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godorovg

  • 2 years ago

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

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

  2. UnkleRhaukus
    • 2 years ago
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    you also have to divide by the original index

  3. godorovg
    • 2 years ago
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    you divide by 3 because of 3x^3

  4. cornitodisc
    • 2 years ago
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    x^n=nx^n-1

  5. godorovg
    • 2 years ago
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    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
    • 2 years ago
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    opps ive got mixed up

  7. godorovg
    • 2 years ago
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    y' 3x^2 is X^2 right?

  8. UnkleRhaukus
    • 2 years ago
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    i should have said times by the index \[y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=anx^{n-1}\]

  9. UnkleRhaukus
    • 2 years ago
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    so sorry if i confused you

  10. godorovg
    • 2 years ago
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    so times by index makes 3x^3 Y' 6x^2 right?

  11. UnkleRhaukus
    • 2 years ago
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    not quite \[y(x)=3x^3\qquad y'(x)=3\times3x^2\]

  12. godorovg
    • 2 years ago
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    here is what I know y' when doing this you lower the power by one and something??

  13. UnkleRhaukus
    • 2 years ago
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    when taking the derivative multiply by the power and then lower the power

  14. godorovg
    • 2 years ago
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    so you are telling that power before lowering divide to y'

  15. godorovg
    • 2 years ago
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    like this 4x^3 so that would be 'y 4 right?

  16. godorovg
    • 2 years ago
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    4x

  17. UnkleRhaukus
    • 2 years ago
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    i ment times when i said divide originally , dont divide when taking the derivative

  18. godorovg
    • 2 years ago
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    can we start over plz..

  19. UnkleRhaukus
    • 2 years ago
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    yes ,

  20. godorovg
    • 2 years ago
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    y' 4x^2 would y'4x

  21. UnkleRhaukus
    • 2 years ago
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    \[y(x)=ax^n\qquad\Rightarrow\qquad y'(x)=n\times ax^{n-1}\] \[y(x)=4x^2\qquad\Rightarrow\qquad y'(x)=2\times 4x^{2-1}\]

  22. godorovg
    • 2 years ago
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    y'(x) would be 8x

  23. UnkleRhaukus
    • 2 years ago
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    thats right now

  24. godorovg
    • 2 years ago
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    okay so the power is what you are times before lowering the power?

  25. UnkleRhaukus
    • 2 years ago
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    that is right

  26. godorovg
    • 2 years ago
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    so for another example 10X^5 y'(x) 50x^4

  27. UnkleRhaukus
    • 2 years ago
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    you got it \(\checkmark\)

  28. godorovg
    • 2 years ago
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    so back to the problem now

  29. UnkleRhaukus
    • 2 years ago
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    \[y(x)=3x^3+2x^2 +x+5\]\[y'(x)=\]

  30. godorovg
    • 2 years ago
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    y'(x) 6x+4X +5

  31. UnkleRhaukus
    • 2 years ago
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    not quite, note that \[x=x^1\]\[5=5x^0\]

  32. godorovg
    • 2 years ago
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    y'(x) 9x+4+5

  33. UnkleRhaukus
    • 2 years ago
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    try again

  34. godorovg
    • 2 years ago
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    y'(x) 6x^2 +4x+5 +1

  35. UnkleRhaukus
    • 2 years ago
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    ok lets do this one term at a time \[(3x^3)^\prime=\]

  36. godorovg
    • 2 years ago
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    y'(x) 9x^2

  37. UnkleRhaukus
    • 2 years ago
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    right, now \[(2x^2)^\prime=\]

  38. godorovg
    • 2 years ago
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    y'(x) 9x^2 +2x

  39. godorovg
    • 2 years ago
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    y'(x) 9x^2 4+5+1

  40. UnkleRhaukus
    • 2 years ago
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    nope

  41. godorovg
    • 2 years ago
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    y'(x) 9x^2 +4X+5

  42. UnkleRhaukus
    • 2 years ago
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    you have the first two terms right,

  43. UnkleRhaukus
    • 2 years ago
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    \[f(x)=x=x^1\] \[f^\prime(x)=1\times x^{1-1}=x^0=1\]

  44. godorovg
    • 2 years ago
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    so y'(x) 9x^2+4x+1+5

  45. UnkleRhaukus
    • 2 years ago
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    the derivative of the last term \[g(x)=5=5x^0\] \[g^\prime(x)=0\times 5x^{0-1}=\]

  46. godorovg
    • 2 years ago
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    is 0

  47. UnkleRhaukus
    • 2 years ago
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    yeah the derivative of a constant is always zero

  48. godorovg
    • 2 years ago
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    so the end number unless it has a power drop off?

