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anonymous

  • one year ago

Learning about derivatives, instantaneous rate of change, ..via differential quotient equation.. and limits equations, I almost have it down now, need a review of the final few posts to check my work.. thanx Given: f[x] = x/e^x^2 f'[x] = Lim (h->0) [(f[x+h] -f[x])/ h] f'[x] = Lim (h->0) [ ((x+h)/(E^(x+h)^2) - (x/e^(x^2)) ) *1/h ]

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  1. anonymous
    • one year ago
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    or at the very least please send me a good link that explains how to deal with canceling variables embedded in a polynomial exponent

  2. jigglypuff314
    • one year ago
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    Do you have to use the definition of the derivative to solve for your given function? derivative of e^u = d/du * e^u for example, derivative of e^(x^3) = (3x^2)*e^(x^3) Here's the proof for the concept if you would like: https://www.khanacademy.org/math/differential-calculus/taking-derivatives/der_common_functions/v/proof-d-dx-e-x-e-x

  3. anonymous
    • one year ago
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    \[f[x] = \frac{x}{e^{x^2}}\] \[ f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] -f[x] }{h} \right) \] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ E^{(x+h)^2} } - \frac{ x}{ E^{(x^2)} } \right) * \frac{1}{h} \right) \]

  4. anonymous
    • one year ago
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    Unfortunately I do..

  5. anonymous
    • one year ago
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    i have really come to admire those rules of the derivative..

  6. amistre64
    • one year ago
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    first principles eh

  7. anonymous
    • one year ago
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    yeah the baby steps..

  8. anonymous
    • one year ago
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    I've been on it for 3 days.. and my OCD wont allow me to let it go

  9. anonymous
    • one year ago
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    I cant friggin sleep, so pissed off.

  10. anonymous
    • one year ago
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    staring at equations for days. Im missing some fundamental cancelation trick of limits.. but all the vids Im watching, I still ahvent found one that deals with a complex rational like this, and a polynomial exponent.

  11. anonymous
    • one year ago
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    I can get there like this.. I know the answer to this is going to be \[ f'[x] = (1-2x^2)e^{x^2}\] and I can get there by.. g[x] = -x^2 g'[x] = -2x h[x] = E^g[x] h'[x] = -2xe^-x^2 f[x] = x h[x] f'[x] = (h'[x]/g'[x]) + (f[x] * g'[x]) but I cant submit this.

  12. anonymous
    • one year ago
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    and I cant reverse the logic, that would allow me to do that with algebra.

  13. Australopithecus
    • one year ago
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    Its alright here, we can try to work through your final limit using this informal proof. \[\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{h}\]

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  14. Australopithecus
    • one year ago
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    Although maybe amistre64 has a better way of solving this limit or problem

  15. anonymous
    • one year ago
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    Is that the first trick of limits, that we can remove x ? \[ f[x] = x/E^{x^2} \] \[ f[x] = x E^{x^2} \]

  16. anonymous
    • one year ago
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    oops .. that 's -x \[ f[x] = x E^{-x^2} \]

  17. Australopithecus
    • one year ago
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    Well this problem demonstrates product rule as well, we already solved part of the derivative, we differentiated x/e^(x^2) in regards to the x term but we still need to differentiate x/e^(x^2) in terms of e^(-x^2) that is why we are left with that limit to solve

  18. Australopithecus
    • one year ago
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    This is where you left off \[e^{-x^2}(xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2}- e^{-x^2}}{h}) + 1)\]

  19. anonymous
    • one year ago
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    oh, I just found your image...

  20. Australopithecus
    • one year ago
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    yeah you could potentially use that to solve your problem

  21. Australopithecus
    • one year ago
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    or the limit

  22. anonymous
    • one year ago
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    whats the strategy to get h out of that numerator ? should that even be our goal from here?

  23. Australopithecus
    • one year ago
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    Well just considering the limit: \[\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2}- e^{-x^2}}{h}\] Follow along with that image I posted you should be able to use it to solve this limit

  24. anonymous
    • one year ago
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    Whats the idea behind the first part of your image where f(x) becomes f(g(x)), is that because the equation x / e^x^2 has been split into two functions? And if so where was the split? was it at the x^2 ?

