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anonymous

  • one year ago

Can anybody explain the steps to the algebra that arrives at the derivative function to f[x] = x/e^(x^2) f'[x] = ???

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  1. Australopithecus
    • one year ago
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    are you using the definition of the derivative or are you allowed to use the rule set?

  2. logan13
    • one year ago
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    35

  3. anonymous
    • one year ago
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    I think at this point anything goes, I'm mainly just trying to understand how the f[x+h] - f[x] /h equation cancels out.. or if there is another alternative approach that gets me there. Its not actually for any test.. just my ability to use and apply it.

  4. logan13
    • one year ago
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    ya me 2

  5. Australopithecus
    • one year ago
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    Here is a list of basic calc rules (calculus 1) If you want to solve the derivative quickly I would suggest learning the rules they are included on this pdf. I can help you use them and break it down for you if you want. Although if you want we can work through the definition of the derivative method, been a long time since I have dont it but shouldn't be too hard.

  6. anonymous
    • one year ago
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    okay I think that means the definition of the derivative for me

  7. anonymous
    • one year ago
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    ah thats awesome.. man that's a lot of rules.. geez, how do you guys memorize this?

  8. Australopithecus
    • one year ago
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    Its actually not that hard just like learning algebra rules. YOu apply them a million times and you can just look at a function and see its derivative

  9. anonymous
    • one year ago
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    yeah, I can see how this would save a lot of fiddly algebra

  10. Australopithecus
    • one year ago
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    So first you need to set up the expression do you know how to? Hint: for f(x+h) you just replace all x terms in the function with (x+h)

  11. anonymous
    • one year ago
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    yes 1 sec .. I can post how far I got

  12. Australopithecus
    • one year ago
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    Doing the definition of this will probably be really dirty actually lol

  13. Australopithecus
    • one year ago
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    but I will help you none the less :D

  14. anonymous
    • one year ago
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    \[\left( \frac{ x+h }{e^{(x+h)^{2}}} - \frac{x}{e^{x^{2}}} \right)* \frac{1}{h}\] \[\left( (x+h)e^{-(x+h)^2}-x e^{-x^{2}} \right) * \frac{1}{h}\] \[\frac{ (x+h)e^{-(x+h)^2}-x e^{-x^{2}} } {h} \] and a few other variations of that

  15. anonymous
    • one year ago
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    thanks I really really appreciate it.. this is going on 14 hours of head scratching for me now

  16. anonymous
    • one year ago
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    Im not sure if I want to keep a denominator or change it somehow, or if there is a good way to get rid of it.

  17. hartnn
    • one year ago
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    you've solved limits questions before? have you come across \(\Large \lim \limits_{x\to 0 } \dfrac{e^x -1}{x}\) ?

  18. anonymous
    • one year ago
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    theres a good question ... hmmm ... its been a while.. about a year, but I think I did that in my high school pre calc..

  19. hartnn
    • one year ago
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    good, remember L'Hopital's rule? that can be easily solved by it...

  20. anonymous
    • one year ago
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    I couldnt tell you off hand how to solve that..

  21. anonymous
    • one year ago
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    I can go look it up and refresh though

  22. hartnn
    • one year ago
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    by L'Hopital's rule \(\Large \lim \limits_{x\to 0 } \dfrac{e^x -1}{x} =\Large \lim \limits_{x\to 0 } \dfrac{\dfrac{d}{dx}[e^x -1]}{\dfrac{d}{dx}[x]} = \\ \Large \lim \limits_{x\to 0 } e^x = e^0 =1\) If you haven't learnt how to take the derivatove, you can always take \(\Large \lim \limits_{x\to 0 } \dfrac{e^x -1}{x} = 1\) as a standard result

  23. hartnn
    • one year ago
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    now with the algebra part, \(\Large (x+1) e^{-(x+1)^2 } - xe^{-x^2} \) expand (x+1)^2 if we look carefully, we can just factor out \(e^{-x^2}\) from that can you try that ?

  24. anonymous
    • one year ago
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    Yeah, I dont think I've covered this before.. but.. it looks like it makes some sense.. there are two opposing functions of some kind that are equal at extreme infitesimal values of x and therefore give the result one. And from this I guess we can simplify the equation. I'll go read up and fill in this blank in my understanding.

