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anonymous

  • 5 years ago

kk then, what is deriative??

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  1. anonymous
    • 5 years ago
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    study of how a function change, or tangent line

  2. amistre64
    • 5 years ago
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    derivative is what is used to describe how one thing changes with respect to another

  3. anonymous
    • 5 years ago
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    hmmm.

  4. anonymous
    • 5 years ago
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    It is like a slope

  5. amistre64
    • 5 years ago
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    in the equation: y = 3x+2, the rate at which y changes with respect to x is '3'

  6. amistre64
    • 5 years ago
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    for every change in 'x', 'y' changes by 3

  7. anonymous
    • 5 years ago
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    hmm then u mean 2y=3x+3+2 ??

  8. amistre64
    • 5 years ago
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    not quite

  9. anonymous
    • 5 years ago
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    or 2y=6x+2?? (the 2 is const in this equetion?

  10. amistre64
    • 5 years ago
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    if you multiply the left by 2 the rights gotta be multiplied by 2 as well

  11. amistre64
    • 5 years ago
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    thats more of a scaling notion

  12. amistre64
    • 5 years ago
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    the derivative has to do with the change in y with respect to x when there is 'no' change in x

  13. anonymous
    • 5 years ago
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    hmmm

  14. amistre64
    • 5 years ago
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    \[\lim_{h -> 0}\frac{(f(x+h)-f(x)}{(x+h)-(x)}\]

  15. amistre64
    • 5 years ago
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    the derivative of: 3x+2 is then (3(x+h)+2) - (3x+2) ----------------- ; when h=0 x+h-x

  16. amistre64
    • 5 years ago
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    3x+3h+2 -3x-2 -------------- ; when h=0 h 3h --- ; when h=0 h 3 ; when h=0

  17. anonymous
    • 5 years ago
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    hmmm

  18. amistre64
    • 5 years ago
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    so the rate of change of a linear function is just the slope of the line

  19. anonymous
    • 5 years ago
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    hmmm

  20. amistre64
    • 5 years ago
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    the rate of change at any given point of a quadratic function is the slope of the line at a given point that is defined as the lim{h ->0} (f(x+h)-f(x)//h)

  21. amistre64
    • 5 years ago
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    x^2+x-3 ; how fast is y changing when x = 4?

  22. anonymous
    • 5 years ago
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    can you answer it? i think i can'T do it yet

  23. amistre64
    • 5 years ago
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    \[\lim_{h -> 0}\frac{ [(x+h)^2 +(x+h)-3]-[x^2+x-3]}{x+h-x}\]

  24. amistre64
    • 5 years ago
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    \[\lim{x^2+h^2 +2xh +x+h-3-x^2-x-3 \over h}\]

  25. anonymous
    • 5 years ago
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    okay give me a min

  26. anonymous
    • 5 years ago
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    getting my pencil and notebook

  27. amistre64
    • 5 years ago
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    that last 3 shoud be +3

  28. anonymous
    • 5 years ago
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    kk

  29. anonymous
    • 5 years ago
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    h^2 + 2xh + h / h

  30. amistre64
    • 5 years ago
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    \[(x^2-x^2)+(x-x)+(-3+3)+(2xh+h) \over h\]

  31. amistre64
    • 5 years ago
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    yes, i missed the ^2 on the h; you are roght

  32. anonymous
    • 5 years ago
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    :D

  33. amistre64
    • 5 years ago
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    \[\lim_{h \rightarrow 0}\left({h^2\over h}+{2xh\over h}\right)\]

  34. anonymous
    • 5 years ago
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    now we give x 3??

  35. amistre64
    • 5 years ago
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    when x=4 :)

  36. amistre64
    • 5 years ago
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    and I think we dropped an h someplace in the writing

  37. anonymous
    • 5 years ago
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    ahh missed 4 xd

  38. anonymous
    • 5 years ago
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    yes why don'T we do it one a more simple func??

