4. f (x) =2.5 4 sin cos x Exercise Set Exercise 2 x 3. f ⇧⇧⇧(x) = Set 4x sin2.5 x+ sin8xx)cos cos x2 x(5 sin10 x(5 91 1 + 5(sin x cos x) cos x) 9. f (x) = 4 csc x cot x + csc x 2. = 5. f ⇧⇧(x) (x) = = x23 + cos 2 8. ff⇧ (x) (x) = = (x + 1) sec x+ tan +2 (sec + 5(sin 1)sin secx x 2tan + 2x sec x sin + x) 1. 4 sin x+ + 2(5cos x xx) + cosx x) sin x(5 sin x) cos x(5 x)(2x) cos x)= (x1 (5 ⇧⇧ ⇧ 4. f (x) = 4 sin x cos x = 5. f (x) = 10. f ⇧(x) = Set sin csc x + x csc x2cot x Exercise 2 x2.5 2 (5 + (5 + sin x) 3. f ⇧(x) = 2.5 4x2 sin x 8x cossin x 2x) Exercise Set 91 + sin x) cos 9. f⇧⇧ (x) = (x10 4 csc x cot x +x cscsin xx(2x + cos x) x2 cos x 2x sin x 6. f (x) = = 22 x) 2+ 2. (x) = = sec cos cos sin43x(sec x(5 sin x) cos x(5 cosx)x)= sec13+x 5(sin 2(tan 2 x tan 1. ff⇧⇧⇧(x) sin+x+ 2x(x cos 11. x) + x)(sec + 2sec tan x +x sin x) (x +xx sin x) 2 xsin = 5. = + sin x x) cos x x)sin x(52 + cossin x x)2x 4. ff ⇧(x) = 4(x x cos 2 x(2x + cos x) 2 sin x ⇧(x) (5 + sin p 172/ 1 – 25 odds 6. f (x) =Set = 10. sin x csc x 2+ x csc x 2cot x Exercise 2.5 sin x) (x23 x + sin x)x2cot2 x 2(x + 2 10 2xx)( 12. x)( csc x cot x) = csc csc 7. fff⇧⇧⇧⇧(x) (x) = sec tan xcsc 2+ secx(cot x x(5 3. (x) = (csc 4x sin x+ 8xx) cos 2. + cos x 1 + 5(sin x cos x) sin x(5 sin x) cos cos x) 3 2 2 x4 sin 2 x + sin xx sin x(2xx + cos x) x3 xcos x x2x sin 2x) = 5. ff⇧⇧⇧⇧(x) 1. = x+ 2 cos cos 11. = (x sec x(sec x) ++(tan x)(sec tan x) = sec + sec tan x 2 2 ⇧(x) = 2 6. f (x) = (5 sin x) (5 + sin x) 2 2sec x 2 2 7. f x tan x 2 ⇧ (x) = sec 2+ 2 cot x) 2 csc x( 2csc x 2 (1 csc x)( csc x) cot x(0 csc x ⇧ (x + sin x) (x + sin x) 8. ff⇧⇧ (x) (x) = = 4(xsin + 1) sec x tan x + (sec x)(2x) = (x + 1) sec x tan x + 2x sec xcsc x + cot x) 4. x cos x 2 13. f (x) = = , but 1 + cot2 x = csc2 x 3. f ⇧(x) = 4x10sin x 8x2 cos x+ csc x)2 2 x)2 3 (1+ (1x + csc ⇧ (x) = (csc 2 2 12. f x)( csc x) (cot x)( csc x cot x) = csc x csc cot x 2. +sin cos ⇧(x) = (x32 + x)xcos x x22+ sin(sec x(2x + cos=x)(x2csc x 1) cos 2x xsin 8. fffff⇧⇧⇧(x) (x)= =sec (x + x+tan x)(2x) + secx x tan +x2x seccsc x x x( 9. csc x1) x4 x 6. (x) = 7. tan xsec 2csc sec ⇧ 2 x x) 2 xxx(5 1=+ 5(sin xcsc xcos 1) x) sin x(5 +cot sin cos cos x) 2 2 2 2 ⇧ (identity), thus cot x csc x = 1, so f (x) = = . (x + sin x) 4. f (x) = 4 sin x cos x (x + 2sin x) 2 = 5. cscx) x)2 x( csc x 1 + csc 2 (1 x)2 cot x(0 csc x cot csc cscx x + cot2 x) (5 +csc sin x) (5(1x) ++sin ⇧⇧⇧ 2 + csc x)( 2 3. (x) = 4x2sin sin cos xcsc 9. fff⇧ (x) (x)= = 4+ csc cot xx cotx)(2x) 13. =(x , but 1 + cot2 x = csc2 x 10. (x) = x1)xsec csc8x + xcsc x2 8. xxtan x(1 + = (x2 + 1) sec= x tan x + 2x (1 sec+xcsc x)2 ⇧ 2+(sec csc x) 1 + 5(sin x cos x) sin x(5 + sin x) cos x(5 cos x) 7. f (x) = (1 sec2 xtan tanx)(sec x sec x)x (sec x)(sec2 x) 2sec x tan x + sec x tan2 x sec3 x x2tan = =x csc 5. ff ⇧⇧⇧(x) (x) = = (x+ + sin2 x)(5 xsin x) sin x(2x + cos x) cos x x) 2x csc 1) 2 csc x = 14. 22 xsin x 2cos 2 2x + +x(sin 2cot 4. f = 4 sin x cos x ⇧⇧⇧(x) 2 10. (x) = sin x csc x + xtan csc x so 11. f (x) sec x(sec x) + (tan x)(sec x tan x) f=⇧ (x) sec=3=x (5 + sec x tan(1 x+ 6. (identity), thus cot x csc xx)= 1, = . 9. f ⇧(x) = 42csc x cot x(x(1 +2 + csc x 2 2+ 22 tan x) 2 + sin x) (x sin x) (1 + csc x) 1 + csc x 8. f (x) = (x + 1) sec x tan x + (sec x)(2x) = (x + 1) sec x tan x + 2x sec x 2 sec x(tan2x + tan x sec2 x) sec x(tan x 1)3 2 2 cot ⇧(x) = (csc (x + sin x) x+ cos sin x(2x +x cos cos xx x2x 1sec +x5(sin cos x) x(5 +2csc sin x) x(5 cos x) = 12. f= f⇧⇧⇧(x) x)( x)+(tan (cot x)( cotx) cscsin x 11. (x)= = sin sec x(sec x)2cos + x)(sec xxcsc tan x) =x) + 3sec tan xx 2 x 2 2= xcsc 10. sin x csc x x csc x cot 2 2= 3 6. f (x) = ⇧ 2 x) = 5. (1 + tan x) (1 + tan x) 2 (1 + tan x)(sec x tan (sec x)(sec x) sec x tan x + 2 2 2 2 7. tan x (5 sec 2 9. f (x) = sec4 x csc x cot x +2+ csc (x+ sin (x + sin sin x)xx) (5 + sin x)2x) sec x tan x sec x = 14. f ⇧ (x) = = 2 2 22 3 x( 2(1 (1 2+ tan + tan2x) 9212. f⇧⇧⇧ (x) = (csc 2 (1 +x)( csc2x)( x) cot xsec cot x) csc 3= csc2=csc x) + (cotx) x)(xx(0 csc x cot x) csccsc x x + cot x) , but 1 + cot2 x = Chapter 2 x(sec 2 ⇧csc 11. fff(x) ==sin sec (tan x)(sec tan x) = x +=sec xxtan csc x xxcotcsc 13. (x) csc2 x 15. = 1 (identity), so f (x) = 0. ⇧⇧(x) 22x + cosx)x+ 2 2 2sin 2x 2 8. = (x + 1) sec tan + (sec x)(2x) = (x + 1) sec x tan x + 2x sec x 10. sin x csc x + x csc x cot 2x 2x (1 + csc x) (1 + csc x) (x + sin x) cos x x(2x + cos x) x cos x 2x sin x 7. ff ⇧ (x) (x) = sec x tan x 2 sec sec x(tan x + tan x sec x) sec x(tan x 1) 6. f=(x) = = x(0 ⇧ csc= 2 csc csc x x)x( 1) x 2 x + cot2 x) 2 cot 2+ 2 2 csccsc ⇧ 2 2 3x 2 csc 2 x( (1 + csc x)( x x) csc 2 (x 2 + sin x) (x sin 2x) 21, csc 2 sin sin (1 + tan x) +cos tan x)xcot 12. f (x) = (csc x)( csc x) + (cot x)( cot x) = csc cot x 2 sin ⇧ 92 (identity), thus cot x csc x = so(1x fx (x) = . x2 19. dy/dx =2sec sin xcot x= xcsc sin x= xcsc cosxfx 2 x 2+ cos x, d2 y/dx =2 13. f⇧⇧⇧(x) (x)= = cot2 xChapter =socsc2 2x 2= 2xcsc 2 3sin 11. (x) = x(sec x) + (tan x)(sec x tan x) = sec x + sec x tan x 16. f sec x tan x sec x 2 tan x sec x = 2 0; also, (x) x tan,2 but x = 11 + (identity), 9. 