  25. Australopithecus
    • one year ago
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    There is something you should note though that will likely come up the proof of d/dx e^x = e^x by definition of e: \[\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1\]

  26. Australopithecus
    • one year ago
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    well e^(-x^2) is a composite function

  27. anonymous
    • one year ago
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    oh is that proof because technically, we have e^0 = (almost 1)-1 and we get a something/something

  28. Australopithecus
    • one year ago
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    if f(x) = e^(x) and g(x) = -x^2 then f(g(x)) = e^(-x^2)

  29. Australopithecus
    • one year ago
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    that is how the value of e is defined

  30. Australopithecus
    • one year ago
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    one of the ways it is defined

  31. anonymous
    • one year ago
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    there was another x in there though.. f[x] = x / E^(x^2) is the x removed, or placed into f or g ?

  32. anonymous
    • one year ago
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    oops.. thats E^-x^2

  33. Australopithecus
    • one year ago
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    we pulled it out of the limit

  34. Australopithecus
    • one year ago
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    because it is technically a constant

  35. anonymous
    • one year ago
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    ah ok.. so that sits.. outside the lim definition still ? or does it now hold some value like 1?

  36. anonymous
    • one year ago
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    or does it cancel to 0 ? I havent clicked yet on how that constant rule is actually applied.

  37. anonymous
    • one year ago
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    just watching another vid on constant rule.

  38. anonymous
    • one year ago
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    lol, man, nobody explains this well.. they just gloss over and say ... 'oh we can pull this out.' and they move on

  39. anonymous
    • one year ago
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    okay, I think it goes outside ..

  40. Australopithecus
    • one year ago
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    no it just sits outside the limit, if you have learned product rule this is a result of it. By product rule: d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x) in this problem assuming f(x) = e^(-x^2) and g(x) = x we solved f(x)g'(x) the limit left represents f'(x)g(x) since g(x) = x we can pull it out of the limit because it is just the original function, the limit itself represents f'(x)

  41. anonymous
    • one year ago
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    but it hangs around..

  42. Australopithecus
    • one year ago
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    the e^(x^2) just cancels out, you could cancel it out right now it wouldn't change your solution

  43. anonymous
    • one year ago
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    so in reference to \[\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1\] To be clear, this means that if we encounter e^x then we have encountered the value 1? And anything attached to that or based on that can be considered to be factored with the value 1? so E^x^2 is actually 1^2 ?

  44. Australopithecus
    • one year ago
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    No it is essentially what defines what the value of e is, it comes up when you try to use the definition of a derivative to find the derivative of e^x http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx

  45. anonymous
    • one year ago
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    hmmm, that's an odd and curious property. :/

  46. Australopithecus
    • one year ago
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    I can help you work through the informal proof if you want.

  47. Australopithecus
    • one year ago
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    @amistre64 , @Hero Might be able to help you more so with this.

  48. anonymous
    • one year ago
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    I think if the problem was just f[x] = e^(x^2) then I would have no problem.. x^2 = 2x and the derivative would be 2x * e^x^2 but I'm not sure what to do about that leading x/ .... maybe converting it into this form and then trying to work it out... f[x] = x e^-(x^2) do we just put the x outside the limit from the start? f'[x] = x Lim (h->0) [ (f[x+h] - f[x] ) /h ] It seems like its not a problem of 2 functions, but of 3 f[x] = x h[x] g[x] = E^h[x] h[x] = -x^2

  49. anonymous
    • one year ago
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    sorry thats x g[x]

  50. anonymous
    • one year ago
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    wish you could back edit these posts

  51. Australopithecus
    • one year ago
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    Not sure what you are confused about here, look what I wrote about product rule. where g(x) = x and f(x) = e^(-x^2) the derivative of f(x)g'(x) is just e^(-x^2) This is shown in your problem, if you multiply e^(-x) by both terms left in the bracket you will get your x*(limit you need to solve) + e^(-x^2) the limit remaining is the derivative of f(x)

  52. Australopithecus
    • one year ago
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    You would have this exact limit if you were trying to take the derivative with the definition using e^(-x^2)

  53. Australopithecus
    • one year ago
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    If it was e^(x) and not e^(-x^2) you would have no problem because you wouldn't need chain rule

  54. Australopithecus
    • one year ago
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    you are correct it is a problem of 3 functions

  55. Australopithecus
    • one year ago
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    this problem requires product rule and chain rule

  56. Australopithecus
    • one year ago
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    but the third function is easily differentiable so its not a problem

  57. Australopithecus
    • one year ago
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    your problem is essentially m(x)*f(g(x))

  58. Australopithecus
    • one year ago
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    this is actually a good thing for you to realize since you need to be able to recognize things like this when taking derivatives

  59. Australopithecus
    • one year ago
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    you need to be able to split a function into multiple functions to take a derivative in most cases

  60. anonymous
    • one year ago
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    ah ok.. so I've been thinking g[x] = -x^2 for all cases of g[x] .. in this situation, and not seeing that g[x] is a term used to express generally a factor in an equation that varies from place to place.. where with Expression A (g[x]) * Expression B (f[x]), is in this case, x is in expression A and hence the derivative of g[x] * f[x] is in this case just E^-x^2 and in the next stage of this equation, g[x] can and will be defined to be something else, as it is applied to the general derivative equations.

