Trigonometric Differentiation Exercises-1

UniversitasSaskatchewanensi

s

DEO

ET

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2003 Doug MacLean

Trigonometric Differentiation — Exercise Set I.

Find the derivative f ′(x) for the following functions. Say what you can about the sign of f ′(x).

(I.1) f(x) = tan3(4x) Solution

(I.2) f(x) = tan((4x)3

)Solution

(I.3) f(x) = cot(√x)

Solution

(I.4) f(x) = sec2((2x + 1)2

)Solution

(I.5) f(x) = csc3 x − sec3 x Solution

(I.6) f(x) = x3 tan(2x)− x2 sec(3x) Solution

(I.7) f(x) = sec2 x cscx Solution

(I.8) f(x) = x tan(

1x

)Solution

(I.9) f(x) = x2 tan(

1x

)Solution

(I.10) f(x) = x3 tan(

1x

)Solution

Trigonometric Differentiation Exercises-2

UniversitasSaskatchewanensi

s

DEO

ET

PAT-

RIÆ

2003 Doug MacLean

Trigonometric Differentiation — Exercise Set II.

Find the derivative f ′(x) for the following functions. Say what you can about the sign of f ′(x).

(II.1) f(x) = x tanx Solution

(II.2) f(x) = tan2 x Solution

(II.3) f(x) = tan3 x Solution

(II.4) f(x) = tan4 x Solution

(II.5) f(x) = tan5 x Solution

(II.6) f(x) = 1− cotx1+ cotx

Solution

(II.7) f(x) = 1+ cotx1− cotx

Solution

(II.8) f(x) = 1− tanx1+ tanx

Solution

(II.9) f(x) = 1+ tanx1− tanx

Solution

(II.10) f(x) = tanx1− tanx

Solution

Trigonometric Differentiation Exercises-3

UniversitasSaskatchewanensi

s

DEO

ET

PAT-

RIÆ

2003 Doug MacLean

Trigonometric Differentiation — Exercise Set III.

Find the derivative f ′(x) for the following functions. Say what you can about the sign of f ′(x).

(III.1) f(x) = sin(sinx) Solution

(III.2) f(x) = tan2(cosx) Solution

(III.3) f(x) = cot(tan3 x) Solution

(III.4) f(x) = csc(cot4 x) Solution

(III.5) f(x) = sec(csc5 x) Solution

(III.6) f(x) = sin(sin(sinx)) Solution

(III.7) f(x) = cos(cosx) Solution

(III.8) f(x) = cos(cos(cosx)) Solution

(III.9) f(x) = tan(cot(tanx)) Solution

(III.10) f(x) = csc(

11+ x2

)Solution

Trigonometric Differentiation Exercises-4

UniversitasSaskatchewanensi

s

DEO

ET

PAT-

RIÆ

2003 Doug MacLean

Solution Set IBack to Questions

(I.1) f(x) = tan3(4x)

Solution: f ′(x) = 3 tan2(4x)(tan 4x)′ = 3 tan2(4x) sec2(4x)(4x)′ = 12 tan2(4x) sec2(4x) ≥ 0

(I.2) f(x) = tan((4x)3

)

Solution: f ′(x) = sec2((4x)3

)((4x)3

)′ = sec2((4x)3

)(3(4x)2

)(4x)′ = sec2

((4x)3

)(3(4x)2

)(4) =

12(4x)2 sec2((4x)3

) ≥ 0

(I.3) f(x) = cot(√x)

Solution: f(x) = cot(x

12

)

f ′(x) = − csc2(x

12

)(x

12

)′ = − csc2(x

12

) 12x−

12 = − 1

2√x

csc2 (√x) < 0

Trigonometric Differentiation Exercises-5

UniversitasSaskatchewanensi

s

DEO

ET

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2003 Doug MacLean

Back to Questions

(I.4) f(x) = sec2((2x + 1)2

)

Solution: f ′(x) = 2 sec((2x + 1)2

)(sec

((2x + 1)2

))′ =2 sec

((2x + 1)2

)sec

((2x + 1)2

)tan

((2x + 1)2

)((2x + 1)2

)′ =2 sec2

((2x + 1)2

)tan

((2x + 1)2

)2(2x + 1)(2x + 1)′ =

2 sec2((2x + 1)2

)tan

((2x + 1)2

)2(2x + 1)(2) = 8(2x + 1) sec2

((2x + 1)2

)tan

((2x + 1)2

)

(I.5) f(x) = csc3 x − sec3 x

Solution: f ′(x) = 3 csc2 x(cscx)′ − 3 sec2 x(secx)′ =3 csc2 x(− cscx cotx)− 3 sec2 x(secx tanx) = −3 csc3 x cotx − 3 sec3 x tanx

(I.6) f(x) = x3 tan(2x)− x2 sec(3x)