4 x x + csc x (1 + csc x) 1 + csc x sec 2 2 8. f (x) = (x + 1) sec x tan x(1++(sec =cos (x3 x+ 1) sec x3 tan x + 2x(1 sec+xcsc= csc x)(2x) x) x) cos x ⇧ 2 2 2 x (x) = ⇧⇧ x cotcsc x( 3 csc csc x( x csc 1) x 2 csccsc 7. fff⇧⇧⇧(x) (x) = = 0. sec x+ xx)(22 x 22sec (12thus csccos csc cot 15. sin xtan + = 12 x) (identity), so ffcsc = 2x2 cot x(0 20.x) 2x d 2x x)] x 2(x) 3+ (identity), cot csc = 1,csc soxx (x) == = . 3 x x) 2x)x 2 x 20. dy/dx = csc x x, y/dx [(csc x)( csc x) + (cot x)( csc x cot = csc + csc x cot ⇧(x) 2 12. f (x) (csc x)( x) + (cot x)( cot x) csc x csc x cot 13. f = = , but 1 2+xcot2 x = csc2 x (1 + tan x)(sec x tan (sec x)(sec x) sec x tan x + sec x tan x sec 10. = sin x csc x + x csc x cot 19. dy/dx = x sin x + cos x, d y/dx = cos x sin x sin x = cos x 2 sin x 2 9. ff ⇧(x) xcsc x)2 2 x = (1 + csc x) 1 +x)csc 14. (x) = = 4 csc x cot x + csc = (1 + (1 + csc 2 2 tan x x) x)(2x) = (x22sin (1 + tan x) ⇧ x sec x 8. f (x) (x)== (x2 + 1) sec x (1 tan+x2tan + +3csc 1) tan +1) 2x x( csc xxx), 17. (because sin(sec x sec x =2 (sin x)(1/ cos x) x=sin tan so sec x csc2 x ⇧csc 2x sec 2 92 Chapter so 2 2= csc ⇧⇧⇧ (x) =1 2 2 x)( (1 + csc csc x) cot x(0 x cot x) csc x( x csc +. cot x) tan2 x2= 1 (identity), 16. f sec x tan x 2 tan x sec x = 2 0; also, f (x)x = sec2 x (identity), thus cot x csc x = 1, so f (x) = = 2 2 + x tan x 2 2 2 2 2 11. f (x) = sec x(sec x) + (tan x)(sec x tan x) = sec x + sec x tan x 21. dy/dx = x(cos x) + (sin x)(1) 3( sin x) = x cos x + 4 sin x, ⇧ 10. xtan cscx, xtan csc x= cotsec x 3x)( sec x(tan x+ xx d+xsec x) x(tan 1)3 xx)(1 20. = (1sin csc xtan cot y/dx [(csc x)(x cos csc += (cot x)] = csc33xx + csc x cot x cot2 x = csc2 x + x)(sec x) (sec x)(sec x) sec xcos tan x2+ csc sec x cot tan xcsc sec 13. dy/dx f=⇧(x) = , but 1 + x + csc x) 1 + x 2 2 = x) 2 x) 2 (12+ csc 2 14. ff⇧⇧ (x) (x) = = 0. + x tan x)(sec = x)(1)] (1tan + csc sec2 x 1= x)tan x) +2(tan tan x) (1 + tan 9. fd (x) =2 (1 csc+xsin cotx) x 2+ +(1csc x tan + x)2x[x(sec + tan x)2xcsc=x f 2⇧⇧⇧(x) = = xx2(11) (because sec2 x y/dx =4(1 x( x =x) x3 sin xx(+35csc cos 2 2 (cos2x)(1)2+ 4 cos csc 2csc 2 2 2 ⇧ 12. f (x) = (csc x)( x) + (cot x)( csc x cot x) = csc x csc x cot x 2 2 (1 + x tan x) (1 + x tan x) (1 + x tan x) 2 2 3 11. sec x(sec x) + (tan x)(sec x tan x) = sec x + sec tan x 19. dy/dx = x sin x + cos x, d y/dx = x cos x sin x sin x = x cos x 2 sin x 92 Chapter 2 (identity), thus cot x csc x = 1, so f (x) = = . 