  61. anonymous
    • one year ago
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    so Ive been scratching my head, trying to interpret the language

  62. Australopithecus
    • one year ago
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    I am not understanding what you are saying. product rule is the following: f(x)*g(x) = f'(x)g(x) + f(x)g'(x) so if you had xln(x) the derivative would be: 1*ln(x) + (1/x)*x

  63. Australopithecus
    • one year ago
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    g(x) = x g'(x) = 1 f(x) = ln(x) f'(x) = 1/x

  64. anonymous
    • one year ago
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    okay, I think I'm going to go read up more on the product rule and the chain rule again... by the way, is this problem going to need Ln[] to finish it off at some point?

  65. Australopithecus
    • one year ago
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    I posted an informal proof you could follow along with

  66. Australopithecus
    • one year ago
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    to solve the limit

  67. anonymous
    • one year ago
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    I was just saying before, that I wasn't understanding what and how g(x) was getting defined. And I just realized it was being defined on the fly, and changing in context of the problem as it evolved.

  68. anonymous
    • one year ago
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    so as you might understand from that last statement.. the informal proof had this mysterious g(x) throughout it.. and it wasn't intuitively clicking how that was defined and applied.

  69. Australopithecus
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac{e^{-(x + h)^2} - e^{-x^2}}{h} = \lim_{h \rightarrow 0} \frac{e^{-(x + h)^2} - e^{-x^2}}{h}*\frac{-(x+h)^2 - x^2}{-(x+h)^2 - x^2}\] do you understand that m(x) = e^(-x^2) is a composite function g(x) = -x^2, f(x) = e^x therefore, m(x) = f(g(x)) = e^(g(x)) = e^(-x^2)

  70. anonymous
    • one year ago
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    I think you've multiplied by 1, using the exponents to construct the 1.. you've chosen -x^2, so it looks like some kind of conjugate, constructed from the exponents.

  71. anonymous
    • one year ago
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    yes, I understand on the m(x) .. thnx

  72. Australopithecus
    • one year ago
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    \[=(\lim_{h \rightarrow 0} \frac{e^{-(-x+h)^2}-e^{-x^2}}{-(x+h)^2 - x^2})*(\lim_{h \rightarrow 0} \frac{-(x+h)^2 - x^2}{h})\]

  73. Australopithecus
    • one year ago
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    oops

  74. Australopithecus
    • one year ago
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    I made a mistake

  75. Australopithecus
    • one year ago
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    that should be +x^2

  76. Australopithecus
    • one year ago
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    because - (-x^2) is + x^2

  77. Australopithecus
    • one year ago
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    you should be able to solve the limit with that

  78. anonymous
    • one year ago
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    okay.. I'll have a go from there.. I got some visitors that just came in.. so give me an hour or so

  79. anonymous
    • one year ago
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    ok cool, they left.. I told em I was busy

  80. anonymous
    • one year ago
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    Are you saying this equation should look like this? \[ \lim_{h \rightarrow 0} \frac{e^{-(x + h)^2} - e^{-x^2}}{h}*\frac{-(x+h)^2 + x^2}{-(x+h)^2 + x^2} \]

  81. Australopithecus
    • one year ago
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    Here \[(\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{-(x+h)^2+ x^2})(\lim_{h \rightarrow 0} \frac{-(x+h)^2+ x^2}{h})\]

  82. anonymous
    • one year ago
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    ah because of the - in the top term .. cool

  83. Australopithecus
    • one year ago
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    yeah then it splits into two limits

  84. Australopithecus
    • one year ago
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    just solve them now they should be solvable

  85. Australopithecus
    • one year ago
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    you will need to expand

  86. Australopithecus
    • one year ago
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    everything

  87. anonymous
    • one year ago
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    that will be amazing

  88. anonymous
    • one year ago
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    Im excited

  89. anonymous
    • one year ago
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    \[ (\lim_{h \rightarrow 0} \frac{ e^{-(x+h)^2}-e^{-x^2} }{ -(x+h)^2+ x^2 } ) (\lim_{h \rightarrow 0} \frac{ -(x+h)^2+ x^2 }{ h } ) \] \[ (\lim_{h \rightarrow 0} \frac{ e^{-x^2-2hx-h^2}-e^{-x^2} }{ -x^2-2hx-h^2+ x^2 } ) (\lim_{h \rightarrow 0} \frac{ -x^2-2hx-h^2+ x^2 }{ h } ) \] so far ...