  25. hartnn
    • one year ago
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    x+h, not x+1 sorry :P

  26. hartnn
    • one year ago
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    i'll type the simplification by the time you try \(\Large (x+h) e^{-x^2 +2xh-h^2 } - xe^{-x^2}\) \(\Large (x+h) e^{-x^2} e^{2xh-h^2 } - xe^{-x^2}\) \(\Large e^{-x^2} [(x+h) e^{2xh-h^2 } - x]\)

  27. anonymous
    • one year ago
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    ahhh! you expanded the (x+h)^2 then used the e^(a + b) = e^a e^b rule to get to stage 2.. then factored out e^-x^2 .. jeez I was having a hell of time working out how to do that.. I got basic algebra holes

  28. hartnn
    • one year ago
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    thats right! :) now forget about e^(-x^2) that will be a constant and will be pulled out of the limit, as we are taking the limit with "h" as variable. \(\Large [(x+h) e^{2xh-h^2 } - x]= x(e^{2xh-h^2 }) + h (e^{2xh-h^2 } )-x \) now we have 3 terms and don't forget the h in the denominator

  29. hartnn
    • one year ago
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    the middle term is interesting \(\huge \dfrac{h(e^{2xh-h^2})}{h} = e^{2xh-h^2}\) since h->0 , you can directly plug in h =0 in that ^^

  30. hartnn
    • one year ago
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    from the other 2 terms, factor out the x \(\Large \dfrac{x(e^{2xh-h^2 }) -x}{h}= x[\dfrac{(e^{2xh-h^2 }) -1}{h} ]\) now doesn't that fraction look very similar to e^x-1 ;) I'll let you try further... will come later to check how much you could solve :)

  31. hartnn
    • one year ago
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    ask if you have any doubts in any explanation above

  32. anonymous
    • one year ago
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    I will. thanks, just digesting it..

  33. anonymous
    • one year ago
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    Im having trouble finishing this with a working function.. One of the problems I have is when I go back to this point.. I dont get the same result when I expand.. instead I get .. \[ \Large \left((x+h) e^{-(x+h)^2 } - xe^{-x^2}\right) * \frac{1}{h}\] \[ \Large \left((x+h) e^{-((x+h)(x+h)) } - xe^{-x^2}\right) * \frac{1}{h}\] \[ \Large \left( (x+h) e^{-x^2-2hx-h^2 } - xe^{-x^2}\right) * \frac{1}{h}\] \[ \Large \left( x e^{-x^2-2hx-h^2 } + h e^{-x^2-2hx-h^2 } - xe^{-x^2}\right) * \frac{1}{h}\] \[ \Large e^{-x^2} \left( x e^{-2hx-h^2 } + h e^{-2hx-h^2 } - x\right) * \frac{1}{h}\] And before I go any further may I confirm if I haven't got it wrong from the start?

  34. Australopithecus
    • one year ago
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    Ok so I solved this problem I didnt use L'Hopital's rule. Not really following how the other guy is using it. This is a pretty weird problem, was kind of fun solving because the weird limit at the end. First off to solve this problem you need to know limit rules Here is a list of them http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html First off, your algebra looks fine. Step 1 cancel out h terms in your equation Step 2 you should notice that one of your terms does not have h in the numerator or denominator thus you can treat it as its own function and apply this rule called the addition law: \[\lim_{h \rightarrow a} (g(x) + f(x)) = \lim_{h \rightarrow a} g(x) + \lim_{h \rightarrow a} f(x)\] Also notice that you can treat x terms such as e^-x^2 and x as constants so thus you can simplify your problem even more using the rule called the constant law: \[\lim_{h \rightarrow a} c*f(x) = c*\lim_{h \rightarrow a} f(x)\] (in this case c would be x and e^-x^2 terms) Step 3 The term with no h in it you can solve its limit leaving you with only two terms, combine those terms factor out all the x terms you can. Then you will be left with a pretty weird limit, but it is solvable. Step 4 once you have the weird limit you are going to have to do some funky math tricks (completing the square and realizing and factoring out a term) to get it in the form of the definition of a derivative from there you can realize that the solution to the limit is the derivative of a specific exponential function. I just used derivative rules to solve the limit but you can use the definition. the link below goes over how to solve exponential functions using the definition of a derivative if you need help let me know: http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx If you need help with this final step let me know and if you are confused with any of the details I would be happy to walk you through it.

  35. Australopithecus
    • one year ago
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    Give me a bit I need to look up a proof for the exponential rule using definition of a derivative

  36. Australopithecus
    • one year ago
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    or you could just be content with d/dx e^(f(x)) = f'(x)e^(f(x))

  37. Australopithecus
    • one year ago
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    Do you follow my steps though? they are kind of arbitrary but yeah

  38. anonymous
    • one year ago
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    thanks, I can see how I can get rid of one h easily.. in the middle term.. the other two, I'm not sure how easily they will go. \[\Large e^{-x^2} \left(\frac{ x e^{-2hx-h^2 }}{h} + \frac{ h e^{-2hx-h^2} }{h} - \frac{x}{h} \right) \] \[\Large e^{-x^2} \left(\frac{ x e^{-2hx-h^2 }}{h} + \frac{ e^{-2hx-h^2} }{1} - \frac{x}{h} \right) \] I feel like I need to do something about getting rid of those exponents already, do you have a trick for reducing xe^... any more ?

