Differentiation of Trigonometric Functions MODULE - V DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS Trigonometry is the branch of Mathematics that has mae itself inispensable for other branches of higher Mathematics may it be calculus, vectors, three imensional geometry, functions-harmonic an simple an otherwise just cannot be processe without encountering trigonometric functions. Further within the specific limit, trigonometric functions give us the inverses as well. The question now arises : Are all the rules of fining the erivatives stuie by us so far appliacable to trigonometric functions? This is what we propose to eplore in this lesson an in the process, evelop the formulae or results for fining the erivatives of trigonometric functions an their inverses. In all iscussions involving the trignometric functions an their inverses, raian measure is use, unless otherwise specifically mentione.. OBJECTIVES After stuing this lesson, you will be able to : fin the erivative of trigonometric functions from first principle; fin the erivative of inverse trigonometric functions from first principle; apply prouct, quotient an chain rule in fining erivatives of trigonometric an inverse trigonometric functions; an fin secon orer erivative of a function. EXPECTED BACKGROUND KNOWLEDGE Knowlege of trigonometric ratios as functions of angles. Stanar limits of trigonometric functions namely. (i) tan lim 0 (ii) lim (iii) limcos (iv) lim 0 0 0 0 Definition of erivative, an rules of fining erivatives of function. MATHEMATICS 5
MODULE - V Differentiation of Trigonometric Functions. DERIVATIVE OF TRIGONOMETRIC FUNCTIONS FROM FIRST PRINCIPLE (i) Let y For a small increment δ in, let the corresponing increment in y be δ y. y+δ y ( +δ) an δ y (+δ) δ δ cos + C+ D C D C D cos δ δy δ cos + δ δ δ δy δ lim lim cos + lim δ δ cos. δ 0 δ 0 δ 0 δ 0 δ lim δ Thus, cos i.e., () cos (ii) Let y cos For a small increment δ y+δ y cos( +δ) an δ y cos( +δ) cos in, let the corresponing increment in y be δ y. δ δ + δ δy δ + δ δ δ δy lim lim + lim δ δ δ 0 δ 0 δ 0 Thus, 5 MATHEMATICS
Differentiation of Trigonometric Functions i.e., ( cos ) (iii) Let y tan For a small increament δ y+ δ y tan( +δ ) in, let the corresponing increament in y be δ y. an δ y tan( +δ) tan (+δ) cos(+δ) cos MODULE - V (+δ) cos.cos(+δ) cos(+δ)cos [(+δ) ] cos(+δ)cos δ cos(+δ) cos δy δ δ δ cos(+δ)cos Thus, or sec δy δ lim lim lim δ δ cos(+δ)cos δ 0 δ 0 δ 0 cos sec δ lim δ δ 0 i.e. (tan) sec y sec (iv) Let For a small increament δ y+δ y sec( +δ) an δ y sec(+δ) sec in, let the corresponing increament in y be δ y. cos(+δ) cos cos cos(+δ) cos(+δ)cos δ δ + cos(+δ)cos MATHEMATICS 53
MODULE - V δ + δ δy lim lim δ cos(+δ)cos δ δ 0 δ 0 δ + δ δy lim lim lim δ cos( +δ)cos δ δ 0 δ 0 δ 0 Differentiation of Trigonometric Functions Thus, cos tan.sec cos cos sec.tan i.e. (sec) sec tan Similarly, we can show that (cot) cosec an (cosec) cosec cot Eample. Fin the erivative of Solution : y cot For a small increament δ y+δ y cot( +δ) cot from first principle. in, let the corresponing increament in y be δ y. an δ y cot(+δ) cot cos(+δ) cos ( +δ) cos(+δ) cos (+δ) ( +δ) [ ( +δ) ] ( +δ) [ δ () δ ] ( +δ) 54 MATHEMATICS