2 xx)(1) +xtan x)(sec tan tanx,x + sec x tan x sec x 21. dy/dx =(1x(cos x)tan + (sin sinx)(sec x) = xxx) cos1) xsec + 4xsin ⇧ 2sec x(tan + sec2x)x)3((sec sec x(tan 12 + csc x 2 x ⇧ 14. fftan (x) = = (1 + csc x)2 = tan xcos ⇧(x) x= =sin 1).2sin 15. f x + x = 1 (identity), so f (x) = 0. = = 2 10. (x) = x csc x + x csc x cot x (1 +2 tan x)x + (1 + tan x) x, 2 17. dy/dx f2⇧(x) == (because sin sec4(1 xcos =x(sin x)(1/ cos x) = cos tan x), so 2(cos 22 sin 2 2 x2 ((1 + tan x) tan x) 22. sin x) + =x x= x++ + 4x2cos 2+ 35csc (1 + csc x)( csc x) cot x(0 csc cot x)xcsc x( csc x + cot y/dx = sin x) (cos x)(1) +[(csc 4csc cos+ xcot = x cos x xx 1(csc +x( xx)( tan x 2 2 2 x)( 2sin 3 x) 12. == csc x) + x)(2x) (cot xx)( x) x2x csc cot x csc ⇧ (x) 20. ffd dy/dx csc x cot = csc x) + (cot x)( csc x cot x cot 2x, d y/dx 2 2x)] = csc3 x + csc 2 2 2 13. (x) = = , but 1 +xcot2 x = csc2 x 2 sec x(tan x + tan x sec sec x(tan 1) (1 tan x)(sec x tan x) (sec x)(sec x) sec x tan x + sec x tan x sec x 19. dy/dx = x sin x + cos x, d y/dx = x cos x sin x sin x = x cos x 2 sin x 2 2 (x + 1) cot x 2 2 2 2 2 sin x x 2 sec(1x+ csc 2tan 2 2 ⇧⇧⇧ 2 x)(sec(1x) 3 (tan sin 2 2 + csc x) x) 2 2 tan x 1 (1 + x tan x[x(sec x) + x)(1)] 11. f (x) = sec x(sec x) + (tan x)(sec x tan x) = sec x + sec x tan x d (cos x) + x)(2x)] sin x) + cos x]==cot 40;sin xso=f(2 x ) cosxx =tan 4(x + 1) sin x 14. 18. fff(x) cos= csc+x x2[x( x)(1/ x) x), 2 [x ⇧y/dx ⇧(cos 16. = (x) = 2=sec tan x2x) sec x (sin 2 tan sec 2sin also, (x) (identity), so2 x 2(1 2= 2x 2 = 2 xsec (x)== =sin =(1 = 2 x) x = 21 (because sec 15. (x) xxcot + cos x(because = 12 tan (identity), so (x) 3csc + tan (1cos +2=fx tan x) x)x + tan x) 20.x)x( 32(1 xx)( 1) csc cos xxcsc cos x2xxcos (1 + csc csc+ x) cot x(0 x3= cot csc x( csc xxcos csc 2 + cot 22. dy/dx = x ( sin x) + (cos x)(2x) + 4 = sin x + x + 4 x, ⇧csc 2(sin 2(1 +3( 2 ⇧⇧ x tan x) (1 + tan x) (1 + x tan x) 21. dy/dx = x(cos x) + x)(1) sin x) = x cos x + 4 sin x, (identity), thus cot x csc x = 1, so f (x) = = . 