  39. Australopithecus
    • one year ago
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    You need to use the limit notation while solving your problem

  40. Australopithecus
    • one year ago
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    so right now, \[\lim_{h \rightarrow 0} e^{-x^2}(\frac{xe^{-2hx-h^2}}{h} + e^{-2hx-h^2} + \frac{x}{h})\]

  41. Australopithecus
    • one year ago
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    now apply the rules I showed you above

  42. anonymous
    • one year ago
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    cool this is looking good, okay I might need a bit to read up on pauls notes there.. and get the gist of how the rules apply on this function as it is.. that might take me a bit.. then I have an appointment at 6 tonight., so it might be a few hours before I have this down.

  43. Australopithecus
    • one year ago
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    Lets just try to get it in the right form first?

  44. anonymous
    • one year ago
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    feels like it's 1 minute from an ah ha! moment though

  45. Australopithecus
    • one year ago
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    then we can try to tackle the limit

  46. Australopithecus
    • one year ago
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    so yeah using proof to solve the limit is apparently very long and tricky according to some people who know what they are talking about. the link I posted wont be helpful in this case I dont think

  47. anonymous
    • one year ago
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    if I understand it right, this function will become several limit functions?

  48. Australopithecus
    • one year ago
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    No you will split this into two functions

  49. anonymous
    • one year ago
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    and conceptually, x and e will become roughly the same very small value

  50. Australopithecus
    • one year ago
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    you have 3 terms, you split them into two functions, then you solve one of the limits, then you combine the function with two terms and simplify

  51. Australopithecus
    • one year ago
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    Can you apply the addition rule to your problem?

  52. anonymous
    • one year ago
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    ...looking at that

  53. Australopithecus
    • one year ago
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    If you want an example I can provide one?

  54. Australopithecus
    • one year ago
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    For example \[\lim_{h \rightarrow 0} (x + \frac{x}{h})\] Applying addition rule, f(x) = x and g(x) = x/h so, \[\lim_{h \rightarrow 0} (x + \frac{x}{h}) = \lim_{h \rightarrow 0} x + \lim_{h \rightarrow 0} \frac{x}{h}\]

  55. Australopithecus
    • one year ago
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    we can also apply the constant rule here by assuming x is a constant because we are taking the limit in terms of h so, \[\lim_{h \rightarrow 0} x + \lim_{h \rightarrow 0} \frac{x}{h} = x* \lim_{h \rightarrow 0} 1 + x*\lim_{h \rightarrow 0} \frac{1}{h} = x(\lim_{h \rightarrow 0} 1 + \lim_{h \rightarrow 0} \frac{1}{h})\] this isnt the best example because there is no limit for 1/h thus the addition rule is not valid but yeah just an example of how to do the manipulatons

  56. Australopithecus
    • one year ago
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    or rather the limit of x/h does not exist

  57. anonymous
    • one year ago
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    ah ok is this where I should be next? \[\lim_{h \rightarrow 0} e^{-x^2} (\lim_{h \rightarrow 0} + \frac{xe^{-2hx-h^2}}{h} + \lim_{h \rightarrow 0} \frac{x}{h}+ \lim_{h \rightarrow 0} e^{-2hx-h^2} )\]

  58. Australopithecus
    • one year ago
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    No sorry that is wrong, lets take a step back, you have three terms, and you want to treat all individual x terms as constants. so first off you want to designate an f(x) term and a g(x) term

  59. Australopithecus
    • one year ago
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    look at the first part of my example

  60. Australopithecus
    • one year ago
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    you want to group the terms with h in the denominator

  61. Australopithecus
    • one year ago
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    using the rule, \[\frac{a}{b} + \frac{c}{b} = \frac{a+c}{b}\]

  62. Australopithecus
    • one year ago
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    then you will have two terms

  63. Australopithecus
    • one year ago
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    Then you can declare them as separate functions and apply the addition rule

  64. anonymous
    • one year ago
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    So first if I understand this right, then that term on the front can be ignored, because it does not have h in it, I think this is what you mean by constant? \[\lim_{h \rightarrow 0} \frac{xe^{-2hx-h^2 }+x}{h} + \lim_{h \rightarrow 0} e^{-2hx-h^2} \]

  65. anonymous
    • one year ago
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    or at least I am guessing for the time being, can be put on the side. and used later.

  66. Australopithecus
    • one year ago
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    You should have e^(-x^2) multiplied by everything but yeah that is right

  67. anonymous
    • one year ago
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    ok, yes, that what I was thinking..

