan
x2+ csc
sec x cot
tan
xcsc
sec
13. dy/dx
f=⇧(x)
=
,
but
1
+
x
+
csc
x)
1
+
x
2
2
= x)
2 x)
2 (12+ csc
2
14. ff⇧⇧ (x)
(x) =
= 0. + x tan x)(sec
= x)(1)]
(1tan
+ csc
sec2 x
1=
x)tan
x) +2(tan
tan
x)
(1 + tan
9. fd
(x)
=2 (1
csc+xsin
cotx)
x 2+
+(1csc
x tan
+
x)2x[x(sec
+ tan
x)2xcsc=x
f 2⇧⇧⇧(x)
=
= xx2(11)
(because sec2 x
y/dx
=4(1
x(
x =x) x3 sin
xx(+35csc
cos
2
2 (cos2x)(1)2+ 4 cos
csc
2csc
2
2
2
⇧
12.
f
(x)
=
(csc
x)(
x)
+
(cot
x)(
csc
x
cot
x)
=
csc
x
csc
x
cot
x
2
2
(1
+
x
tan
x)
(1
+
x
tan
x)
(1
+
x
tan
x)
2
2
3
11.
sec
x(sec
x)
+
(tan
x)(sec
x
tan
x)
=
sec
x
+
sec
tan
x
19.
dy/dx
=
x
sin
x
+
cos
x,
d
y/dx
=
x
cos
x
sin
x
sin
x
=
x
cos
x
2
sin
x
92
Chapter 2
(identity),
thus
cot
x
csc
x
=
1,
so
f
(x)
=
=
.
2 xx)(1)
+xtan
x)(sec
tan
tanx,x + sec x tan x sec x
21. dy/dx
=(1x(cos
x)tan
+ (sin
sinx)(sec
x) = xxx)
cos1)
xsec
+ 4xsin
⇧ 2sec x(tan
+
sec2x)x)3((sec
sec
x(tan
12 + csc x
2 x
⇧
14.
fftan
(x)
=
= (1 + csc x)2
=
tan
xcos
⇧(x)
x=
=sin
1).2sin
15.
f
x
+
x
=
1
(identity),
so
f
(x)
=
0.
=
=
2
10.
(x)
=
x
csc
x
+
x
csc
x
cot
x
(1
+2 tan
x)x +
(1 +
tan
x) x, 2
17. dy/dx
f2⇧(x) ==
(because
sin
sec4(1
xcos
=x(sin
x)(1/
cos
x)
= cos
tan
x),
so
2(cos
22 sin
2
2 x2 ((1 +
tan
x)
tan
x)
22.
sin
x)
+
=x
x=
x++
+
4x2cos
2+
35csc
(1
+
csc
x)(
csc
x)
cot
x(0
csc
cot
x)xcsc
x(
csc
x + cot
y/dx
=
sin
x)
(cos
x)(1)
+[(csc
4csc
cos+
xcot
=
x
cos
x xx
1(csc
+x(
xx)(
tan
x
2
2
2 x)(
2sin
3 x)
12.
==
csc
x)
+ x)(2x)
(cot
xx)(
x)
x2x
csc
cot
x csc
⇧ (x)
20. ffd
dy/dx
csc
x
cot
=
csc
x)
+
(cot
x)(
csc
x
cot
x cot
2x, d y/dx
2
2x)] = csc3 x + csc
2
2
2
13.