Differentiation of Trigonometric Functions [( +δ) δ] ( +δ) MODULE - V δy [( +δ) δ] δ δ ( +δ) δ y [(+δ) δ ] +δ an lim lim lim δ 0δ δ 0 δ (+δ ) δ 0 ( +δ) or Hence. ( ).cosec (cot ) cosec [(+δ) δ] lim δ (+δ) δ 0 Eample. Fin the erivative of cosec from first principle. Solution : Let y cosec an y+δ y cosec( +δ) δ y cosec(+δ) cosec cosec( +δ ) + cosec cosec(+δ ) + cosec cosec(+δ) cosec cosec(+δ ) + cosec ( +δ) cosec(+δ ) + cosec (+δ) cosec(+δ ) + cosec [ ( +δ ) ] cos δ δ + cosec( ) cosec ( +δ + ) ( +δ) [ ] δ cos + δy lim lim δ cosec(+δ ) + cosec] δ 0 δ 0 δ/ δ/ [( +δ).] cos or ( (cosec)() MATHEMATICS 55
MODULE - V (cosec) (coseccot) Thus, ( cosec) ( cosec) ( coseccot) Eample.3 Fin the erivative of Solution : Let y sec an y+δ y sec ( +δ) then, δ y sec ( +δ) sec Differentiation of Trigonometric Functions sec from first principle. cos cos ( +δ) cos (+δ)cos [( +δ + ][( +δ )] cos ( +δ)cos (+δ)δ cos (+δ)cos δ y (+δ)δ δ cos ( +δ)cos. δ Now, δ y ( +δ)δ lim lim δ 0δ δ 0 cos (+δ)cos. δ cos cos cos tan.sec cos cos sec(sec.tan) sec (sec tan ) CHECK YOUR PROGRESS.. Fin the erivative from first principle of the following functions with respect to : (a) cosec (b) cot (c) cos () cot (e) cosec (f). Fin the erivative of each of the following functions : (a) (b) cosec (c) tan 56 MATHEMATICS
Differentiation of Trigonometric Functions. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS You have learnt how we can fin the erivative of a trigonometric function from first principle an also how to eal with these functions as a function of a function as shown in the alternative metho. Now we consier some more eamples of these erivatives. Eample.4 Fin the erivative of each of the following functions : (i) (ii) tan (iii) 3 cosec(5 ) MODULE - V Solution : (i) Let y, t, where t By chain Rule, Hence, cost t t, we have t () cos (ii) Let y tan an t cos t (). cos t cos By chain rule, tan t where t sec t t an t, we have t sec t t sec sec Hence, ( tan ) Alternatively : Let y tan (iii) Let 3 y cosec(5 ) sec sec 3 3 3 cosec(5 )cot(5 ) [5 ] 3 3 5 cosec(5 )cot(5 ) or you may solve it by substituting 3 t 5 MATHEMATICS 57
MODULE - V Differentiation of Trigonometric Functions Eample.5 Fin the erivative of each of the following functions : (i) 4 y (ii) y + cos Solution : 4 y (i) (ii) 4 4 () + ( ) (Ug prouct rule) 4 3 (cos) + (4 ) 4 3 cos+ 4 3 [cos+ ] y + cos cos cos tan sec sec Alternatively : You may fin the erivative by ug quotient rule Let y + cos (+ cos) () + cos (+ cos) ( ) (+ cos)(cos) (+ cos) ( ) 58 cos + cos + (+ cos ) cos+ (+ cos) (+ cos) sec cos MATHEMATICS
Differentiation of Trigonometric Functions Eample.6 Fin the erivative of each of the following functions w.r.t. : (i) cos (ii) 3 MODULE - V Solution : (i) Let y cos t where t cos Ug chain rule t t an t, we have t cos. ( ) t (ii) Let 3 y cos 3 / 3 ( ) ( ) 3 cos 3 3 cos Thus, 3 3 cos Eample.7 Fin, when (i) y + (ii) y a( cost), a(t + t) Solution : We have, (i) y + + + MATHEMATICS 59
MODULE - V Differentiation of Trigonometric Functions + ( cos)(+ ) ( )(cos) (+ ) + cos (+ ) + (+ ) Thus, / + + (+ ) + + Alternatively, it is more convenient to fin the erivative of such square root function by rationalig the enominator. y + cos sec tan sectan sec cos cos + (ii) a(t+ t), y a( cost) Ug chain rule, a(+ cost), a(t) t t t, we have t a(t) a(+ cost) t t cos t tan t cos 60 MATHEMATICS
