Trigonometric Differentiation Exercises-1 Sa Trigonometric Differentiation — Exercise Set I. sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean Find the derivative f (x) for the following functions. Say what you can about the sign of f (x). (I.1) (I.2) (I.3) (I.4) (I.5) (I.6) (I.7) (I.8) (I.9) (I.10) f (x) = tan3 (4x) f (x) = tan (4x)3 f (x) = cot √ x f (x) = sec2 (2x + 1)2 f (x) = csc3 x − sec3 x f (x) = x 3 tan(2x) − x 2 sec(3x) f (x) = sec2 x csc x 1 f (x) = x tan x 1 f (x) = x 2 tan x 1 3 f (x) = x tan x Solution Solution Solution Solution Solution Solution Solution Solution Solution Solution Trigonometric Differentiation Exercises-2 Sa Trigonometric Differentiation — Exercise Set II. sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean Find the derivative f (x) for the following functions. Say what you can about the sign of f (x). (II.1) (II.2) (II.3) (II.4) (II.5) (II.6) (II.7) (II.8) (II.9) (II.10) f (x) = x tan x Solution f (x) = tan2 x Solution f (x) = tan3 x Solution f (x) = tan4 x Solution f (x) = tan5 x Solution f (x) = 1 − cot x 1 + cot x Solution f (x) = 1 + cot x 1 − cot x Solution f (x) = 1 − tan x 1 + tan x Solution f (x) = 1 + tan x 1 − tan x Solution f (x) = tan x 1 − tan x Solution Trigonometric Differentiation Exercises-3 Sa Trigonometric Differentiation — Exercise Set III. sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean Find the derivative f (x) for the following functions. Say what you can about the sign of f (x). (III.1) (III.2) (III.3) (III.4) (III.5) (III.6) (III.7) (III.8) (III.9) (III.10) f (x) = sin(sin x) Solution f (x) = tan2 (cos x) Solution f (x) = cot(tan3 x) Solution f (x) = csc(cot4 x) Solution f (x) = sec(csc5 x) Solution f (x) = sin(sin(sin x)) f (x) = cos(cos x) Solution Solution f (x) = cos(cos(cos x)) Solution f (x) = tan(cot(tan x)) Solution 1 f (x) = csc 1 + x2 Solution Trigonometric Differentiation Exercises-4 f (x) = tan3 (4x) Solution: (I.2) f (x) = 3 tan2 (4x)(tan 4x) = 3 tan2 (4x) sec2 (4x)(4x) = 12 tan2 (4x) sec2 (4x) ≥ 0 f (x) = tan (4x)3 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 Solution: (I.3) f (x) = cot f (x) = − csc 2 √ x 1 f (x) = cot x 2 Solution: x 1 2 PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean Back to Questions (I.1) Sa Solution Set I sk DEO sis iversitas Un x 1 2 2 = − csc x 1 2 1 √ 1 1 x − 2 = − √ csc2 x <0 2 2 x Trigonometric Differentiation Exercises-5 (I.4) f (x) = sec2 (2x + 1)2 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 Solution: (I.5) f (x) = csc3 x − sec3 x Solution: f (x) = 3 csc2 x(csc x) − 3 sec2 x(sec x) = 3 csc2 x(− csc x cot x) − 3 sec2 x(sec x tan x) = −3 csc3 x cot x − 3 sec3 x tan x (I.6) f (x) = x 3 tan(2x) − x 2 sec(3x) Solution: f (x) = (x 3 ) tan(2x) + x 3 (tan(2x)) − (x 2 ) sec(3x) − x 2 (sec(3x)) = (3x 2 ) tan(2x) + x 3 (sec2 (2x))(2x) − (2x) sec(3x) − x 2 (sec(3x) tan(3x))(3x) = (3x 2 ) tan(2x) + x 3 (sec2 (2x))(2) − (2x) sec(3x) − x 2 (sec(3x) tan(3x))(3) = (3x 2 ) tan(2x) + 2x 3 (sec2 (2x)) − 2x sec(3x) − 3x 2 (sec(3x) tan(3x)) Sa Back to Questions sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean Trigonometric Differentiation Exercises-6 (I.7) Sa Back to Questions sk f (x) = sec2 x csc x f (x) = sec2 x csc x + sec2 x (csc x) = 2 sec x (sec x) csc x + sec2 x(− csc x cot x) = sin x 1 − 2 sec x(sec x tan x) csc x − sec2 x csc x cot x = 2 sec2 x tan x csc x − sec2 x csc x cot x = 2 sec2 x cos x sin x 1 cos x 1 = 2 sec2 x − sec x csc2 x = 2 sec3 x − sec x csc2 x = sec x 2 sec2 x − csc2 x sec2 x sin x sin x cos x 1 f (x) = x tan x Solution: Back to Questions 1 1 1 1 2 1 f (x) = (x) tan + x tan + x sec = tan = x x x x x 1 1 1 1 −1 1 tan + x sec2 − sec2 = tan 2 x x x x x x (I.9) PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean Solution: (I.8) DEO sis iversitas Un f (x) = x 2 tan 1 x 1 1 1 1 2 2 2 1 + x tan + x sec Solution: f (x) = (x ) tan = 2x tan = x x x x x 1 −1 1 2 2 1 2 1 2x tan + x sec − sec = 2x tan x x x2 x x 2 Trigonometric Differentiation Exercises-7 (I.10) Sa Back to Questions sk 1 f (x) = x tan x 1 1 1 1 3 2 3 2 1 + x tan + x sec Solution: f (x) = (x ) tan = 3x tan = x x x x x −1 1 1 2 3 2 1 2 2 1 + x sec − x sec = 3x tan 3x tan x x x2 x x 3 PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean 3 DEO sis iversitas Un Trigonometric Differentiation Exercises-8 Back to Questions (II.1) f (x) = x tan x Solution: (II.2) f (x) = tan2 x Solution: (II.3) f (x) = (x) tan x + x(tan x) = tan x + x sec2 x f (x) = 2 tan x(tan x) = 2 tan x sec2 x f (x) = tan3 x Solution: Back to Questions f (x) = 3 tan2 x(tan x) = 3 tan2 x sec2 x > 0 (II.4) f (x) = tan4 x Solution: f (x) = 4 tan3 x(tan x) = 4 tan3 x sec2 x Sa Solution Set II sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean Trigonometric Differentiation Exercises-9 (II.5) sk f (x) = tan5 x f (x) = 5 tan4 x(tan x) = 5 tan4 x sec2 x > 0 f (x) = Solution: 1 − cot x 1 + cot x f (x) = (1 − cot x) (1 + cot x) − (1 − cot x)(1 + cot x) = (1 + cot x)2 −(− csc2 x)(1 + cot x) − (1 − cot x)(− csc2 x) (1 + cot x) + (1 − cot x) 2 csc2 x 2 = csc x = >0 (1 + cot x)2 (1 + cot x)2 (1 + cot x)2 (II.7) DEO PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean Solution: (II.6) Sa Back to Questions sis iversitas Un f (x) = 1 + cot x 1 − cot x Solution: f (x) = (1 + cot x) (1 − cot x) − (1 + cot x)(1 − cot x) − csc2 x(1 − cot x) − (1 + cot x)(−(− csc2 x)) = = (1 − cot x)2 (1 − cot x)2 − csc2 x (1 − cot x) + (1 + cot x) 2 csc2 x = − <0 (1 − cot x)2 (1 − cot x)2 Trigonometric Differentiation Exercises-10 Solution: f (x) = sec2 x Back to Questions (1 − tan x) (1 + tan x) − (1 − tan x)(1 + tan x) − sec2 x(1 + tan x) − (1 − tan x)(sec2 