  49. UnkleRhaukus
    • 2 years ago
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    thats right

  50. UnkleRhaukus
    • 2 years ago
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    \[f(x)=x^2+5\qquad f'(x)=2x\]\[g(x)=x^2+9\qquad g'(x)=2x\]\[h(x)=x^2−71\qquad f'(x)=2x\]

  51. UnkleRhaukus
    • 2 years ago
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    that last one is ment to be h'

  52. godorovg
    • 2 years ago
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    cool love it so far because in math you can't drop anything, in calc 1 you can this is way cool..

  53. godorovg
    • 2 years ago
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    do you have time to do one more?

  54. UnkleRhaukus
    • 2 years ago
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    yeah can you get the answer to you original question now?

  55. UnkleRhaukus
    • 2 years ago
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    \[y(x)=3x^3+2x^2 +x+5\] \[y~'(x)=\]

  56. godorovg
    • 2 years ago
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    y'(x) 9x^2+4x+1

  57. UnkleRhaukus
    • 2 years ago
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    \[\Large\color\red\checkmark\]

  58. godorovg
    • 2 years ago
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    okay any other tips here?

  59. UnkleRhaukus
    • 2 years ago
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    \[y'(x)=\frac{\text dy(x)}{\text dx}\]

  60. godorovg
    • 2 years ago
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    explain a little??

  61. UnkleRhaukus
    • 2 years ago
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    its just a different notation, they men the same thing

  62. UnkleRhaukus
    • 2 years ago
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    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. godorovg
    • 2 years ago
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    dx(x)^1?dx

  64. godorovg
    • 2 years ago
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    so the power on the is 1 right?

  65. godorovg
    • 2 years ago
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    just to let know, I am in college not high school

  66. UnkleRhaukus
    • 2 years ago
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    \[y(x)=x=x^1\] \[\frac{\text dy(x)}{\text dx}=y'(x)=1\times x^{1-1}=x^0=1\]

  67. UnkleRhaukus
    • 2 years ago
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    im not sure what you mean by this dx(x)^1?dx

  68. godorovg
    • 2 years ago
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    I meant to use / not?

  69. UnkleRhaukus
    • 2 years ago
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    dx(x)^1/dx \[=\large \frac{\text d(x)^1}{\text dx}\]?

  70. godorovg
    • 2 years ago
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    yes that is what I meant to type

  71. UnkleRhaukus
    • 2 years ago
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    \[\frac{\text d(x)}{\text dx}=\frac{\text d(x^1)}{\text dx}=1\times x^0=1\]

  72. UnkleRhaukus
    • 2 years ago
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    im not sure if i am answer your question?

  73. godorovg
    • 2 years ago
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    okay now for the last question here how do this with y"(x) sin

  74. godorovg
    • 2 years ago
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    yes you have

  75. UnkleRhaukus
    • 2 years ago
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    should we evaluate y' in the first question when x=3 before that?

  76. godorovg
    • 2 years ago
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    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
    • 2 years ago
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    \[y(x)=3x^3+2x^2+x+5\] \[y'(x)=9x^2+4x+1\] \[y'(3)=9(3)^2+4(3)+1=9\times3\times3+4\times3+1\]

  78. godorovg
    • 2 years ago
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    y'(3) 81x^2+12+1

  79. godorovg
    • 2 years ago
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    12x

  80. UnkleRhaukus
    • 2 years ago
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    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. godorovg
    • 2 years ago
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    sorry i mess me up by not putting the like this (3)^2

  82. godorovg
    • 2 years ago
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    so the answer is F'(3) 81x^2 + 12X +1

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