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(Solved) Math 2211 3.5 Review Name________________________ Just Relax and Think Positively! I Know you CAN Do This! Differentiate: f(x) = 30 3 2. F(x) = 4 8...


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Math 2211 3.1−3.5 Review Name________________________ Just Relax and Think Positively!!!! I Know you CAN Do This!!!
Differentiate:
1. f(x) = √30
3 2. F(x) = 4 8
3. f(t) = 1 6 − 3 4 + 2 4. h(x) = (x − 2)(2x + 3)
5. B(y) = cy −6
7 6. h(x) = √ 7 + √ 2
7. y = √(x – 1)
8. f(x) = 2 −3+1
2
1 2 9. y = (√ + 3 ) √ 10. y = ex + 1 + 1
11. g(x) = √ 12. y = 1+
+1 13. y = 3+−2 14. y = (−1)2
15. y = 2cscx + 5cosx
16. f(x) = √ sin x
17. G(x) = ex(tanx − x) 18. y = 1+sin +cos 19. f(t) =
20. y = cot 1−sec tan 21. y = x3√ 2 + 5
22. F(x) = (4x – x2)100
23. f(x) = (1 + x4)2/3
3 24. f(t) = √1 + tan 1 2 25. y = ( 3 − 5) (3 3 + 5 2 − 7)
26. y = sin((3 2 + 2 + 3)4 )
27. y = 5sin√ − 3
28. y = 6 (4 3 − 7 + 5)
29. y = e−2t cos 4t
30. h(t) = (t4 − 1)3 (t3 + 1)4
31. y = 101− 2 2 5 32. G(y) = (+1) 33. f(t) = √ 2 +4 34. y = esec 3x
35. y = 2x√ 2 + 1
36. Find an equation of the tangent line to the curve y = x4 + 2x2 − x at the point (1, 2).
37. Find an equation of the normal line to the curve y = (1 + 2x)2 at the point (1, 9).
38. Find an equation of the tangent line to the curve y = x − √ at the point (1, 0).
39. Find an equation of the tangent and normal lines to the curve y = (2 + x)e−x at the point (0, 2).
3 40. Find the first and second derivatives of the function G( r) = √ + √ . 41. The equation of motion of a particle is s = t4 − 2t3 + t2 – t, where s is in meters and t is in seconds.
a. Find the velocity and acceleration as functions of t.
b. Find the acceleration after 1 second.
42. For what values of x does the graph of f(x) = x3 + 3x2 + x + 3 have a horizontal tangent?
43. Find an equation of the tangent line to the curve y = (1 + 2x)10 at the point (0, 1).
44. Find an equation of the tangent line to the curve y = √1 + 3 at the point (2, 3). 45. Find f '(x) and f ''(x): f(x) = 2−1
46. Find f '(x) and f ''(x): f(x) = cos(x2)
47. Find f '(x) and f ''(x): f(x) = cos2x
48. Find f '(x) and f ''(x): f(x) = Find dy/dx by implicit differentiation:
49. 2√ + √ = 3
50. 2x3 + x2y − xy3 = 2
51. y5 + x2y3 = 1 + y 2 52. Use implicit differentiation to find an equation of the tangent line to the curve
x2 + 2xy – y2 + x = 2 at the point (1, 2) Math 2211 1. 0 2. 6x7
3. 3t5 −12t3 +1
4. 4x − 1
5. −6cy−7
6.
7. 7
2
3
2 2 5⁄2 + 7 −5⁄7
1 1⁄2 − 2 −1⁄2 8. 3x−2 − 2x−3
1 2 9. 1 + 3 −5⁄6 − 3 −5⁄3
10. ex+1
1 11. 1⁄2 + ( −1⁄2 )
2 12.
13.
14. (+1)2
−2 3 −3 2 −3
( 3 + −2)2
−−1
( − 1)3 15. −2cscxcotx − 5sinx
16. √cosx + 2√ 17. ( 2 − 1 + − )
18. (+)2 19. − 2 + 3.1-3.5 Review SOLUTIONS 20. (1−)
2 21. 4 ( 2 + 5)−1⁄2 + 3 2 ( 2 + 5)1⁄2
22. 100(4x − x2)99 (4 − 2x)
23.
24. 8 3
3 3 √1+ 4 2 3 3 √(1+)2 25. −70 −6 + 51 −4 + 12 −3 − 5 −2
26. cos(3 2 + 2 + 3)4 [4(3 2 + 2 + 3)3 (6 + 2)]
27. 5√−3
2√−3 28. 6(12 2 − 7) [tan(4 3 − 7 + 5)]5 2 (4 3 − 7 + 5)
29. −2 −2 (24 + 4)
30. 12 2 ( 4 − 1)2 ( 3 + 1)3 (2 4 + − 1)
31. −2x(ln10)101−
32.
33. 2 5 9 (+2)
(+1)6
4 − 2
2 1⁄2 ( 2 +4)3⁄2 34. 3 3 33
35. 2(2 2 +1)
√ 2 +1 36. y = 7x 5
37. y = − 1
12 x+ 1 1 2 2 38. y = − 109
12 39. Equation of the tangent line: y = −x + 2; Equation of the normal line: y = x + 2.
1 1 40. G1(r) = 2 −1⁄2 + 3 −2⁄3; 1 G11(r) = − 4 −3⁄2 − 41. a v(t) = 4t3 − 6t2 +2t − 1; a(t) = 12t2 − 12t + 2
1 42. −1 ± √6
3 2 −5⁄3 9 b. a(1) = 2m/s2 43. y = 20x + 1
44. y = 2x – 1
45. f1(x) = − 2 −1
( 2 −1)2 ; ; ”()= 2 3 +6
( 2 −1)3 46. y'= −2xsin(x2); y'' = −4x2cos(x2) − 2 sin(x2) 47. y' = −2cosxsinx; y'' = −2cos2x + 2sin2x 48. y'= ∙ ;
49. –
50. 2√
√ −6 2 −2+ 3 2 −3 2
2 51. 2( − 2 )
5 4 +3 2 2 − 7 3 2 2 52. y = − y''= ∙ + ∙ ∙ 2

 


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