13. ff⇧ (x) = 0. , but 1 + cot x = csc2 x (x) = 22 3 2 2x)( csc22 x) += 22 csc x 2cot x)] = 2 2(x 2 csc 22x 3 csc 2 = 20. dy/dx = csc x2 x)[2x cot x,2xcot d2 y/dx = [(csc (cot x)( csc x x+ csc x2cot (3 cot xsec +x) 1) x] (x + 1) cot x csc x x cot 6x+cot xx) 3(x + 21)xcsc2 x sec x(tan xx)( + tan x) sec x(tan xx) 1) (1 + x)csc 1csc + csc x cot (1(cos + csc (1 2 2 = 12. fdy/dx (csc csc x) + (cot x)( csc x cot csc x x tan x = 1). ⇧ (x) 2 2 2 2 23. = (sin x)( sin x) + x)(cos x) = cos x sin x, 2 ⇧ fd = = x [x =x = (2 x ) cos x2 4(x2 + 21) sin x . sin0. x sin x =2(x) = 4+ (cosx)xx)+ x)(2x)] 2[x( x) x] 15. cos = 1(sin (identity), so (x)2sin = ⇧y/dx 2 2+ csc+xx(cos 1) y/dx = x( (cos cos + cosxx4=sin (3 + x)+ 16. fdf(x) (x)==2sin 2 tan sec x tan sec x 2 tan xcot sec xf2xtan 2 5csc 0; also, fcsc (x)x =(3sec cot x x)tan x = 1 (identity), so (1+ + tan2xx) (1 ⇧== x)x23sin 2x)(1) x sin 3x (identity), thus cot x csc x = 1, so f (x) = = .3 2 2 2 cos x cos 2 x)] 2 2(sin 2+sin 2csc 17. f (x) = (because x sec x = (sin x)(1/ cos x) = tan x), so 2 d y/dx = (cos x)( sin x) (cos x)( sin x) [(sin x)(cos x) + x)(cos = 4 sin x cos x (1 + tan x)(sec x tan x) (sec x)(sec x) sec x tan x + sec x tan x sec x ⇧ (1 + csc x) 1 + x (1 csc x)((sin x) 3( cot x(0 cotxx) 21. dy/dx =1(x x(cos x)cot x)(1) sin x) csc = xxcos + 42sincsc x, x( csc x csc x + cot x) 1) x csc f⇧(x) (x)= = 0. +2 x+tan x+ 2 = 2(cos sin x sin x = cot x), so2 14. = = , but2 1 + cot2 x = csc2 x 13. fdy/dx (x) = =sin 18. f (because cos x csc x x)(1/ x) 2 x) ⇧= ⇧ 2 2 23. = (sin x)( sin + (cos x)(cos x) = cos x sin x, 2 2 2 2 15. fdy/dx f (x) cos x = 1(1 16. (x) = == 2(1 tan sec x2+ 2 tan sec x f2= 2x + 2x = 2x+ 0;x+tan also, (x) tan x) xx) (1 x)csc 22. xsec (xxx+cot sin x) +(1 (cos x)(2x) + 4 so cos x(x) =+=(tan x0.sin 4tan cos 2fx, + csc (1 + x) = sec 1x tan x = 1 (identity), so xx 2sin 3 xcos sec + tan x)(sec x)(identity), tan x[x(sec df2⇧2⇧y/dx =3x( sin x) +2x (cos x)(1) 4 cos xx) =2cos x3 x sinx)(1)] x +cos 5tan cos x+ sec 2x 2 tan 2 2 2+ (x) = = = =2=sec34xsin x2cos2 x(because sec2 x + x)(sec tan x) (sec x)(sec x) sec x x x tan x 2 (1 csc x( csc 1) csc 2 sin tan x 2 2 2 2 2 24. dy/dx = sec x, d y/dx = 2 sec x tan x ⇧(x) 2 2 ⇧ 2 f 0. d y/dx = (cos x)( x) + (cos x)( sin x) [(sin x)(cos x) + (sin x)(cos x)] 2 2 2 sec= x(tan x[x + cot tan x cot sec x) sec x(tan 1) (3 cot x)[2x x(12 + (x + 1) csc x]xsin (x + 1)x)cot xtan csc 6x cotxxx 2xxcot x +3(x +x1) csc2 x xx tan x) (1 + x tan x) (1 + tan x) 14. fdf⇧(x) = = y/dx = (cos x) + (sin x)(2x)] + 2[x( x) + cos x] 4 sin x = (2 ) cos 4(x 1) sin thus x csc x = 1, so f (x) = = . 