(x)
=
=
, but
1 +xcot2 x = csc2 x
2
sec
x(tan
x
+
tan
x
sec
sec
x(tan
1)
(1
tan
x)(sec
x
tan
x)
(sec
x)(sec
x)
sec
x
tan
x
+
sec
x
tan
x
sec
x
19. dy/dx
=
x
sin
x
+
cos
x,
d
y/dx
=
x
cos
x
sin
x
sin
x
=
x
cos
x
2
sin
x
2
2
(x
+
1)
cot
x
2
2
2
2
2
sin
x
x
2 sec(1x+ csc
2tan
2 2
⇧⇧⇧
2 x)(sec(1x)
3 (tan sin
2
2
+
csc
x)
x)
2
2
tan
x
1
(1
+
x
tan
x[x(sec
x)
+
x)(1)]
11.
f
(x)
=
sec
x(sec
x)
+
(tan
x)(sec
x
tan
x)
=
sec
x
+
sec
x
tan
x
d
(cos
x)
+
x)(2x)]
sin
x) +
cos
x]==cot
40;sin
xso=f(2
x ) cosxx =tan
4(x + 1) sin
x
14.
18.
fff(x)
cos=
csc+x
x2[x(
x)(1/
x)
x),
2 [x
⇧y/dx
⇧(cos
16. =
(x)
=
2=sec
tan
x2x)
sec
x (sin
2 tan
sec
2sin
also,
(x)
(identity),
so2 x
2(1
2=
2x
2 =
2 xsec
(x)==
=sin
=(1
= 2 x) x = 21 (because
sec
15.
(x)
xxcot
+
cos
x(because
=
12 tan
(identity),
so
(x)
3csc
+
tan
(1cos
+2=fx
tan
x)
x)x
+
tan
x)
20.x)x(
32(1
xx)(
1)
csc
cos
xxcsc
cos
x2xxcos
(1
+
csc
csc+
x)
cot
x(0
x3=
cot
csc
x(
csc
xxcos
csc
2 + cot
22.
dy/dx
=
x
(
sin
x)
+
(cos
x)(2x)
+
4
=
sin
x
+
x
+
4
x,
⇧csc
2(sin
2(1 +3(
2
⇧⇧
x
tan
x)
(1
+
tan
x)
(1
+
x
tan
x)
21.
dy/dx
=
x(cos
x)
+
x)(1)
sin
x)
=
x
cos
x
+
4
sin
x,
(identity),
thus
cot
x
csc
x
=
1,
so
f
(x)
=
=
.
13. ff⇧ (x)
= 0.
, but
1 + cot x = csc2 x
(x)
=
22
3 2
2x)( csc22 x) +=
22 csc x 2cot x)] =
2
2(x
2 csc
22x
3 csc
2 =
20.
dy/dx
=
csc
x2 x)[2x
cot
x,2xcot
d2 y/dx
=
[(csc
(cot
x)(
csc
x x+ csc
x2cot
(3
cot
xsec
+x)
1)
x]
(x
+ 1)
cot
x csc
x x cot
6x+cot
xx)
3(x
+ 21)xcsc2 x
sec
x(tan
xx)(
+
tan
x)
sec
x(tan
xx)
1)
(1
+
x)csc
1csc
+
csc
x cot
(1(cos
+
csc
(1
2
2 =
12.
fdy/dx
(csc
csc
x)
+
(cot
x)(
csc
x
cot
csc
x
x
tan
x
=
1).
⇧ (x)
2
2
2
2
23.
=
(sin
x)(
sin
x)
+
x)(cos
x)
=
cos
x
sin
x,
2
⇧
fd
= = x [x
=x = (2 x ) cos x2 4(x2 + 21) sin x .
sin0.
x
sin
x
=2(x)
= 4+
(cosx)xx)+
x)(2x)]
2[x(
x)
x]
15.
cos
=
1(sin
(identity),
so
(x)2sin
=
⇧y/dx
2
2+
csc+xx(cos
1)
y/dx
=
x(
(cos
cos
+
cosxx4=sin
(3 +
x)+
16. fdf(x)
(x)==2sin
2 tan
sec
x
tan
sec
x
2 tan
xcot
sec
xf2xtan
2 5csc
0; also, fcsc
(x)x =(3sec cot
x x)tan x = 1 (identity), so
(1+
+
tan2xx)
(1
⇧== x)x23sin
2x)(1)
x sin
3x
(identity),
thus
cot
x
csc
x
=
1,
so
f
(x)
=
=
.3 2
2
2
cos
x
cos
2 x)]
2
2(sin
2+sin
2csc
17.
