AP C ALCULUS D ERIVATIVES OF THE T RIGONOMETRIC F UNCTIONS 1. lim (x2 + cos x) = 1 x→0 −1 + 2. lim (sin x − cos x) = 2 x→π/3 √ 2 2 sin 5t =− 3. lim t π t→π/4 √ 3 sin(cos x) = sin 1 sec x √ sin x 2 2 = 5. lim 3π x→π/4 3x 4. lim x→0 6. lim x→π/4 1 tan x = 4x π sin2 x sin x = lim sin x = 0 x→0 x x→0 x 7. lim 8. tan 3x lim x→0 3 tan 2x = = 9. 10. 11. 12. sin 3x cos 3x lim x→0 3 sin 2x cos 2x lim x→0 sin 3x cos 2x cos 3x 3 sin 2x = sin 3x 3x cos 2x lim 3x 3 sin 2x x→0 cos 3x 2x 2x = 1 2 dy = − sin x − 2 sec2 x dx dy = cos x − sin x dx dy dx dy dx = (x)(− csc x cot x) + (csc x)(1) = csc x − x csc x cot x = (csc x)(− csc2 x) + (cot x)(− csc x cot x) = − csc3 x − cot2 x csc x = − csc x(csc2 x + cot2 x) 1 13. 14. 15. 16. dy dx = (1 + cos x)(cos x) − (sin x)(− sin x) (1 + cos x)2 = cos x + cos2 x + sin2 x (1 + cos x)2 = 1 + cos x (1 + cos x)2 = 1 1 + cos x dy x sec2 x − tan x = dx x2 dy dx dy dx = (sin x + cos x)(1) − x(cos x − sin x) (sin x + cos x)2 = sin x + cos x − x cos x + x sin x (sin x + cos x)2 = (x−3 )(sin x sec2 x + tan x cos x) + (sin x tan x)(−3x−4 ) = sin x sec2 x + tan x cos x −3 sin x tan x + x3 x4 x sin x sec2 x + x tan x cos x − 3 sin x tan x x4 £ 2 ¤ (sec x) (x )(sec2 x) + (tan x)(2x) − (x2 tan x)(sec x tan x) sec2 x = 17. dy dx = x2 sec3 x + 2x sec x tan x − x2 sec x tan2 x sec2 x = 18. Slope of tangent ¯ dy ¯¯ dy = x(− sin x) + cos x −→ = −1 −→ mT = −1 dx dx ¯x=π Point Given as (π, −π) Equation of tangent y + π = −(x − π) 19. Slope of tangent ¯ dy ¯¯ dy = sec2 x −→ = 2 −→ mT = 2 dx dx ¯x=π/4 Point Given as ³π 4 ´ ,1 Equation of tangent ³ π´ y−1=2 x− 4 2 20. Slope of tangent g 0 (x) = 2 cos x −→ g 0 ³π ´ 6 = √ 3 −→ mT = √ 3 Point ³π´ ³π ´ g = 1 −→ ,1 6 6 Equation of tangent √ ³ π´ y−1= 3 x− 6 21. For f to have a horizontal tangent, f 0 (x) = 0. f 0 (x) = 1 + 2 cos x f 0 (x) = 0 when 1 + 2 cos x = 0 −→ x = 4π 2π + 2kπ or x = + 2kπ 3 3 3