17. (identity), (because sin sec x = (sin x)(1/ cos = x), so x sin xx)(1 = 2 f⇧ (x) = x)12+ csc x 2 . x)=2x sec2 x =2 2 sin 2 = 1 + x tan x 22 2(1x+ tan x== 21). 16. ftan (x) sec x tan xx) sec 2 tan =+ 0; tan also, f (x) = sec(3x cot tanx) x = 1 (identity), so (1 tan +x) tan x)x232 sin(1x2++csc (3+ 4(1 cot 3x 22. dy/dx = x2tan ( + sin x) + (cos x)(2x) cos x = 2x cos x + 4 cos x, cos x cos 2 2 2 x)2 2sec x(tan 2 x x 2 + 2then x(tan x x 2f ⇧= sec sec2so x tan2 x 1 (1 tan + xx, tan x)(sec x) x[x(sec x)2+1) (tan x)(1)] ⇧⇧ sec 25. Let f= (x) = (x) = sec x.x x= (sin 24. dy/dx = x, d y/dx 2 sec xsec tan 17. (because sin xtan = tan x), ff(x) (x) = 0.sec 23. f= dy/dx = (sin x)( sin x) +tan (cos x)(cos x) = cos x2 cos sin2x) x, 2 2 x)(1/ = = =3 x (because sec2 x 2 x 22+ 2 (x) = 21 (1 x)(sec x x) (sec x)(sec x) sec x tan x + sec x tan xx)x22 )sec + x tan 2 2 2sin x 2 ⇧ (x + 1) cot xx) d y/dx = [x (cos x) + (sin x)(2x)] + 2[x( sin x) + cos x] 4 sin x = (2 cos x 4(x + 1) ⇧(x) (1 + tan (1 + tan x) (1 + x tan x) (1 + x tan (1 + x tan x) 15. f = sin x + cos x = 1 (identity), so f (x) = 0. 14. (x) = = = 18. ff2(x) = (because x),2 so 2 2 2x csc x =2(cos x)(1/ sin x) = cot 2 tan cos 2 2 (1 (1x) x) + tan x) ⇧ sin x tanx)]x= 4 sin x1cos x + xcot tan (tan x)(1)] dtan (cos x)( + (cos x)( sin x)x) +[(sin x)(cos x) +(1sec (sin x)(cos 3tan x fx)(sec x(x) == 1). xtan ⇧+x) 2x[x(sec ⇧ y/dx (a) f (0) = 0 and (0) = 1, so y 0 = (1)(x 0), y = x. 25. Let f = tan x, then f (x) = sec x. f (x) = = = (because sec2 x 17. f (x) = 2 sin2 x sec x = (sin = tan x), so 22(because ⇧ 2 x)(1/ 2+ x tan x) 2 x cos2x) 2 (1 2 2 2 22fx] 2 2x cot 2 2 (x (1 + x tan x) + x tan x) (1 sin sin x 15. dy/dx f ⇧(x) = sin x + cos x = 1 (identity), so (x) = 0. (3 cot x)[2x cot x + 1) csc (x + 1) cot x csc x 6x cot x x 3(x + 1) csc x 1 + tan x 23. = (sin x)( sin x) + (cos x)(cos = cos x sin x, sec x(tan tan x sec x) sec x(tan x 1) (x) . 16. tan ff ⇧(x) 2 sec tan sec2⌃x 2 ⌥ 2 tan x sec2 x 2=2 ⌃3 2 3 = 0;=2 also, f (x) = sec2 x tan2 x = 1 (identity), so 2 = = ⌃ (x ⌥2 + ⌥ 2 x1) 2 xx 2= x= cot 2 = 2 (tan (3 cot x) ⌥1). ⌥ x) ⌥ x)(1)] ⌥ x tan2 x (3 cot 1x)2 cos x cos x 24. dy/dx = sec x, d y/dx 2 sec x tan x sec (1 + x tan x)(sec tan x[x(sec x) + (1 + tan x) (1 + tan x) ⇧ 2 2 ⇧ 2 18. d(b) (because cos x csc x (cos x)(1/ sin = cot x), so (a) (0) = 0 and f (0) = 1, so y 0 = (1)(x 0), y = x. y/dx = (cos x)( sin x) + (cos x)( sin x) [(sin x)(cos x) + (sin x)(cos x)] = 4 sin x cos x = 1 and f = 2, so y 1 = 2 x , y = 2x + 1. ff⇧⇧(x) (x)f= = = = (because sec x = 0. 