f
(x)
=
(because
x
sec
x
=
(sin
x)(1/
cos
x)
=
tan
x),
so
2
d
y/dx
=
(cos
x)(
sin
x)
(cos
x)(
sin
x)
[(sin
x)(cos
x)
+
x)(cos
=
4
sin
x cos x
(1
+
tan
x)(sec
x
tan
x)
(sec
x)(sec
x)
sec
x
tan
x
+
sec
x
tan
x
sec
x
⇧
(1
+
csc
x)
1
+
x
(1
csc
x)((sin
x) 3(
cot x(0
cotxx)
21. dy/dx
=1(x
x(cos
x)cot
x)(1)
sin x) csc
= xxcos
+ 42sincsc
x, x( csc x csc x + cot x)
1)
x csc
f⇧(x)
(x)=
=
0.
+2 x+tan
x+
2 =
2(cos
sin
x
sin
x = cot x), so2
14.
=
= , but2 1 + cot2 x = csc2 x
13.
fdy/dx
(x)
=
=sin
18.
f
(because
cos
x
csc
x
x)(1/
x)
2 x)
⇧=
⇧
2
2
23.
=
(sin
x)(
sin
+
(cos
x)(cos
x)
=
cos
x
sin
x,
2
2
2
2
15. fdy/dx
f (x)
cos
x
=
1(1
16.
(x) =
==
2(1
tan
sec
x2+
2 tan
sec
x f2=
2x + 2x
= 2x+
0;x+tan
also,
(x)
tan
x) xx)
(1
x)csc
22.
xsec
(xxx+cot
sin
x)
+(1
(cos
x)(2x)
+
4 so
cos
x(x)
=+=(tan
x0.sin
4tan
cos
2fx,
+
csc
(1
+
x) = sec 1x tan x = 1 (identity), so
xx
2sin
3 xcos
sec
+
tan
x)(sec
x)(identity),
tan
x[x(sec
df2⇧2⇧y/dx
=3x(
sin
x)
+2x
(cos
x)(1)
4 cos
xx)
=2cos
x3 x
sinx)(1)]
x +cos
5tan
cos
x+ sec
2x
2 tan
2
2 2+
(x) =
=
=
=2=sec34xsin x2cos2 x(because
sec2 x
+
x)(sec
tan
x)
(sec
x)(sec
x)
sec
x
x
x
tan
x
2 (1
csc
x(
csc
1)
csc
2 sin
tan
x
2
2
2
2
2
24.
dy/dx
=
sec
x,
d
y/dx
=
2
sec
x
tan
x
⇧(x)
2
2
⇧
2
f
0.
d
y/dx
=
(cos
x)(
x)
+
(cos
x)(
sin
x)
[(sin
x)(cos
x)
+
(sin
x)(cos
x)]
2
2
2
sec=
x(tan
x[x
+ cot
tan
x cot
sec
x)
sec
x(tan
1)
(3
cot
x)[2x
x(12 +
(x
+
1)
csc
x]xsin
(x
+
1)x)cot
xtan
csc
6x
cotxxx
2xxcot
x +3(x
+x1) csc2 x
xx
tan
x)
(1
+
x
tan
x)
(1
+
tan
x)
14.
fdf⇧(x)
=
=
y/dx
=
(cos
x)
+
(sin
x)(2x)]
+
2[x(
x)
+
cos
x]
4
sin
x
=
(2
)
cos
4(x
1)
sin
thus
x
csc
x
=
1,
so
f
(x)
=
=
.