  90. Australopithecus
    • one year ago
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    cancel stuff out

  91. anonymous
    • one year ago
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    \[(\lim_{h \rightarrow 0} \frac{ e^{-x^2-2hx-h^2}-e^{-x^2} }{ -2hx-h^2 } ) (\lim_{h \rightarrow 0} \frac{ -2hx-h^2 }{ h } )\] ... I think I'm going the wrong way with this.. should it be canceling within the exponent of the numerator somehow?

  92. anonymous
    • one year ago
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    this looks like it will zero out again. maybe if I multiply by log[] / log[] ? I dont know how else to get access to those exponents.. I can split them if they are E^(a + b) but E^(a - b) is a little trickier.

  93. anonymous
    • one year ago
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    I'll read on L hopitals. and the chain rule.. I cant help wondering if is requires some kind of Log[] / Log [] to pull that exponent out down out of the stratosphere. so we can get access to the terms.

  94. anonymous
    • one year ago
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    those negative exponents are hard to work with.. I know we can use a square root rule, for square roots, and Im not sure about higher degrees, but does that same idea apply to a polynomial in the exponent?

  95. anonymous
    • one year ago
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    maybe we can treat them like exponential functions in the global scope, and just throw away the low degree terms.

  96. anonymous
    • one year ago
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    or maybe we run the limit just on the exponent, and then after knowing which way the exponent goes.. we can then say what it will do as x when in e^(x)

  97. anonymous
    • one year ago
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    can you give me a screen shot, wolfram alpha crashes my computer if I go anywhere near it.. I think the people who run my course put some widget into chrome that kills it.

  98. anonymous
    • one year ago
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    if not that's okay.

  99. Australopithecus
    • one year ago
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  100. Australopithecus
    • one year ago
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    oh crud I just realized

  101. Australopithecus
    • one year ago
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    we can solve this

  102. Australopithecus
    • one year ago
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    really easily using the definition of e

  103. anonymous
    • one year ago
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    damn, how the hell does it get that?

  104. Australopithecus
    • one year ago
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    pull e^(-x^2) out of the limit what do you get?

  105. Australopithecus
    • one year ago
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    I spaced this is easily solvable using the definition of e

  106. anonymous
    • one year ago
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    do we pull e^(-x^2) from the top and bottom? or just the top?

  107. Australopithecus
    • one year ago
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    what?

  108. Australopithecus
    • one year ago
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    just factor out e^(-x^2)

  109. Australopithecus
    • one year ago
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    and you end up with the definition of e which equals 1 because -2xh - h^2 is essentially the same as just h, as h approaches 0

  110. anonymous
    • one year ago
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    zero out h?

  111. Australopithecus
    • one year ago
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    No I am saying, \[\lim_{h \rightarrow 0} \frac{e^{-2x - h^2} -1}{-2xh - h^2} = \lim_{h \rightarrow 0} \frac{e^{h(-2x - h)} -1}{h(-2x - h)} =\lim_{h \rightarrow 0} \frac{e^{h} -1}{h} = 1\]

  112. Australopithecus
    • one year ago
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    so your problem is solved

  113. Australopithecus
    • one year ago
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    just factor out e^(-x^2) and the rest turns to 1

  114. Australopithecus
    • one year ago
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    plug it back into the original equation you had in place of the limit and it is solved

  115. Australopithecus
    • one year ago
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    You have to accept the assumption that \[h \approx h(2x - h)\]

  116. Australopithecus
    • one year ago
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    if you multiply something by a really small number you are going to get a really small number

