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Derivatives of Trigonometric Functions section 3.9 Consider the unit circle ; so we have OC = 1 in the picture below. △ △ Because of similarity between the two triangles OCD and OAB we have CD AB = OC OA ⇒ CD sin θ = 1 cos θ ⇒ CD = tan θ D B θ O A C Because of the equality CD = tan θ , the axis that passes through the points C and D is called the tangent axis. Now assume for a moment that the radius of the circle is r. The sector OBC faces the polar angle θ. angle area of associated sector 2π πr2 θ ? 1 ⇒ (θ)(πr2 ) 1 = θr2 2π 2 ?= area of sector with polar angle θ = 12 θr2 Note: This formula is true only if θ is measured in radians. Now let us assume for a moment that the above circle has radius 1. Then by putting r = 1 in the above formula we have: 1 area of sector OBC = θ 2 Now note that area of triangle OAB < area of sector OBC < area of triangle OCD 1 1 1 (OA)(AB) < θ < (OC)(CD) 2 2 2 1 1 1 (cos θ)(sin θ) < θ < (1)(tan θ) 2 2 2 multiply by 2 and divide by sin θ cos θ < θ 1 < sin θ cos θ Now let θ → 0+ and use the Sandwich Theorem to get lim θ→0+ θ =1 sin θ Then lim θ→0+ sin θ 1 1 = lim θ = + θ θ→0 limθ→0+ sin θ 2 θ sin θ = 1 =1 1 (∗) Now we calculate the left-hand limit: limθ→0− sin θ θ = limx→0+ sin(−x) −x = limx→0+ − sin x −x = limx→0+ sin x x = 1 change of variable x = −θ using the identity sin(−x) = − sin x using the identity (*) So , lim− θ→0 sin θ =1 θ (∗∗) From both (∗) and (∗∗) we have sin θ θ→0 θ lim =1 Example (section 3.9 exercise 33). Calculate lim θ→0 tan θ θ Solution. sin θ tan θ sin θ 1 = lim cos θ = lim θ→0 θ θ→0 θ θ→0 θ cos θ lim = lim θ→0 sin θ 1 1 lim = (1)( ) = 1 θ θ→0 cos θ 1 sin( x2 ) 1 x→∞ sin( x ) Example (section 3.9 exercise 36). Calculate lim 3 Solution. Using the change of variable y = 1 x we can write sin( x2 ) sin(2y) lim = lim = lim 1 x→∞ sin( ) y→0 y→0 sin(y) x sin(2y) y sin y y = lim 2 sin(2y) 2y y→0 sin y y = (2)(1) =2 1 We now calculate the derivative of the sine function: For this we use sin′ (x) = = = sin(x+∆x)−sin x ∆x ∆x→0 lim 2 ∆x→0 ∆x lim lim ∆x→0 [ sin sin( ∆x 2 ) ( ∆x 2 ) ( ∆x ) 2 ( cos x + ∆x 2 ( lim cos x + ∆x→0 )] ∆x 2 using a+b sin a − sin b = 2 sin( a−b 2 ) cos( 2 ) ) = (1) cos(x + 0) = cos x So d dx sin x = cos x Now we find the derivative of the cosine function: (π ) d d cos x = sin − x dx dx 2 = sin′ (π = cos (π ){ }′ − x π2 − x 2 2 = − cos ) − x {−1} (π 2 ) −x = − sin x So d dx cos x = − sin x 4 Now the derivative of the tangent function : tan′ x = d sin x dx cos x = {sin x}′ {cos x}−{sin x}{cos x}′ cos2 x = {cos x}{cos x}−{sin x}{− sin x} cos2 x = cos2 x+sin2 x cos2 x d dx = 1 cos2 x = sec2 x tan x = sec2 x Now the derivative of the cotangent function : cot′ x = d cos x dx sin x = {cos x}′ {sin x}−{cos x}{sin x}′ sin2 x = {− sin x}{sin x}−{cos x}{cos x} sin2 x = −(cos2 x+sin2 x) sin2 