17. (identity),
(because
sin
sec
x
=
(sin
x)(1/
cos
=
x),
so
x
sin xx)(1
=
2
f⇧ (x)
= x)12+ csc x 2
.
x)=2x sec2 x =2 2 sin
2 = 1 + x tan x
22
2(1x+ tan
x== 21).
16. ftan
(x)
sec
x tan
xx)
sec
2 tan
=+
0; tan
also,
f (x) = sec(3x cot
tanx)
x = 1 (identity), so
(1
tan
+x)
tan
x)x232 sin(1x2++csc
(3+ 4(1
cot
3x
22. dy/dx = x2tan
( +
sin
x)
+
(cos
x)(2x)
cos
x
=
2x
cos
x
+
4
cos
x,
cos
x
cos
2
2 2 x)2 2sec x(tan
2 x
x
2 +
2then
x(tan
x
x 2f ⇧=
sec
sec2so
x tan2 x
1
(1 tan
+
xx,
tan
x)(sec
x)
x[x(sec
x)2+1)
(tan x)(1)]
⇧⇧ sec
25.
Let
f=
(x)
=
(x)
= sec
x.x x= (sin
24.
dy/dx
=
x,
d
y/dx
2 sec
xsec
tan
17.
(because
sin
xtan
= tan
x),
ff(x)
(x)
=
0.sec
23. f=
dy/dx
=
(sin
x)(
sin
x)
+tan
(cos
x)(cos
x)
= cos
x2 cos
sin2x)
x,
2
2 x)(1/
=
=
=3 x
(because sec2 x
2 x
22+
2 (x) =
21
(1
x)(sec
x
x)
(sec
x)(sec
x)
sec
x
tan
x
+
sec
x
tan
xx)x22 )sec
+
x
tan
2
2
2sin x
2
⇧
(x
+
1)
cot
xx)
d
y/dx
=
[x
(cos
x)
+
(sin
x)(2x)]
+
2[x(
sin
x)
+
cos
x]
4
sin
x
=
(2
cos
x
4(x
+
1)
⇧(x)
(1
+
tan
(1
+
tan
x)
(1
+
x
tan
x)
(1
+
x
tan
(1
+
x
tan
x)
15.
f
=
sin
x
+
cos
x
=
1
(identity),
so
f
(x)
=
0.
14.
(x)
=
=
=
18. ff2(x)
=
(because
x),2 so 2 2
2x csc x =2(cos x)(1/ sin x) = cot
2 tan cos
2 2 (1
(1x)
x)
+ tan
x)
⇧ sin
x
tanx)]x= 4 sin x1cos x
+
xcot
tan
(tan
x)(1)]
dtan
(cos
x)(
+ (cos
x)(
sin
x)x) +[(sin
x)(cos
x) +(1sec
(sin
x)(cos
3tan
x fx)(sec
x(x)
==
1).
xtan
⇧+x)
2x[x(sec
⇧ y/dx
(a)
f
(0)
=
0
and
(0)
=
1,
so
y
0
=
(1)(x
0),
y
=
x.
25.
Let
f
=
tan
x,
then
f
(x)
=
sec
x.
f
(x)
=
=
=
(because
sec2 x
17. f (x) = 2
sin2 x sec x =
(sin
= tan
x), so
22(because
⇧ 2 x)(1/
2+ x tan x)
2 x cos2x)
2 (1
2 2
2
22fx]
2 2x cot
2
2 (x
(1
+
x
tan
x)
+
x
tan
x)
(1
sin
sin
x
15. dy/dx
f ⇧(x)
=
sin
x
+
cos
x
=
1
(identity),
so
(x)
=
0.
(3
cot
x)[2x
cot
x
+
1)
csc
(x
+
1)
cot
x
csc
x
6x
cot
x
x
3(x
+
1)
csc
x
1
+
tan
x
23.
=
(sin
x)(
sin
x)
+
(cos
x)(cos
=
cos
x
sin
x,
sec
x(tan
tan
x
sec
x)
sec
x(tan
x
1)
(x)
.