  117. anonymous
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac {e^{-(x + h)^2} - e^{-x^2}} {-(x+h)^2 + x^2} \] \[\lim_{h \rightarrow 0} \frac {e^{-(x^2 + 2hx +h^2)} - e^{-x^2}} {-(x^2+2hx +h^2) + x^2} \] \[\lim_{h \rightarrow 0} \frac {e^{-x^2 - 2hx -h^2} - e^{-x^2}} {-x^2-2hx -h^2 + x^2} \] \[\lim_{h \rightarrow 0} \frac {e^{-x^2}*(1/e^{2hx})*(1/e^{h^2}) - e^{-x^2}} {-x^2-2hx -h^2 + x^2} \] \[ e^{-x^2} \lim_{h \rightarrow 0} \frac { (1/e^{2hx})*(1/e^{h^2}) - e^{-x^2}} {-2hx -h^2 } \]

  118. Australopithecus
    • one year ago
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    the rest is equal to 1 by the definition of the value of e

  119. anonymous
    • one year ago
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    so e^2hx is .. e^2*0*x = e^0 = 1 ? you mean?

  120. anonymous
    • one year ago
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    but dont we still have the problem of -2hx - h^2 = -2*0*x - 0*0 = 0 ?

  121. anonymous
    • one year ago
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    in the denominator

  122. Australopithecus
    • one year ago
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    no, I am simply saying \[\lim_{h \rightarrow 0}\frac{e^{-2hx - h^2}-1}{-2hx - h^2} = \lim_{h \rightarrow 0}\frac{e^{h}-1}{h} = 1\]

  123. Australopithecus
    • one year ago
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    I can't make it any more clear than this

  124. Australopithecus
    • one year ago
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    \[-2xe^{-x^2}\] is what you get after canceling out h and solving the limit on the left

  125. anonymous
    • one year ago
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    sorry, you need to be that clear with me.. :)

  126. Australopithecus
    • one year ago
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    do you follow though?

  127. anonymous
    • one year ago
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    yeah ... actually yeah.. I can feel the air in my lungs again .

  128. Australopithecus
    • one year ago
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    ok so the problem is solved are you content?

  129. Australopithecus
    • one year ago
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    you can do the final bit of algebra now

  130. anonymous
    • one year ago
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    -2hx-h^2 is equivalent to h in this situation

  131. Australopithecus
    • one year ago
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    I think this is what the other guy was trying to point out to us earlier

  132. Australopithecus
    • one year ago
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    yes

  133. Australopithecus
    • one year ago
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    in the original question but I hadn't really made that connection as of yet

  134. anonymous
    • one year ago
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    yeah.. ah ah..! okay.. thats' what that pauls website was trying to say too I think.. they mentioned this.. but it went over my head

  135. Australopithecus
    • one year ago
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    so you can solve this now?

  136. anonymous
    • one year ago
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    I think so.. , so I just have to reduce to I find that ahh.. property/ equation, Im not sure what you call that class of equations.. signatures or something? and then from there.. we have it.. just pull all the terms back in that were placed outside of the limit.. and you have your equation.

  137. Australopithecus
    • one year ago
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    Just try to write out the solution

  138. anonymous
    • one year ago
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    \[ e^{-x^2} \lim_{h \rightarrow 0} \frac { e^{-2hx-h^2} - e^{-x^2}} {-2hx -h^2 } = \lim_{h \rightarrow 0} \frac { e^{h} -1} {h } = 1 \] amI on the right track here?

  139. Australopithecus
    • one year ago
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    not really

  140. Australopithecus
    • one year ago
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    the limit becomes 1 not the entire expression

  141. anonymous
    • one year ago
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    \[f[x] = \frac{x}{e^{x^2}} \] \[f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] -f[x] }{h} \right)\] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ e^{(x+h)^2} } - \frac{ x}{ e^{x^2} } \right) * \frac{1}{h} \right) \] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( (x+h) e^{-(x+h)^2} - x e^{-x^2} \right) * \frac{1}{h} \right) \] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( x e^{-x^2-2hx-h^2} + h e^{-x^2-2hx-h^2} - x e^{-x^2} \right) * \frac{1}{h} \right) \] \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \left( x e^{-2hx-h^2} + h e^{-2hx-h^2} - x \right) * \frac{1}{h} \right) \right) \] \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} }{h} + \frac{ h e^{-2hx-h^2} }{h} - \frac{ x }{h} \right) \right) \] \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} + \frac{ e^{-2hx-h^2} }{1} \right) \right) \] \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + \lim_{h \rightarrow 0} \frac{ e^{-2hx-h^2} }{1} \right) \] Limit of 2nd term = 1 \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + 1 \right) \] Factor out x \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{-2hx-h^2} - 1 }{h} \right) + 1 \right) \] Very close to the definition for e \[\lim_{h \rightarrow 0} \frac{e^h -1}{ h} = 1 \] \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(-2x-h)} - 1 }{h} * \frac{-2x-h}{-2x-h} \right) + 1 \right) \] \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(-2x-h)} - 1 }{h(-2x-h)} * \lim_{h \rightarrow 0} ({-2x-h}) \right) + 1 \right) \] So by definition of E and the constant law \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h} - 1 }{h} * \lim_{h \rightarrow 0} ({-2x-h}) \right) + 1 \right) \] So by definition of E and the constant law \[ f'[x] = x e^{-x^2} \left( 1 * ({-2x-0}) + 1 \right) \] \[ f'[x] = x e^{-x^2} \left( -2x+1 \right) \] \[ f'[x] = e^{-x^2} \left( -2x^2+x \right) \] \[ f'[x] = x e^{-x^2}-2x^2e^{-x^2} \]