  39. amistre64
    • 5 years ago
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    \[\lim_{h ->0}({2xh\over h}+{h\over h}+{h^2\over h})\]

  40. amistre64
    • 5 years ago
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    the simple function was the linear lol

  41. amistre64
    • 5 years ago
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    now when h=0 we get the equation of the derivative: 2x+1

  42. amistre64
    • 5 years ago
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    at x=4; y is changing at a rate of 9 with respect to each x

  43. amistre64
    • 5 years ago
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    the quick way after being taught all the limit definitions and the long way; is to use the power rule: x^2 +x -3 -> 2x +1

  44. anonymous
    • 5 years ago
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    hmm ^^ so that can we use this for another thing?

  45. anonymous
    • 5 years ago
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    with another func i mean

  46. anonymous
    • 5 years ago
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    (don't got the power rule can you explain?)

  47. amistre64
    • 5 years ago
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    yes; the derivative of position tells you velocity at a given point; the derivative of velocity tells you the acceleration at any given point

  48. amistre64
    • 5 years ago
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    the power rule is just the 'pattern' that occurs from doing derivatives the long way..

  49. anonymous
    • 5 years ago
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    oO

  50. amistre64
    • 5 years ago
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    \[Cx^n \implies C*nx^{n-1} \] \[5x^3 \implies 5*3x^{3-1} \iff 15x^2\]

  51. amistre64
    • 5 years ago
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    which way would you do derivatives? the limit way or the power rule?

  52. amistre64
    • 5 years ago
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    \[\begin{array}c 5x^4&+x^2&-4x^0\\20x^3&+2x&+0\end{array}\]

  53. anonymous
    • 5 years ago
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    hmm i think limit way

  54. amistre64
    • 5 years ago
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    ok; do that the limit way... \[\frac{[5(x+k)^4+(x+h)^2-4]-[5x^4+x^2-4]}{h}\]

  55. anonymous
    • 5 years ago
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    which equetion we doing??

  56. amistre64
    • 5 years ago
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    f(x) = \(5x^4 +x^2 -4\) the limit way

  57. amistre64
    • 5 years ago
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    except for it should say 5(x+h)^4 lol

  58. amistre64
    • 5 years ago
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    the power rule is quicker, easier, and just as precise

  59. anonymous
    • 5 years ago
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    ok :=)

  60. anonymous
    • 5 years ago
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    then do it in power rule ^^ if its that trustworthy

  61. amistre64
    • 5 years ago
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    the power rule is the 'patter' that you discover after doing about a thousand limit ways

  62. anonymous
    • 5 years ago
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    it was Cx^n = C x nx^n-1

  63. amistre64
    • 5 years ago
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    yes

  64. anonymous
    • 5 years ago
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    ok

  65. anonymous
    • 5 years ago
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    then for 5(x+4)^4 its >>>> 5 x 4(x)^3 ???

  66. anonymous
    • 5 years ago
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    hey you there?

  67. amistre64
    • 5 years ago
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    ok.... you trying to use the x+h in that?

  68. anonymous
    • 5 years ago
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    no i tried to use power rule that i did C.nx^n-1

  69. amistre64
    • 5 years ago
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    5x^4 ; try it on that one ...

  70. anonymous
    • 5 years ago
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    oka wait

  71. anonymous
    • 5 years ago
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    5 x 4(x)^3

  72. anonymous
    • 5 years ago
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    5 . 4.x^3

  73. anonymous
    • 5 years ago
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    true?

  74. amistre64
    • 5 years ago
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    true; now put it all together; 20x^3

  75. anonymous
    • 5 years ago
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    yess i got it??

  76. amistre64
    • 5 years ago
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    you did; the power rule takes a complicated form and reduces it to simple math

  77. anonymous
    • 5 years ago
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    yes its really usefull :) to you, am i good at maths?

  78. amistre64
    • 5 years ago
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    you can be good at maths if you wanna be :)

  79. anonymous
    • 5 years ago
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    :) really want to be ^^ i want to learn highschool units before i go to hichschool ^^

  80. anonymous
    • 5 years ago
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    really thanks amistre :)

  81. amistre64
    • 5 years ago
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    youre welcome :) keep up the practics

  82. anonymous
    • 5 years ago
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    normally just searching questions on Hz. google xd

  83. amistre64
    • 5 years ago
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    some good free practice can be had at interactmath.com

  84. anonymous
    • 5 years ago
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    ok ty

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