Differentiation of Trigonometric Functions Eample.8 Fin the erivative of each of the following functions at the inicate points : (i) y + ( 5) at π (ii) y cot+ sec + 5 at π/6 Solution : (i) y + ( 5) π At, cos () + ( 5) ( 5) cos+ 4( 5) cosπ+ 4( π 5) + 4π 0 4π (ii) y cot+ sec + 5 π At, 6 cosec sec(sectan) + cosec + sec tan π π π cosec + sec tan 6 6 6 4 4+ 3 3 MODULE - V 8 4 + 3 3 Eample.9 If y (a+y), prove that ( + ) a y a Solution : It is given that y y (a+y) or (a+ y) Differentiating w.r.t. on both sies of () we get...() (a+ y)cosy ycos(a+ y) (a+ y) (a+ y y) or (a+ y) MATHEMATICS 6
MODULE - V or ( + ) a y a Differentiation of Trigonometric Functions Eample.0 If y + +...to infinity, prove that Solution : We are given that cos y y + +...toinfinity or y + y or y + y Differentiating with respect to, we get y cos+ or cos Thus, y (y ) cos CHECK YOUR PROGRESS.. Fin the erivative of each of the following functions w.r.t : (a) y 34 (b) y cos5 (c) y tan () y (e) y (f) y tan (g) y π cot3 (h) y sec0 (i) y cosec. Fin the erivative of each of the following functions : sec (a) f() sec + () ( ) + cos (b) f() cos (c) f() f() + cos(e) f() cosec (f) f() cos3 (g) f() 3 3. Fin the erivative of each of the following functions : (a) 3 y (b) y cos (c) 4 y tan () 4 y cot (e) 5 y sec (f) 3 y cosec (g) y sec (h) y sec+ tan sec tan 6 MATHEMATICS
Differentiation of Trigonometric Functions 4. Fin the erivative of the following functions at the inicate points : π + π (a) y cos( +π /), (b) y, 3 cos 4 MODULE - V 5. If y tan+ tan+ tan +..., to infinity Show that (y ) sec. 6. If cosy cos(a + y), prove that cos (a+ y). a.3 DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS FROM FIRST PRINCIPLE We now fin erivatives of stanar inverse trignometric functions, first principle. (i) We will show that by first principle the erivative of ( ) ( ) w.r.t. is given by cos, tan, by Let y. Then y an so +δ (y+δy) Asδ 0, δy 0. Now, δ (y+δy) y or (y+δy) y δ (y+δy) y δy δy δ (y+δy) y δy lim lim δy 0 δy δ 0δ [On iviing both sies by δ ] [ δy 0when δ 0] cos y+ δy δy lim δy 0 δy ( cosy) cos y ( y) ( ) MATHEMATICS 63
MODULE - V ( ) ( ) ( cos ) (ii) ( ) For proof procee eactly as in the case of (iii) Now we show that, ( tan ) Differentiation of Trigonometric Functions +. Let y tan.then tany an so +δ tan(y+δy) Asδ 0, alsoδy 0 Now, δ tan(y+δy) tany tan(y+δy) tany δy δy δ tan(y+δy) tany δy lim lim δy 0 y δ 0 δ δ [ δy 0when δ 0] (y+δy) y lim δy δy 0 cos(y+δy) cosy (y+δy)cosy cos(y+δy)y lim δy 0 δ y.cos(y+δy)cosy (y+δy y) lim δy 0δ y.cos(y+δy)cosy δy lim δy 0 δ y cos(y+δy)cosy sec y cos y sec y + tan y + + ( tan ) + (iv) ( cot ) For proof procee eactly as in the case of tan. 64 MATHEMATICS
Differentiation of Trigonometric Functions (v) We have by first principle - (sec ) ( - ) Let y sec. Then sec y an so +δ sec(y+δy). As δ 0,also δy 0. Now δ sec(y+δy) secy. sec(y+δy) secy δy δy δ MODULE - V sec(y+δy) secy δy lim lim δy 0 y δ 0 δ δ [ δy 0when δ 0] y y y + δ δ lim ( ) δy 0 δ y.cosycos y+δy y y y + δ δ lim δy 0 cosycos( y+δy) δy y secytany cosycosy secytany secy sec y ( ) ( ) ( ) - (vi) ( ) sec cosec. ( - ) For proof procee as in the case of sec. Eample. Fin erivative of ( ) Solution : Let y from first principle. y Now, ( +δ ) (y +δy) ( ) ( ) +δ y+δy y δ δ MATHEMATICS 65