16. tan
ff ⇧(x)
2 sec
tan
sec2⌃x 2 ⌥ 2 tan
x sec2 x 2=2 ⌃3
2 3 = 0;=2 also, f (x)
= sec2 x tan2 x = 1 (identity),
so
2 =
=
⌃ (x
⌥2 +
⌥
2 x1)
2 xx
2=
x=
cot
2 =
2 (tan
(3
cot
x)
⌥1).
⌥ x)
⌥ x)(1)]
⌥ x tan2 x (3 cot 1x)2
cos
x
cos
x
24.
dy/dx
=
sec
x,
d
y/dx
2
sec
x
tan
x
sec
(1
+
x
tan
x)(sec
tan
x[x(sec
x)
+
(1
+
tan
x)
(1
+
tan
x)
⇧
2
2
⇧
2
18. d(b)
(because
cos
x
csc
x
(cos
x)(1/
sin
=
cot
x),
so
(a)
(0)
=
0
and
f
(0)
=
1,
so
y
0
=
(1)(x
0),
y
=
x.
y/dx
=
(cos
x)(
sin
x)
+
(cos
x)(
sin
x)
[(sin
x)(cos
x)
+
(sin
x)(cos
x)]
=
4
sin
x
cos
x
=
1
and
f
=
2,
so
y
1
=
2
x
,
y
=
2x
+
1.
ff⇧⇧(x)
(x)f=
=
=
=
(because
sec
x
= 0. 3 cot x
2 sin x 4 sin x
22
2 2
+ xxtan
x tanfx)
(12+x x tan
16. f (x) = 422sec
x tan x sec x4 (1
2 tan
secx)
x=
2 3 =(10;2+also,
(x)2 = sec
tanx)
x = 1 (identity), so
3 x2
(x
+
1)
cot
x
2
2
2
2
2
2
2
2
⇧
cos
cos
x
⇧
2
⌃
⌥
⌃
⌥
⌃
⌥
(3
cot
x)[2x
cot
x
(x
+
1)
csc
x]
(x
+
1)
cot
x
csc
x
6x
cot
x
2x
cot
x
3(x
+ 1) csc2 x
tan
x
=
1).
15.
f
(x)
=
sin
x
+
cos
x
=
1
(identity),
so
f
(x)
=
0.
25.
Let
f
(x)
=
tan
x,
then
f
(x)
=
sec
x.
⇧
18. fff(x)
⇧(x)=
⌥sec
⌥ ⌃2 seccos
⌥ =so
2⌥ x csc x = (cos x)(1/
⌃
⌥2 x,xd2 y/dx(because
⌃⌥ sin⌥x)⌥ = cot x),
0.
tan
⇧2 =
(x)f=
=
.
24.
dy/dx
=
x
tan
x
⌥
⌥
⌥
3
cot
x
=
1
and
f
=
2,
so
y
1
=
2
x
,
y
=
2x
+
1.
(b)
2 x)(1/ cos x) = tan x), so
17. (c)
f (x) f= 4
sec2,xcot
=x)
(3 cot x)2
= x1 (because
and f4⇧ sin 2x(3=
so
y(sin
+ 1 = 22 4x +
,y=
2x
1.
2 +
1(x
+
x
tan
2
2 2 sin x
2
2
2
2
4
4
4
2
sin xx csc x
(3 tan
x)[2x
cot x (x + 1) 2csc x] (x + 1) cot
6x cot x 2x cot
x 3(x
+ 1) csc x
+cot
1)
cot
x⇧ (0)
⇧⇧
xtan
(a)
(0)
=
and
fxx)(sec
1,2so
0csc
=xx(1)(x
y x)(1)]
=
x.x)
2 = sec2 x
2 (cos0),
16.
=
sec
sec
x=2 x)
tan
=
2sin
0;2x),
also,
f (x)
tan22x = 1 (identity),
so
(x)ff=
=
=
.