x) = = (1 + tan x)2 (1 + tan x)2 −(1 + tan x) − (1 − tan x) −2 sec2 x = (1 + tan x)2 (1 + tan x)2 (II.9) f (x) = f (x) = 1 + tan x 1 − tan x (1 + tan x) (1 − tan x) − (1 + tan x)(1 − tan x) sec2 x(1 − tan x) − (1 + tan x)(− sec2 x) = = (1 − tan x)2 (1 − tan x)2 2 sec2 x (1 − tan x)2 (II.10) f (x) = tan x 1 − tan x Solution: f (x) = sec2 x (tan x) (1 − tan x) − tan x(1 − tan x) sec2 x(1 − tan x) − tan x(− sec2 x) = = >0 (1 − tan x)2 (1 − tan x)2 (1 − tan x)2 Sa (II.8) 1 − tan x f (x) = 1 + tan x sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean Trigonometric Differentiation Exercises-11 Back to Questions (III.1) Solution: (III.2) Solution: f (x) = sin(sin x) f (x) = (cos(sin x))(sin x) = (cos(sin x))(cos x) = cos x cos(sin x) f (x) = tan2 (cos x) f (x) = 2 tan(cos x)(tan(cos x)) = 2 tan(cos x)(sec2 (cos x))(cos x) = 2 tan(cos x)(sec2 (cos x))(− sin x) = −2 sin x tan(cos x) sec2 (cos x) (III.3) Solution: f (x) = cot(tan3 x) f (x) = − csc2 (tan3 x)(tan3 x) = − csc2 (tan3 x)(3 tan2 x(tan x) ) = − csc2 (tan3 x)(3 tan2 x(sec2 x) = −3 tan2 x sec2 x csc2 (tan3 x) < 0 (III.4) Solution: f (x) = csc(cot4 x) f (x) = − csc(cot4 x) cot(cot4 x)(cot4 x) = − csc(cot4 x) cot(cot4 x)(4 cot3 x)(cot x) = − csc(cot4 x) cot(cot4 x)(4 cot3 x)(− csc2 x) = 4 cot3 x csc2 x csc(cot4 x) cot(cot4 x) Sa Solution Set III sk DEO PAT- ET RIÆ atc h sis iversitas Un e w ane n 2003 Doug MacLean Trigonometric Differentiation Exercises-12 (III.5) Solution: Solution: f (x) = sec(csc5 x) f (x) = sec(csc5 x) tan(csc5 x)(csc5 x) = sec(csc5 x) tan(csc5 x)(5 csc4 x)(csc x) = f (x) = sin(sin(sin x)) f (x) = cos(sin(sin x))(sin(sin x)) = cos(sin(sin x))(cos(sin x))(sin x) = cos(sin(sin x))(cos(sin x))(cos x) = cos x cos(sin x) cos(sin(sin x)) (III.7) Solution: (III.8) Solution: sk DEO PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean sec(csc5 x) tan(csc5 x)(5 csc4 x)(− csc x cot x) = −5 csc5 x cot x sec(csc5 x) tan(csc5 x) (III.6) Sa Back to Questions sis iversitas Un f (x) = cos(cos x) f (x) = − sin(cos x)(cos x) = − sin(cos x)(− sin x) = sin x sin(cos x) f (x) = cos(cos(cos x)) f (x) = − sin(cos(cos x))(cos(cos x)) = − sin(cos(cos x))(− sin(cos x))(cos x) = − sin(cos(cos x))(− sin(cos x))(− sin x) = − sin x sin(cos x) sin(cos(cos x)) Trigonometric Differentiation Exercises-13 (III.9) Sa Back to Questions sk PAT- ET RIÆ atc h e w ane n 2003 Doug MacLean f (x) = tan(cot(tan x)) f (x) = sec2 (cot(tan x))(cot(tan x)) = sec2 (cot(tan x))(− csc2 (tan x))(tan x) = sec2 (cot(tan x))(− csc2 (tan x))(sec2 x) = − sec2 x csc2 (tan x) sec2 (cot(tan x)) < 0 (III.10) Solution: 1 f (x) = csc 1 + x2 1 1 1 1 2 −1 2 −2 f (x) = − csc ) = − csc ) (2x) = cot (1 + x cot (−1)(1 + x 1 + x2 1 + x2 1 + x2 1 + x2 1 1 −4x csc cot (1 + x 2 )2 1 + x2 1 + x2 DEO sis iversitas Un
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