## Answers to ExercisesforSection 3: Derivatives of Trigonometric Functions

1. $f'(x) = \cos x + \sin x$

3. $g'(x) = (\cos x)(\tan x) + (\sin x)(\sec^2x)$

5. $h'(x) = -2\cosec x \cotan x - \sec x \tan x + 3$

7. $r'(x) = \cos x - x \sin x + 2x$

9. $s'(x) = (2x-1)\tan x + (x^2-x+1)\sec^2x$

11. $t'(x) = \frac{-(\cosec^2x)(\sec x)(\tan x) - \sec x}{(1 + \sec x)^2}$

13. $k'(x) = -2\cos x \sin x$

15. $j'(x) = 2\sec^2x \tan x$

17. $u'(x) = -(2x - 1)\sin(x^2 - x)$

19. $v'(x) = (2.2x^{1.2}+1.2)\sec(x^{2.2}+1.2x-1)\tan(x^{2.2}+1.2x-1)$

21. $w'(x) = \sec x \tan x \tan(x^2 - 1) + 2x \sec x\ \sec^2(x^2 - 1)$

23. $y'(x) = e^x(-\sin(e^x) + \cos x - \sin x)$

25. $z'(x) = \sec x$

27. $z'(x) = \sec x$

31.$e^{-2x}[-2\sin(3x) + 3\cos(3x)]$

33. $1.5[\sin(3x)]^{-0.5}\cos(3x)$

35. $\sec(x^3/x^2-1) \tan(x^3/x^2-1) (x^4-3x^2) / (x^2-1)^2$

37. $(1/x)\cotan(2x-1) - 2\ln \|x\| \cosec^2(2x-1)$

39. $c'(t) = (3.5)(2\pi)\cos[2\pi(t-0.75)];\ \ c'(0.75) \approx 21.99$ per year$\approx 42¢$ per week

41. (a) $d(t) = 5\cos(2\pit/13.5)+10$   (b) $d'(t) = -(10\pi/13.5)\sin(2\pit/13.5); d'(7) \approx 0.270.$ At noon, the tide was rising at a rate of $0.270$ feet per hour.

43. $c'(t) = 1.035^t[\ln \|1.035\|(0.8\sin(2\pit) + 10.2) + 1.6\pi \cos(2\pit)];\ \ c'(1) = 1.035[10.2\ln \|1.035\|+ 1.6\pi] \approx 5.5656$ per year, or $0.11$ per week.

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 Stefan Waner (matszw@hofstra.edu) Steven R. Costenoble (matsrc@hofstra.edu)
Last Updated: September, 1996