18. fLet
(x)
=
(because
cos
x
=
x)(1/
==sec
cot
2sec
x
tan
x
1cot
tanxthen
tan
x[x(sec
+3 (tan
⌃
⌃ 2(1
⌥0xxcot
⌥yxxsec
⌃
3⌥xtan
2 x)
17.
ff ⇧(x)
(because
sin
x=
(sin
x)(1/
cos
x)
=
x),
so so⌥
25.
(x)
=⌥3+
tan
x,
f ⇧ (x)
=
sec
x.cot
2
cos
x
cos
(3
x)
(3
x)
⌥
⌥
ff(c)
=
=
(because
sec
x
⇧
x
⇧ (x)f= 1 + x tan
2 y+1=2
2
2
1 andf⌃⇧f(x)
=
2,x)so
,y=
+ x)1.
(x)f (x)
=⌃0.=⌥sin=x, then
26. Let
cos
x.1)
x2 tan
(1
(1 cot
+ x2 tan
⌥(1=+
2 x⌥+
2 2x
(3
x)[2x
(x
+x[x(sec
csc22x]x) ⌃+(x(tan
+
1) cot
x+
6x
cot
2x
x x)
3(x2 + 1) csc2 x
2 x
2x
2
4 + cot
4 tan
4 x cscsec
2tan
⌥
⌥ x)(1)]
⌥xx tan
⇧ 2
⇧cot⌥
x
1
(1
x
tan
x)(sec
x)
f
(x)
=
=
.2
⇧
⇧
=
1
and
f
=
2,
so
y
1
=
2
x
,
y
=
2x
+
1.
(b)
f
tan
= 1).
f (x)x
2 (because sec x
(a)
f=
(0)
0 and f (0) =4 1, so y(3 0cot
= 2(1)(x
0), y4= x. =
x)2
x)
4=tan
2 x tan x)2 = (1(3+ x cot
2
x
(1
+
x
tan
x)
(1
+
tan
x)
⇧
⇧ 1, sin
(0) =
= sin
0 and
f(because
(0)f =
socos
yx sec
0), cos
y =x)
x.= tan x), so
17.
(x)2ff=
(sin x)(1/
26. f(a)
Let
x, then
(x) =
x.0x==(1)(x
2 ⌥
tan
x(x)
=1
1).
+
x tan
x x ⌃⌥⌥
⌥
⌃
⌃ ⌥⌥
⌃
⌃⌥ ⌥ ⌥ ⌥
(x
+
1) cot
⌥
⌥x), so⌥
⌥
⇧
18. (b)
f(c)
(x) f= (1 +=x
cos
sin
2+(cos
2 +
1tan
and
f (because
so
ycscsox1 y=
=
21+=
xx)(1/
, y x)
= ,=
2x
=
1x)(sec
and
f2⇧ x)= 2,
=xx[x(sec
2,
2 xx)(1)]
+
y cot
= 2x
sec
x+1.tan21.
x
1
tan
x)
(tan
3
cot
x
⇧
⇧
⇧ (⌥) =
4 24= 00 and
(⌥)
and
0 2=
( 1)(x
y4= =x + 2⌥. 2
f(b)
(x)f=
=
(because sec2 x
(a)
(0)
=4 1, 1,
so4 so
y y0=
(1)(x
0), y4⌥),
= x.
(x=
+ 1)
cotffx(0)
2
22
2
2
2
2
2
(1
+
x
tan
x)
(1
+
x
tan
x)
(1
+
x
tan
x)
18. f ⇧(x) =⌃ (3 ⌥ cot x)[2x (because
cos+x1)
csc
x =x](cos(xx)(1/
x)
cot x⌃ (x⌥
csc
+ 1)sin
cot
csccotxx), so
6x cot x 2x cot x 3(x + 1) csc2 x
⌃
⌥x=
2 = ⌥
3 cot x
f (x)
= ⌥
.