Solution: f ′(x) = (x3)′ tan(2x)+ x3(tan(2x))′ − (x2)′ sec(3x)− x2(sec(3x))′ =(3x2) tan(2x)+ x3(sec2(2x))(2x)′ − (2x) sec(3x)− x2(sec(3x) tan(3x))(3x)′ =(3x2) tan(2x)+ x3(sec2(2x))(2)− (2x) sec(3x)− x2(sec(3x) tan(3x))(3) =(3x2) tan(2x)+ 2x3(sec2(2x))− 2x sec(3x)− 3x2(sec(3x) tan(3x))

Trigonometric Differentiation Exercises-6

UniversitasSaskatchewanensi

s

DEO

ET

PAT-

RIÆ

2003 Doug MacLean

Back to Questions

(I.7) f(x) = sec2 x cscx

Solution: f ′(x) =(sec2 x

)′cscx + sec2 x (cscx)′ = 2 secx (secx)′ cscx + sec2 x(− cscx cotx) =

2 secx(secx tanx) cscx − sec2 x cscx cotx = 2 sec2 x tanx cscx − sec2 x cscx cotx = 2 sec2 xsinxcosx

1sinx

−sec2 x

1sinx

cosxsinx

= 2 sec2 x1

cosx− secx csc2 x = 2 sec3 x − secx csc2 x = secx

(2 sec2 x − csc2 x

)

(I.8) f(x) = x tan(

1x

)

Solution: Back to Questions

f ′(x) = (x)′ tan(

1x

)+ x

(tan

(1x

))′= tan

(1x

)+ x sec2

(1x

)(1x

)′=

tan(

1x

)+ x sec2

(1x

) −1x2

= tan(

1x

)− 1x

sec2(

1x

)

(I.9) f(x) = x2 tan(

1x

)

Solution: f ′(x) = (x2)′ tan(

1x

)+ x2

(tan

(1x

))′= 2x tan

(1x

)+ x2 sec2

(1x

)(1x

)′=

2x tan(

1x

)+ x2 sec2

(1x

) −1x2

= 2x tan(

1x

)− sec2

(1x

)

Trigonometric Differentiation Exercises-7

UniversitasSaskatchewanensi

s

DEO

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2003 Doug MacLean

Back to Questions

(I.10) f(x) = x3 tan(

1x

)

Solution: f ′(x) = (x3)′ tan(

1x

)+ x3

(tan

(1x

))′= 3x2 tan

(1x

)+ x3 sec2

(1x

)(1x

)′=

3x2 tan(

1x

)+ x3 sec2

(1x

) −1x2

= 3x2 tan(

1x

)− x sec2

(1x

)

Trigonometric Differentiation Exercises-8

UniversitasSaskatchewanensi

s

DEO

ET

PAT-

RIÆ

2003 Doug MacLean

Solution Set IIBack to Questions

(II.1) f(x) = x tanx

Solution: f ′(x) = (x)′ tanx + x(tanx)′ = tanx + x sec2 x

(II.2) f(x) = tan2 x

Solution: f ′(x) = 2 tanx(tanx)′ = 2 tanx sec2 x

(II.3) f(x) = tan3 x

Solution: Back to Questions

f ′(x) = 3 tan2 x(tanx)′ = 3 tan2 x sec2 x > 0

(II.4) f(x) = tan4 x

Solution: f ′(x) = 4 tan3 x(tanx)′ = 4 tan3 x sec2 x

Trigonometric Differentiation Exercises-9

UniversitasSaskatchewanensi

s

DEO

ET

PAT-

RIÆ

2003 Doug MacLean

Back to Questions

(II.5) f(x) = tan5 x

Solution: f ′(x) = 5 tan4 x(tanx)′ = 5 tan4 x sec2 x > 0

(II.6) f(x) = 1− cotx1+ cotx

Solution: f ′(x) = (1− cotx)′(1+ cotx)− (1− cotx)(1+ cotx)′

(1+ cotx)2=

−(− csc2 x)(1+ cotx)− (1− cotx)(− csc2 x)(1+ cotx)2

= csc2 x(1+ cotx)+ (1− cotx)

(1+ cotx)2= 2 csc2 x(1+ cotx)2

> 0

(II.7) f(x) = 1+ cotx1− cotx

Solution:

f ′(x) = (1+ cotx)′(1− cotx)− (1+ cotx)(1− cotx)′

(1− cotx)2= − csc2 x(1− cotx)− (1+ cotx)(−(− csc2 x))

(1− cotx)2=

− csc2 x(1− cotx)+ (1+ cotx)