  142. anonymous
    • one year ago
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    SOOO close.. but I factored an extra x in there somehow.. it must be something to do with how I pulled the variables out.. I guess you need to make sure that there are brackets around the different levels. And hence why they're such sticklers for making sure the lim h->0 is carried through on all the stages.

  143. anonymous
    • one year ago
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    btw.. thank you, thank you, to everyone who helped me get this far..

  144. anonymous
    • one year ago
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    GIVEN \[ f[x] = \frac{x}{e^{x^2}} \] FIND \[ f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] -f[x] }{h} \right) \] Substitute function \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ e^{(x+h)^2} } - \frac{ x}{ e^{x^2} } \right) * \frac{1}{h} \right) \] Move into numerator \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( (x+h) e^{-(x+h)^2} - x e^{-x^2} \right) * \frac{1}{h} \right) \] Expand so we can isolate the first term e^-x^2 \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( x e^{-x^2-2hx-h^2} + h e^{-x^2-2hx-h^2} - x e^{-x^2} \right) * \frac{1}{h} \right) \] Pull e^(-x^2) out of the limit. \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \left( x e^{-2hx-h^2} + h e^{-2hx-h^2} - x \right) * \frac{1}{h} \right) \right) \] Distribute 1/h \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} }{h} + \frac{ h e^{-2hx-h^2} }{h} - \frac{ x }{h} \right) \right) \] Simplify and combine like terms \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} + \frac{ e^{-2hx-h^2} }{1} \right) \right) \] Isolate the 2nd term into it’s own limit and remove from the main limit. \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + \lim_{h \rightarrow 0} \frac{ e^{-2hx-h^2} }{1} \right) \] Simplify 1/e^(h=0) = 1 \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + 1 \right) \] Factor out x \[ f'[x] = e^{-x^2} \left( x \left( \lim_{h \rightarrow 0} \left( \frac{ e^{-2hx-h^2} - 1 }{h} \right) \right) + 1 \right) \] Very close to definition for e \[ \lim_{h \rightarrow 0} \frac{e^h -1}{ h} = 1 \] neutralize the limit, by making h term = h term in numerator and denominator so we have equivalence to E definition. \[ f'[x] = x e^{-x^2} \left( x \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(-2x-h)} - 1 }{h} * \frac{-2x-h}{-2x-h} \right) \right) + 1 \right) \] Simplify and isolate the new term we used to create a 1. \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( x \left( \frac{ e^{h(-2x-h)} - 1 }{h(-2x-h)} * \lim_{h \rightarrow 0} ({-2x-h}) \right) \right) + 1 \right) \] So by definition of E and the constant law we have two final limits. \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( x \left( \frac{ e^{h} - 1 }{h} * \lim_{h \rightarrow 0} ({-2x-h}) \right) \right) + 1 \right) \] Now we just simplify our expression to the derivative. \[ f'[x] = e^{-x^2} \left( \left( x \left( 1 * (-2x-0) \right) \right) + 1 \right) \] \[ f'[x] = e^{-x^2} \left( \left( x \left( -2x \right) \right) + 1 \right) \] \[ f'[x] = e^{-x^2} \left( -2x^2 + 1 \right) \] OMG!! I dont believe it! Has it been 4 days and 40+ hours!! THATS IT!! \[ f'[x] = e^{-x^2} \left( 1-2x^2 \right) \] \[ f'[x] = e^{-x^2}-2x^2e^{-x^2} \]

  145. hartnn
    • one year ago
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    excellent work hughfuve ^^ ! :)

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