anonymous 4 years ago 3x^3+2x^2 +X+5 the derivative evalauate y' when x=3 need a little help bit rusty

1. anonymous

in order to the first you lower the power by one like this 3x^2+2x^1+5 that is der right?

2. UnkleRhaukus

you also have to divide by the original index

3. anonymous

you divide by 3 because of 3x^3

4. anonymous

x^n=nx^n-1

5. anonymous

so if I divide by 3, than that would make this x^2 instead of 3x^2 I am on the right track?

6. UnkleRhaukus

opps ive got mixed up

7. anonymous

y' 3x^2 is X^2 right?

8. UnkleRhaukus

i should have said times by the index $y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=anx^{n-1}$

9. UnkleRhaukus

so sorry if i confused you

10. anonymous

so times by index makes 3x^3 Y' 6x^2 right?

11. UnkleRhaukus

not quite $y(x)=3x^3\qquad y'(x)=3\times3x^2$

12. anonymous

here is what I know y' when doing this you lower the power by one and something??

13. UnkleRhaukus

when taking the derivative multiply by the power and then lower the power

14. anonymous

so you are telling that power before lowering divide to y'

15. anonymous

like this 4x^3 so that would be 'y 4 right?

16. anonymous

4x

17. UnkleRhaukus

i ment times when i said divide originally , dont divide when taking the derivative

18. anonymous

can we start over plz..

19. UnkleRhaukus

yes ,

20. anonymous

y' 4x^2 would y'4x

21. UnkleRhaukus

$y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=n\times ax^{n-1}$ $y(x)=4x^2\qquad\Rightarrow\qquad y'(x)=2\times 4x^{2-1}$

22. anonymous

y'(x) would be 8x

23. UnkleRhaukus

thats right now

24. anonymous

okay so the power is what you are times before lowering the power?

25. UnkleRhaukus

that is right

26. anonymous

so for another example 10X^5 y'(x) 50x^4

27. UnkleRhaukus

you got it $$\checkmark$$

28. anonymous

so back to the problem now

29. UnkleRhaukus

$y(x)=3x^3+2x^2 +x+5$$y'(x)=$

30. anonymous

y'(x) 6x+4X +5

31. UnkleRhaukus

not quite, note that $x=x^1$$5=5x^0$

32. anonymous

y'(x) 9x+4+5

33. UnkleRhaukus

try again

34. anonymous

y'(x) 6x^2 +4x+5 +1

35. UnkleRhaukus

ok lets do this one term at a time $(3x^3)^\prime=$

36. anonymous

y'(x) 9x^2

37. UnkleRhaukus

right, now $(2x^2)^\prime=$

38. anonymous

y'(x) 9x^2 +2x

39. anonymous

y'(x) 9x^2 4+5+1

40. UnkleRhaukus

nope

41. anonymous

y'(x) 9x^2 +4X+5

42. UnkleRhaukus

you have the first two terms right,

43. UnkleRhaukus

$f(x)=x=x^1$ $f^\prime(x)=1\times x^{1-1}=x^0=1$

44. anonymous

so y'(x) 9x^2+4x+1+5

45. UnkleRhaukus

the derivative of the last term $g(x)=5=5x^0$ $g^\prime(x)=0\times 5x^{0-1}=$

46. anonymous

is 0

47. UnkleRhaukus

yeah the derivative of a constant is always zero

48. anonymous

so the end number unless it has a power drop off?

49. UnkleRhaukus

thats right

50. UnkleRhaukus

$f(x)=x^2+5\qquad f'(x)=2x$$g(x)=x^2+9\qquad g'(x)=2x$$h(x)=x^2−71\qquad f'(x)=2x$

51. UnkleRhaukus

that last one is ment to be h'

52. anonymous

cool love it so far because in math you can't drop anything, in calc 1 you can this is way cool..

53. anonymous

do you have time to do one more?

54. UnkleRhaukus

yeah can you get the answer to you original question now?

55. UnkleRhaukus

$y(x)=3x^3+2x^2 +x+5$ $y~'(x)=$

56. anonymous

y'(x) 9x^2+4x+1

57. UnkleRhaukus

$\Large\color\red\checkmark$

58. anonymous

okay any other tips here?

59. UnkleRhaukus

$y'(x)=\frac{\text dy(x)}{\text dx}$

60. anonymous

explain a little??

61. UnkleRhaukus

its just a different notation, they men the same thing

62. UnkleRhaukus

the second has a bit more information because it explicitly say that we are to take the derivative of the function with respect to x, ie we look at the power of the x terms,

63. anonymous

dx(x)^1?dx

64. anonymous

so the power on the is 1 right?

65. anonymous

just to let know, I am in college not high school

66. UnkleRhaukus

$y(x)=x=x^1$ $\frac{\text dy(x)}{\text dx}=y'(x)=1\times x^{1-1}=x^0=1$

67. UnkleRhaukus

im not sure what you mean by this dx(x)^1?dx

68. anonymous

I meant to use / not?

69. UnkleRhaukus

dx(x)^1/dx $=\large \frac{\text d(x)^1}{\text dx}$?

70. anonymous

yes that is what I meant to type

71. UnkleRhaukus

$\frac{\text d(x)}{\text dx}=\frac{\text d(x^1)}{\text dx}=1\times x^0=1$

72. UnkleRhaukus

73. anonymous

okay now for the last question here how do this with y"(x) sin

74. anonymous

yes you have

75. UnkleRhaukus

should we evaluate y' in the first question when x=3 before that?

76. anonymous

y'(x) 9x^2+4x+1 so plug in 3 so we have y'(x) 9x^2 (3) +4x(3) +1 which is y'(x) 27x^2+ 12X +1

77. UnkleRhaukus

$y(x)=3x^3+2x^2+x+5$ $y'(x)=9x^2+4x+1$ $y'(3)=9(3)^2+4(3)+1=9\times3\times3+4\times3+1$

78. anonymous

y'(3) 81x^2+12+1

79. anonymous

12x

80. UnkleRhaukus

when you evaluate a function at a particular x value you substitute this value for x in the equation, there should be no x in left when evaluated

81. anonymous

sorry i mess me up by not putting the like this (3)^2

82. anonymous

so the answer is F'(3) 81x^2 + 12X +1