(1− cotx)2= − 2 csc2 x

(1− cotx)2< 0

Trigonometric Differentiation Exercises-10

UniversitasSaskatchewanensi

s

DEO

ET

PAT-

RIÆ

2003 Doug MacLean

(II.8) f(x) = 1− tanx1+ tanx

Solution: Back to Questions

f ′(x) = (1− tanx)′(1+ tanx)− (1− tanx)(1+ tanx)′

(1+ tanx)2= − sec2 x(1+ tanx)− (1− tanx)(sec2 x)

(1+ tanx)2=

sec2 x−(1+ tanx)− (1− tanx)

(1+ tanx)2= −2 sec2 x(1+ tanx)2

(II.9) f(x) = 1+ tanx1− tanx

f ′(x) = (1+ tanx)′(1− tanx)− (1+ tanx)(1− tanx)′

(1− tanx)2= sec2 x(1− tanx)− (1+ tanx)(− sec2 x)

(1− tanx)2=

2 sec2 x(1− tanx)2

(II.10) f(x) = tanx1− tanx

Solution:

f ′(x) = (tanx)′(1− tanx)− tanx(1− tanx)′

(1− tanx)2= sec2 x(1− tanx)− tanx(− sec2 x)

(1− tanx)2= sec2 x(1− tanx)2

> 0

Trigonometric Differentiation Exercises-11

UniversitasSaskatchewanensi

s

DEO

ET

PAT-

RIÆ

2003 Doug MacLean

Solution Set IIIBack to Questions

(III.1) f(x) = sin(sinx)

Solution: f ′(x) = (cos(sinx))(sinx)′ = (cos(sinx))(cosx) = cosx cos(sinx)

(III.2) f(x) = tan2(cosx)

Solution: f ′(x) = 2 tan(cosx)(tan(cosx))′ = 2 tan(cosx)(sec2(cosx))(cosx)′ =2 tan(cosx)(sec2(cosx))(− sinx) = −2 sinx tan(cosx) sec2(cosx)

(III.3) f(x) = cot(tan3 x)

Solution: f ′(x) = − csc2(tan3 x)(tan3 x)′ = − csc2(tan3 x)(3 tan2 x(tanx)′) =− csc2(tan3 x)(3 tan2 x(sec2 x) = −3 tan2 x sec2 x csc2(tan3 x) < 0

(III.4) f(x) = csc(cot4 x)

Solution: f ′(x) = − csc(cot4 x) cot(cot4 x)(cot4 x)′ = − csc(cot4 x) cot(cot4 x)(4 cot3 x)(cotx)′ =− csc(cot4 x) cot(cot4 x)(4 cot3 x)(− csc2 x) = 4 cot3 x csc2 x csc(cot4 x) cot(cot4 x)

Trigonometric Differentiation Exercises-12

UniversitasSaskatchewanensi

s

DEO

ET

PAT-

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2003 Doug MacLean

Back to Questions

(III.5) f(x) = sec(csc5 x)

Solution: f ′(x) = sec(csc5 x) tan(csc5 x)(csc5 x)′ = sec(csc5 x) tan(csc5 x)(5 csc4 x)(cscx)′ =sec(csc5 x) tan(csc5 x)(5 csc4 x)(− cscx cotx) = −5 csc5 x cotx sec(csc5 x) tan(csc5 x)

(III.6) f(x) = sin(sin(sinx))

Solution: f ′(x) = cos(sin(sinx))(sin(sinx))′ = cos(sin(sinx))(cos(sinx))(sinx)′ =cos(sin(sinx))(cos(sinx))(cosx) = cosx cos(sinx) cos(sin(sinx))

(III.7) f(x) = cos(cosx)

Solution: f ′(x) = − sin(cosx)(cosx)′ = − sin(cosx)(− sinx) = sinx sin(cosx)

(III.8) f(x) = cos(cos(cosx))

Solution: f ′(x) = − sin(cos(cosx))(cos(cosx))′ = − sin(cos(cosx))(− sin(cosx))(cosx)′ =− sin(cos(cosx))(− sin(cosx))(− sinx) = − sinx sin(cosx) sin(cos(cosx))

Trigonometric Differentiation Exercises-13

UniversitasSaskatchewanensi

s

DEO

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2003 Doug MacLean

Back to Questions

(III.9) f(x) = tan(cot(tanx))

f ′(x) = sec2(cot(tanx))(cot(tanx))′ = sec2(cot(tanx))(− csc2(tanx))(tanx)′ =sec2(cot(tanx))(− csc2(tanx))(sec2 x) = − sec2 x csc2(tanx) sec2(cot(tanx)) < 0

(III.10) f(x) = csc(

11+ x2

)

Solution:f ′(x) = − csc

(1

1+ x2

)cot

(1

1+ x2

)((1+ x2)−1

)′ = − csc(

11+ x2

)cot

(1

1+ x2

)((−1)(1+ x2)−2(2x)

)=

−4x(1+ x2)2

csc(

11+ x2

)cot

(1

1+ x2

)