x d dx = −1 sin2 x cot x = − csc2 x 5 = − csc2 x Now the derivative of the secant function : sec′ x = d dx ( 1 cos x ) = {1}′ {cos x}−{1}{cos x}′ cos2 x = {0}{cos x}−{1}{− sin x} cos2 x = sin x cos2 x d dx = 1 sin x cos x cos x = sec x tan x sec x = sec x tan x Now the derivative of the cosecant function : csc′ x = d 1 dx sin x = {1}′ {sin x}−{1}{sin x}′ sin2 x = {0}{sin x}−{1}{cos x} sin2 x = − cos x sin2 x d dx = − sin1 x cos x sin x = − csc x cot x csc x = − csc x cot x 6 Putting all these together: d dx sin x d dx cos x = − sin x d dx tan x = sec2 x d dx cot x = − csc2 x d dx sec x = sec x tan x d dx csc x = − csc x cot x = cos x Note: These formulae are true only if the unit of measurement is radian. And , using the chain rule , we have the following formulae if interim variables are involved: d dx sin u (cos u) du dx d dx cos u = (− sin u) du dx d dx tan u = (sec2 u) du dx d dx cot u = (− csc2 u) du dx d dx sec u = (sec u tan u) du dx d dx csc u = (− csc u cot u) du dx = 7 Example. Find the equation of the tangent line to the curve y = sec x 1+tan x at the point where x=0. Solution. y′ = = {sec x}′ {1 + tan x} − {sec x}{1 + tan x}′ (1 + tan x)2 {sec x tan x}{1 + tan x} − {sec x}{sec2 x} (1 + tan x)2 ⇒ x=0 slope = (0)(1) − (1)(1) = −1 (1)2 Also note that at x = 0 we have y = 1. Therefore: tangent line ⇒ y − 1 = (−1)(x − 0) dy Example (from the textbook). Find dx for Solution. Let u = 4x. Then y = (tan u) 2 ⇒ dy dx = ⇒ y = −x + 1 y = tan2 (4x) { }{ } dy du du dx { }{ } du = 2(tan u)(sec2 u) dx { }{ } 2 = 2(tan u)(sec u) 4 = 8 tan(4x) sec2(4x) 8 ✓ power rule Example (from the textbook - pages 203 and 204). Find dy dx for x2 tan y + y sin x = 5 Solution. This is an implicit differentiation. If the “prime notation” is meant to denote differentiation with respect to x , then we have { } { } {x2 }′ {tan y} + {x2 }{tan y}′ + {y}′ {sin x} + {y}{sin x}′ = {5}′ } { } { {2x}{tan y} + {x2 }{(sec2 y) y ′ } + {y ′ }{sin x} + {y}{cos x} = 0 factor out the term y′ ⇒ ⇒ y′ = (2x tan y + y cos x) + (x2 sec2 y + sin x)y ′ = 0 −(2x tan y + y cos x) (x2 sec2 y + sin x) dy Example (section 3.9 exercise 30 - modified). Find dx for ✓ y = 1+tan3 (3x2 −4) x2 sin(x3 ) . Solution. Set u = 3x2 −4 and t = x3 . If the “prime notation” is meant to denote differentiation 9 with respect to x , then we have y = dy dx = 1+(tan(u))3 x2 sin t so then: {1+(tan u)3 }′ {x2 sin t}−{1+(tan u)3 }{x2 sin t}′ (x2 sin t)2 { 3(tan u)2 (tan u)′ }{ } { }{ = x4 sin2 t { }{ 3(tan u)2 (sec2 u)(u′ ) } { }{ x2 sin t − 1+(tan u)3 = {2x}{sin t}+{x2 }{(cos t)(t′ )} } x4 sin2 t { }{ } { }{ x2 sin t − 1+(tan u)3 3(tan u)2 (sec2 u)(6x) = } {2x}{sin t}+{x2 }{(cos t)(3x2 )} x4 sin2 t { } { 18x3 tan2 u sec2 u sin t = {x2 }′ {sin t}+{x2 }{sin t}′ x2 sin t − 1+(tan u)3 } − }{ 1+tan3 u x4 sin2 t where u = 3x2 − 4 and t = x3 10 2x sin t+3x4 cos t }