3 cot x 2 sin x 4 sin x 22 2 2 + xxtan x tanfx) (12+x x tan 16. f (x) = 422sec x tan x sec x4 (1 2 tan secx) x= 2 3 =(10;2+also, (x)2 = sec tanx) x = 1 (identity), so 3 x2 (x + 1) cot x 2 2 2 2 2 2 2 2 ⇧ cos cos x ⇧ 2 ⌃ ⌥ ⌃ ⌥ ⌃ ⌥ (3 cot x)[2x cot x (x + 1) csc x] (x + 1) cot x csc x 6x cot x 2x cot x 3(x + 1) csc2 x tan x = 1). 15. f (x) = sin x + cos x = 1 (identity), so f (x) = 0. 25. Let f (x) = tan x, then f (x) = sec x. ⇧ 18. fff(x) ⇧(x)= ⌥sec ⌥ ⌃2 seccos ⌥ =so 2⌥ x csc x = (cos x)(1/ ⌃ ⌥2 x,xd2 y/dx(because ⌃⌥ sin⌥x)⌥ = cot x), 0. tan ⇧2 = (x)f= = . 24. dy/dx = x tan x ⌥ ⌥ ⌥ 3 cot x = 1 and f = 2, so y 1 = 2 x , y = 2x + 1. (b) 2 x)(1/ cos x) = tan x), so 17. (c) f (x) f= 4 sec2,xcot =x) (3 cot x)2 = x1 (because and f4⇧ sin 2x(3= so y(sin + 1 = 22 4x + ,y= 2x 1. 2 + 1(x + x tan 2 2 2 sin x 2 2 2 2 4 4 4 2 sin xx csc x (3 tan x)[2x cot x (x + 1) 2csc x] (x + 1) cot 6x cot x 2x cot x 3(x + 1) csc x +cot 1) cot x⇧ (0) ⇧⇧ xtan (a) (0) = and fxx)(sec 1,2so 0csc =xx(1)(x y x)(1)] = x.x) 2 = sec2 x 2 (cos0), 16. = sec sec x=2 x) tan = 2sin 0;2x), also, f (x) tan22x = 1 (identity), so (x)ff= = = . 18. fLet (x) = (because cos x = x)(1/ ==sec cot 2sec x tan x 1cot tanxthen tan x[x(sec +3 (tan ⌃ ⌃ 2(1 ⌥0xxcot ⌥yxxsec ⌃ 3⌥xtan 2 x) 17. ff ⇧(x) (because sin x= (sin x)(1/ cos x) = x), so so⌥ 25. (x) =⌥3+ tan x, f ⇧ (x) = sec x.cot 2 cos x cos (3 x) (3 x) ⌥ ⌥ ff(c) = = (because sec x ⇧ x ⇧ (x)f= 1 + x tan 2 y+1=2 2 2 1 andf⌃⇧f(x) = 2,x)so ,y= + x)1. (x)f (x) =⌃0.=⌥sin=x, then 26. Let cos x.1) x2 tan (1 (1 cot + x2 tan ⌥(1=+ 2 x⌥+ 2 2x (3 x)[2x (x +x[x(sec csc22x]x) ⌃+(x(tan + 1) cot x+ 6x cot 2x x x) 3(x2 + 1) csc2 x 2 x 2x 2 4 + cot 4 tan 4 x cscsec 2tan ⌥ ⌥ x)(1)] ⌥xx tan ⇧ 2 ⇧cot⌥ x 1 (1 x tan x)(sec x) f (x) = = .2 ⇧ ⇧ = 1 and f = 2, so y 1 = 2 x , y = 2x + 1. (b) f tan = 1). f (x)x 2 (because sec x (a) f= (0) 0 and f (0) =4 1, so y(3 0cot = 2(1)(x 0), y4= x. = x)2 x) 4=tan 2 x tan x)2 = (1(3+ x cot 2 x (1 + x tan x) (1 + tan x) ⇧ ⇧ 1, sin (0) = = sin 0 and f(because (0)f = socos yx sec 0), cos y =x) x.