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Section 3.3 Derivatives of Trigonometric Functions 2015 Kiryl Tsishchanka Derivatives of Trigonometric Functions THE DERIVATIVE OF THE SINE AND COSINE FUNCTIONS: We have d (sin x) = cos x dx or (sin x)′ = cos x d (cos x) = − sin x dx or (cos x)′ = − sin x and Proof: We have sin(x + h) − sin x h→0 h (sin x)′ = lim [We use sin(α + β) = sin α cos β + cos α sin β] sin x cos h − sin x cos x sin h sin x cos h + cos x sin h − sin x = lim + = lim h→0 h→0 h h h sin x(cos h − 1) cos h − 1 sin h sin h = lim = sin x lim + cos x · + cos x lim h→0 h→0 h→0 h h h h = sin x · 0 + cos x · 1 = cos x In the same way we prove that (cos x)′ = − sin x. EXAMPLE: If f (x) = 3 sin x − 4 cos x, then f ′ (x) = (3 sin x − 4 cos x)′ = 3(sin x)′ − 4(cos x)′ = 3 cos x − 4(− sin x) = 3 cos x + 4 sin x EXAMPLE: If f (x) = sin x, then f ′ (x) = cos x f ′′ (x) = (cos x)′ = − sin x f ′′′ (x) = (− sin x)′ = − cos x f ′′′′ (x) = (− cos x)′ = −(− sin x) = sin x Therefore f (x) = f (4) (x) = f (8) (x) = f (12) (x) = f (16) (x) = . . . = sin x f ′ (x) = f (5) (x) = f (9) (x) = f (13) (x) = f (17) (x) = . . . = cos x f ′′ (x) = f (6) (x) = f (10) (x) = f (14) (x) = f (18) (x) = . . . = − sin x f ′′′ (x) = f (7) (x) = f (11) (x) = f (15) (x) = f (19) (x) = . . . = − cos x For instance, it immediately follows from here that f (2010) (x) = f (2008+2) (x) = f (4·502+2) = f ′′ (x) = − sin x 1 Section 3.3 Derivatives of Trigonometric Functions 2015 Kiryl Tsishchanka DIFFERENTIATION RULES: c′ = 0 (xn )′ = nxn−1 [cf (x)]′ = cf ′ (x) (sin x)′ = cos x (csc x)′ = − csc x cot x [f (x) ± g(x)]′ = f ′ (x) ± g ′ (x) (cos x)′ = − sin x (sec x)′ = sec x tan x (tan x)′ = sec2 x (cot x)′ = − csc2 x [f (x)g(x)]′ = f ′ (x)g(x) + f (x)g ′ (x) ′ f (x) f ′ (x)g(x) − f (x)g ′ (x) = g(x) [g(x)]2 sec x (sec x)′ ??? ′ EXAMPLE: If f (x) = , then f (x) = WRONG!!! 1 + tan x (1 + tan x)′ ′ sec x (sec x)′ (1 + tan x) − sec x(1 + tan x)′ ′ f (x) = = 1 + tan x (1 + tan x)2 = sec x tan x(1 + tan x) − sec x sec2 x sec x tan x(1 + tan x) − sec x(1′ + (tan x)′ ) = (1 + tan x)2 (1 + tan x)2 sec x(tan x + tan2 x − sec2 x) sec x(tan x − 1) = = [tan2 x − sec2 x = −1] = 2 (1 + tan x) (1 + tan x)2 EXAMPLES: Differentiate 1. f (x) = 3x − 5 2. f (x) = √ x 1 3. f (x) = √ x √ 4. f (x) = −3x−8 + 2 x 5. f (x) = (x3 + 7x2 − 8)(2x−3 + x−4 ) 6. f (x) = 3x + 2 x+1 (x−5 + 1) 7. f (x) = x3 sin x − 5 cos x 8. f (x) = sin x + sec x 1 + x tan x 9. f (x) = sin x 1 + cos x 2 Section 3.3 Derivatives of Trigonometric Functions 2015 Kiryl Tsishchanka Solutions: 1. (3x − 5)′ = (3x)′ − 5′ = 3x′ − 5′ = 3 · 1 − 0 = 3 ′ 1 √ 1 2. ( x)′ = x1/2 = x1/2−1 = x−1/2 2 2 ′ ′ ′ 1 1 1 1 3. √ = x−1/2 = − x−1/2−1 = − x−3/2 = 1/2 x 2 2 x √ 1 4. (−3x−8 + 2 x)′ = −3(x−8 )′ + 2(x1/2 )′ = −3(−8)x−8−1 + 2 x−1/2 = 24x−9 + x−1/2 2 51 . [(x3 + 7x2 − 8)(2x−3 + x−4 )]′ = (x3 + 7x2 − 8)′ (2x−3 + x−4 ) + (x3 + 7x2 − 8)(2x−3 + x−4 )′ = (3x2 + 14x)(2x−3 + x−4 ) + (x3 + 7x2 − 8)(−6x−4 − 4x−5 ) 52 . [(x3 + 7x2 − 8)(2x−3 + x−4 )]′ = (2 + 15x−1 + 7x−2 − 16x−3 − 8x−4 )′ = 2′ + 15(x−1 )′ + 7(x−2 )′ − 16(x−3 )′ − 8(x−4 )′ = 0 + 15 · (−1)x−2 + 7 · (−2)x−3 − 16 · (−3)x−4 − 8 · (−4)x−5 = −15x−2 − 14x−3 + 48x−4 + 32x−5 ′ ′ 3x + 2 3x + 2 3x + 2 −5 −5 (x + 1) = (x−5 + 1)′ (x + 1) + x+1 x+1 x+1 3x + 2 (3x + 2)′ (x + 1) − (3x + 2)(x + 1)′ −5 (x + 1) + (x−5 + 1)′ = (x + 1)2 x+1 3(x + 1) − (3x + 2) −5 x−5 + 1 3x + 2 3x + 2 −6 = (−5x ) = (−5x−6 ) (x + 1) + + (x + 1)2 x+1 (x + 1)2 x+1 6. 7. (x3 sin x − 5 cos x)′ = (x3 sin x)′ − 5(cos x)′ = (x3 )′ sin x + x3 (sin x)′ − 5(cos x)′ = 3x2 sin x + x3 cos x + 5 sin x 8. 9. sin x + sec x 1 + x tan x sin x 1 + cos x ′ ′ = (sin x + sec x)′ (1 + x tan x) − (sin x + sec x)(1 + x tan x)′ (1 + x tan x)2 = [(sin x)′ + (sec x)′ ](1 + x tan x) − (sin x + sec x)(x′ tan x + x(tan x)′ ) (1 + x tan x)2 = (cos x + sec x tan x)(1 + x tan x) − (sin x + sec x)(tan x + x sec2 x) (1 + x tan x)2 = cos x(1 + cos x) − sin x(− sin x) (sin x)′ (1 + cos x) − sin x(1 + cos x)′ = (1 + cos x)2 (1 + cos x)2 = cos x + 1 1 cos x + cos2 x + sin2 x = = 2 2 (1 + cos x) (1 + cos x) 1 + cos x 3 Section 3.3 Derivatives of Trigonometric Functions 2015 Kiryl Tsishchanka Appendix cos h − 1 0 A (cos h − 1)(cos h + 1) A cos2 h − 1 lim = = lim = lim h→0 h→0 h→0 h(cos h + 1) h 0 h(cos h + 1) A −(1 − cos2 h) h→0 h(cos h + 1) = lim T − sin2 h h→0 h(cos h + 1) = lim A = lim h→0 C sin h − sin h · h cos h + 1 sin h − sin h · lim h→0 h h→0 cos h + 1 = lim − sin h h→0 cos h + 1 C = 1 · lim DSP = = − sin 0 1 + cos 0 0 1+1 =0 4