⇧ ⇧⌃ ⌥⌥
⌥
tan
x
=
1).
⌃
⌥
⌃
⌥
2
26. (c)
Let ff(x)⌥= sin=x,
cos
(3
x)
⇧ f f⇧(x)
1 then
⌥=
1x.
1= 22 x
1 + y = , y⌥x=+
1 fand
2,cot
so
y2 (+ 11)(x
+ 1 1. ⌥ (3 12 cot x)2 2
2=
2 2x
2
(b)
f
(⌥)
=
0
and
(⌥)
=
1,
so
y
0
=
⌥),
⌥.
cscy x] (x= + 1) cot
6x cot
(c)
f (3 =cot x)[2x
and cot
f x 4(x
= + 1)
, so
x x 2x+cot x. 3(x + 1) csc x
4x x csc , xy =
f ⇧ (x) = 4 42
= 22
.
4
4
2 x
2
2
4 2 (3 2cot x)2
(x + 1) cot
(3 2 cot x)2
⇧
(a)
f
(0)
=
0
and
f
(0)
=
1,
so
y
0
=
(1)(x
0),
y
=
x.
⌃
⌥
⌃
⌥
⌃
⌥
18. f (x) = ⌥
(because
cos1x csc x = (cos
⌥= cos
1 x)(1/1 sin x) =⌥cot x), so1
⌥
1
26. Let
=3 sincot
x,1 then
x. so y
xand ff⇧⇧(x)
(c) ffIf(x)
xx x sin
, yx=so y ⇧⇧x+ y = 2 cos
+ x. .
27. (a)
y 4= x=sin x2 then
y ⇧ =4sin=x2 + 2x,cos
x2 and 2y ⇧⇧=
=
2
cos
4 2x
2 cot x csc
2 cot22x 3(x2 + 1) csc2 x
(3 cot x)[2x cot x (x + 1) csc x] (x2 + 1)
6x2cot x 4 2x
⇧
f(b)
(x)f=(⌥) = 0 and f⇧ ⇧ (⌥) = 1, so y 0 = 2( 1)(x ⌥), y = x + ⌥.=
.
(a) f (0) = 0 and f (0) = 1, so y(3 0cot
= (1)(x
0), y = x.
x)
(3 cot x)2
(4)
⇧⇧
⇧
⇧⇧
⇧⇧
Di⇤erentiate
get xy x+
y x=so y2 cos
27. (b)
(a) If
thenresult
y ⌃= of
sinpart
x + x(a)
costwice
x andmore
y =to2 cos
sin
+ yx.= 2 cos x.
⌃y⌥=⌥ x sin1x the
⌥⌥
1
1
1 ⌃
⌥⌥
1
⌥
1
⇧
⇧ f
(c) ff (⌥) ==
and
=
, 0so=y ( 1)(x= ⌥), y x= x +, y⌥.=
x
+
.
(b)
0
and
f
(⌥)
=
1,
so
y
4
4
4
2(a)
2 x, to
⇧
⇧⇧
28. (a)
y = cos x 2then
= of
sinpart
x and
y ⇧⇧twice
= cos
so2yget
+yy(4)=+( y ⇧⇧cos
(cos
(b) IfDi⇤erentiate
the yresult
more
= x)22+cos
x.4 x)2= 0; if2 y = sin x then y ⇧ = cos x
⇧⇧
and y ⇧⇧⌃=
⌥ ⌥ sin x
1 so y +⇧ ⇧y⌃ ⌥=⌥( sin
1 x) + (sin x)1 =⇧⇧ 0. 1 ⌃
⌥⌥
1
⌥
1
27. (c)
(a) fIf y = x
thenfy = sin=x + x, cos
=sin x and
so yx and y == 2 cosxx x sin
, y x=so y ⇧⇧x+ y = 2 cos
+ x. .