MODULE - V δy ( ) y cos y+ δ +δ δy lim lim lim (+δ) δy δ δ 0 δy 0 δ 0 Differentiation of Trigonometric Functions cosy cosy y 4. Eample. Fin erivative of Solution : Let y w.r.t. by elta metho. y..() Also (y+δ y) +δ..() From () an (), we get or or (y+δy) y +δ ( +δ )( +δ + ) δy δy cos y+ +δ + δ +δ + δy δ y cos y+ δ +δ + δy δy δy cos y + δ δy +δ + δy δy δy lim lim cos y lim + δ δy δ 0 δy 0 δy 0 or lim +δ + δ 0 cosy or ( δy 0 as δ 0) y cosy 66 MATHEMATICS
Differentiation of Trigonometric Functions CHECK YOUR PROGRESS.3. Fin by first principle that erivative of each of the following : (i) cos (iv) tan cos (ii) tan (v) (iii) cos (vi) tan MODULE - V.4 DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS In the previous section, we have learnt to fin erivatives of inverse trignometric functions by first principle. Now we learn to fin erivatives of inverse trigonometric functions by alternative methos. We start with stanar inverse trignometric functions,cos,... (i) Derivative of Solution : Let y y (i) Differentiating w.r.t. y cosy or y y...[ug (i) ] y Hence, [ ] Similarly we can show that (ii) Derivative of [cos ] tan Solution : Let tan y tany Differentiating w.r.t. y, sec y MATHEMATICS 67
MODULE - V Differentiation of Trigonometric Functions an sec y [ We have written tan y in terms of ] + tan y Hence, ( tan ) + +. + Similarly, ( cot ) (iii) Derivative of sec Solution : Let sec y sec y an secytany sec ytan y secy[ ± sec y] ± sec y sec y sec y π Note : (i) When >, sec y is + ve an tan y is + ve, y 0, π (ii) When <, sec y is ve an tan y is ve, y, π Hence, (sec ) Similarly (cosec ) Eample.3 Fin the erivative of each of the following : (i) (ii) cos Solution : (i) Let y (iii) (cosec ) 68 MATHEMATICS
Differentiation of Trigonometric Functions ( ) ( ) MODULE - V / (ii) Let y cos ( ) ( ) (iii) Let ( ) cos y (cosec ) () 4 4 ( ) (cosec ) cosec (cosec ) cosec (cosec ) cosec Eample.4 Fin the erivative of each of the following : (i) Solution : Let (i) cos tan + cos y tan + tan π π + cos (ii) ( ) MATHEMATICS 69
MODULE - V π tan tan 4 π 4 / (ii) y ( ) Let y ( ) Differentiation of Trigonometric Functions cos( ) ( ) cos( ) cos( ) Eample.5 Show that the erivative of tan w.r.t + is. Solution : Let Let tan θ y tan an z + tanθ y tan tan θ an z tan (tan θ) an z ( θ) θ an z θ θ an z θ tanθ + tan θ θ (By chain rule) θ z 70 CHECK YOUR PROGRESS.4 Fin the erivative of each of the following functions w.r.t. an epress the result in the simplest form (-3) :. (a) (b) cos (c) cos cos. (a) tan (cosec cot) (b) cot (sec + tan) (c) tan cos+ MATHEMATICS
Differentiation of Trigonometric Functions 3. (a) (cos ) (b) sec(tan ) 3 () cos (4 3) (e) cot + + 4. Fin the erivative of : tan + tan w.r.t. tan. (c) ( ) MODULE - V.5 SECOND ORDER DERIVATIVES We know that the secon orer erivative of a function is the erivative of the first erivative of that function. In this section, we shall fin the secon orer erivatives of trigonometric an inverse trigonometric functions. In the process, we shall be ug prouct rule, quotient rule an chain rule. Let us take some eamples. Eample.6 Fin the secon orer erivative of (i) (ii) cos (iii) cos Solution : (i) Let y Differentiating w.r.t. both sies, we get cos Differentiating w.r.t both sies again, we get (ii) Let y cos y (cos) y Differentiating w.r.t. both sies, we get ( ) cos. + cos + Differentiating w.r.t. both sies again, we get y ( cos) + (.cos+ ).cos y (.cos ) + MATHEMATICS 7