= tan x), so 17. (x)2ff= (sin x)(1/ 26. f(a) Let x, then (x) = x.0x==(1)(x 2 ⌥ tan x(x) =1 1). + x tan x x ⌃⌥⌥ ⌥ ⌃ ⌃ ⌥⌥ ⌃ ⌃⌥ ⌥ ⌥ ⌥ (x + 1) cot ⌥ ⌥x), so⌥ ⌥ ⇧ 18. (b) f(c) (x) f= (1 +=x cos sin 2+(cos 2 + 1tan and f (because so ycscsox1 y= = 21+= xx)(1/ , y x) = ,= 2x = 1x)(sec and f2⇧ x)= 2, =xx[x(sec 2, 2 xx)(1)] + y cot = 2x sec x+1.tan21. x 1 tan x) (tan 3 cot x ⇧ ⇧ ⇧ (⌥) = 4 24= 00 and (⌥) and 0 2= ( 1)(x y4= =x + 2⌥. 2 f(b) (x)f= = (because sec2 x (a) (0) =4 1, 1, so4 so y y0= (1)(x 0), y4⌥), = x. (x= + 1) cotffx(0) 2 22 2 2 2 2 2 (1 + x tan x) (1 + x tan x) (1 + x tan x) 18. f ⇧(x) =⌃ (3 ⌥ cot x)[2x (because cos+x1) csc x =x](cos(xx)(1/ x) cot x⌃ (x⌥ csc + 1)sin cot csccotxx), so 6x cot x 2x cot x 3(x + 1) csc2 x ⌃ ⌥x= 2 = ⌥ 3 cot x f (x) = ⌥ . ⇧ ⇧⌃ ⌥⌥ ⌥ tan x = 1). ⌃ ⌥ ⌃ ⌥ 2 26. (c) Let ff(x)⌥= sin=x, cos (3 x) ⇧ f f⇧(x) 1 then ⌥= 1x. 1= 22 x 1 + y = , y⌥x=+ 1 fand 2,cot so y2 (+ 11)(x + 1 1. ⌥ (3 12 cot x)2 2 2= 2 2x 2 (b) f (⌥) = 0 and (⌥) = 1, so y 0 = ⌥), ⌥. cscy x] (x= + 1) cot 6x cot (c) f (3 =cot x)[2x and cot f x 4(x = + 1) , so x x 2x+cot x. 3(x + 1) csc x 4x x csc , xy = f ⇧ (x) = 4 42 = 22 . 4 4 2 x 2 2 4 2 (3 2cot x)2 (x + 1) cot (3 2 cot x)2 ⇧ (a) f (0) = 0 and f (0) = 1, so y 0 = (1)(x 0), y = x. ⌃ ⌥ ⌃ ⌥ ⌃ ⌥ 18. f (x) = ⌥ (because cos1x csc x = (cos ⌥= cos 1 x)(1/1 sin x) =⌥cot x), so1 ⌥ 1 26. Let =3 sincot x,1 then x. so y xand ff⇧⇧(x) (c) ffIf(x) xx x sin , yx=so y ⇧⇧x+ y = 2 cos + x. . 27. (a) y 4= x=sin x2 then y ⇧ =4sin=x2 + 2x,cos x2 and 2y ⇧⇧= = 2 cos 4 2x 2 cot x csc 2 cot22x 3(x2 + 1) csc2 x (3 cot x)[2x cot x (x + 1) csc x] (x2 + 1) 6x2cot x 4 2x ⇧ f(b) (x)f=(⌥) = 0 and f⇧ ⇧ (⌥) = 1, so y 0 = 2( 1)(x ⌥), y = x + ⌥.= . (a) f (0) = 0 and f (0) = 1, so y(3 0cot = (1)(x 0), y = x. x) (3 cot x)2 (4) ⇧⇧ ⇧ ⇧⇧ ⇧⇧ Di⇤erentiate get xy x+ y x=so y2 cos 27. (b) (a) If thenresult y ⌃= of sinpart x + x(a) costwice x andmore y =to2 cos sin + yx.= 2 cos x. ⌃y⌥=⌥ x sin1x the ⌥⌥ 1 1 1 ⌃ ⌥⌥ 1 ⌥ 1 ⇧ ⇧ f (c) ff (⌥) == and = , 0so=y ( 1)(x= ⌥), y x= x +, y⌥.= x + . (b) 0 and f (⌥) = 1, so y 4 4 4 2(a) 2 x, to ⇧ ⇧⇧ 28. (a) y = cos x 2then = of sinpart x and y ⇧⇧twice = cos so2yget +yy(4)=+( y ⇧⇧cos (cos (b) IfDi⇤erentiate the yresult more = x)22+cos x.4 x)2= 0; if2 y = sin x then y ⇧ = cos x ⇧⇧ and y ⇧⇧⌃= ⌥ ⌥ sin x 1 so y +⇧ ⇧y⌃ ⌥=⌥( sin 1 x) + (sin x)1 =⇧⇧ 0. 1 ⌃ ⌥⌥ 1 ⌥ 1 27. (c) (a) fIf y = x thenfy = sin=x + x, cos =sin x and so yx and y == 2 cosxx x sin , y x=so y ⇧⇧x+ y = 2 cos + x. .