MODULE - V Differentiation of Trigonometric Functions (iii) Let y cos Differentiating w.r.t. both sies, we get Differentiating w.r.t. both sies, we get / ( ) ( ) ( ) ( ) 3/ y y Eample.7 If y ( ) 3/ ( ) 3/, show that ( ) y y 0 enote the secon an first, orer erivatives of y w.r.t.. Solution : We have, y Differentiating w.r.t. both sies, we get, where y an y respectively or or ( ) y 0 (squaring both sies) Differentiating w.r.t. both sies, we get ( ) ( ) ( ) y y + y 0 or ( ) y y y 0 or ( ) y y 0 7 CHECK YOUR PROGRESS.5. Fin the secon orer erivative of each of the following : (a) (cos) (b) tan MATHEMATICS
Differentiation of Trigonometric Functions. If y ( ), show that ( ) 3. If y (), prove that 4. If y + tan, show that y y. y tan ycos 0 + +. y cos y 0 + MODULE - V (i) (iii) LET US SUM UP () cos (ii) (tan) sec (iv) (cos) (cot) cosec (v) (sec) sectan (vi) If u is a erivabale function of, then (i) (iii) (v) (i) u (u) cosu (ii) u (tanu) sec u (iv) u (secu) secutanu (vi) ( ) (ii) (cosec) coseccot u (cosu) u u (cotu) cosec u u (coseu) cosecucotu (cos ) (iii) (tan ) (iv) + (cot ) + (v) (sec ) If u is a erivable function of, then u (i) ( u) u (vi) (ii) (cosec ) u (cos u) u (iii) u (tan u) + u (iv) u (cot u) + u (v) u (sec u) u u (vi) u (cosec u) u u The secon orer erivative of a trignometric function is the erivative of their first orer erivatives. MATHEMATICS 73
MODULE - V SUPPORTIVE WEB SITES Differentiation of Trigonometric Functions http://www.wikipeia.org http://mathworl.wolfram.com TERMINAL EXERCISE. If. Evaluate, 3. If 3 y tan, fin. 4 4 π + cos at an 0. 5 y + cos (+ ) 3 ( ), fin. 4. If + y sec +, then show that + 0 5. If 3 3 acos θ,y a θ, then fin +. 6. If y + + +..., fin. 7. Fin the erivative of 8. If y cos(cos), prove that y cot y. 0 +. 9. If y tan show that (+ ) y + y 0. w.r.t. cos 0. If y (cos ), show that. ( )y y 0 74 MATHEMATICS
Differentiation of Trigonometric Functions ANSWERS MODULE - V CHECK YOUR PROGRESS. () (a) coseccot (b) cosec (c) cos () cosec (e) cosec cot (f). (a) (b) cosec cot (c) tansec CHECK YOUR PROGRESS.. (a)cos4 (b) 55 sec (c) (e) cos (f) sec (g) 3πcosec 3 (h)0sec0tan0 (i) coseccot () cos sectan. (a) (sec + ) () cos (+ ) (b) ( cos) (c) cos+ (e) cosec( cot) (f) coscos3 33 3. (a) 3 cos (g) 3cos3 3 (b) (c) 3 4tan sec () 3 4cot cosec (e) 5 5sec tan (f) 3 3cosec cot (g) (h) sec( sec + tan) 4. (a) (b) + CHECK YOUR PROGRESS.3. (i) 4 cos (ii) sec tan (iii) ( ) (iv) 4 + tan (v) (+ ) (vi) ( + ) CHECK YOUR PROGRESS.4. (a) 4 (b) 4 (c) MATHEMATICS 75
MODULE - V. (a) ( ) cos cos 3. (a) (b) Differentiation of Trigonometric Functions (c) (b) ( sec tan ) + (c) 3 () (e) (+ ) 4. ( ) + tan CHECK YOUR PROGRESS.5. (a) coscos(cos) (cos) (b) (+ ) + tan (+ ) TERMINAL EXERCISE. 3 tan sec + 3 tan. 0, 0 5(3 ) 3. 5 3( ) 3 (4+ ) 5. secθ 6. y 7. 76 MATHEMATICS