  68. Australopithecus
    • one year ago
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    so, \[e^{-x^2}(\lim_{h \rightarrow 0} \frac{xe^{-2hx - h^2} +x}{h} + \lim_{h \rightarrow 0} e^{-2hx - h^2})\]

  69. Australopithecus
    • one year ago
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    You can solve one of your terms leaving you with only one limit to solve

  70. Australopithecus
    • one year ago
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    factor out x of the last remaining limit

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

  72. Australopithecus
    • one year ago
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    yeah now you can solve the limit of e^(-2hx - h^2)

  73. anonymous
    • one year ago
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    should I be thinking something along the lines of dropping that second part of that exponent.. (my signs might be off) but something like \[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{-2hx}/e^{- h^2}) \]

  74. Australopithecus
    • one year ago
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    for the second limit all you have to do is sub 0 into h

  75. Australopithecus
    • one year ago
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    the last limit you need to manipulate it to look like the definition of a derivative

  76. anonymous
    • one year ago
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    -2x h - h*h = -2x *0 - 0*0 = 0 ? or do think of this as 2x and 0's cancel out in some way ?

  77. Australopithecus
    • one year ago
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    no that is right you will have e^0 for the second limit

  78. Australopithecus
    • one year ago
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    the reason we cant solve the first limit is because h is in the denominator and dividing by 0 is undefined

  79. anonymous
    • one year ago
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    oh cool, seemed too easy

  80. Australopithecus
    • one year ago
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    so you have the first term of the derivative but to find the second you have to manipulate the last remaining limit to look like the definition of a derivative, hint complete the square in the exponential and factor

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

  82. Australopithecus
    • one year ago
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    yeah so, you have the lim h->0 e^0 = 1

  83. Australopithecus
    • one year ago
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    do you know how to complete the square?

  84. anonymous
    • one year ago
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    ah right I just spotted that ... \[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} 1) \]

  85. Australopithecus
    • one year ago
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    you should remove the limit notation from the second one

  86. Australopithecus
    • one year ago
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    since it is redundant

  87. anonymous
    • one year ago
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    yeah, bit rusty on it.. there's a couple of ways I recall. not sure if these terms divide nicely..

  88. Australopithecus
    • one year ago
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    the last limit is a bit tricky

  89. Australopithecus
    • one year ago
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    do you have any idea how to make it look like the definition of a deriative

  90. anonymous
    • one year ago
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    I have an idea of how it should look when done..

  91. Australopithecus
    • one year ago
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    So the last limit you want it in the form of \[\frac{f(x+h) - f(x)}{h}\]

  92. Australopithecus
    • one year ago
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    well when you are done you wll have a function in terms of x that is the derivative you are looking for

  93. anonymous
    • one year ago
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    \[f'[x] = ​(1 - 2*x^2)* e^{-x^2]} \]

  94. Australopithecus
    • one year ago
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    huh?

  95. anonymous
    • one year ago
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    when I cheat and ask mathematica to give me the derivative of f[x] that's what it gives me.

  96. Australopithecus
    • one year ago
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    ok so you have this right now \[e^{-x^2}(\lim_{h \rightarrow 0} \frac{e^{-2hx-h^2}-1}{h} + 1)\]

  97. Australopithecus
    • one year ago
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    yeah that is the derivative notice when you expand you get e^(-x^2) which is showing up in the problem atm

  98. anonymous
    • one year ago
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    yes

  99. Australopithecus
    • one year ago
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    oops

  100. Australopithecus
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac{x(e^{-2hx-h^2}-1)}{h}\] notice that x is a constant

  101. Australopithecus
    • one year ago
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    this is the limit you have to solve

  102. Australopithecus
    • one year ago
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    I forgot to write HINT: notice x is a constant

  103. Australopithecus
    • one year ago
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    you need to rewrite the limit so it looks like the definition of a derivative

  104. Australopithecus
    • one year ago
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    you can achieve this by using completing the square and factoring

  105. Australopithecus
    • one year ago
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    out a term

  106. Australopithecus
    • one year ago
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    its actually kind of tricky

  107. Australopithecus
    • one year ago
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    then once you have it in that form you can tell what functions derivative is the solution to the limit

  108. anonymous
    • one year ago
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    yeah I have a dumb question.. usually I complete the square on a nicely formed quadratic ax^2 + bx + c ... I'm having a bit of problem seeing where that is

  109. Australopithecus
    • one year ago
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    You need to complete the square in the exponential

  110. Australopithecus
    • one year ago
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    so you need to complete the square with the following: -2hx-h^2

  111. Australopithecus
    • one year ago
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    then after you complete the square with the exponential you will need to remember this rule: \[e^{a+b} = e^a(e^b)\] and also that if: \[e^{a}e^{-a} = e^{0} = 1\]

  112. Australopithecus
    • one year ago
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    so an example if I had 4ab + 4b^2 I could complete the square by writing (a + 2b)^2 - a^2

  113. Australopithecus
    • one year ago
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    This is actually fairly tricky algebra

  114. Australopithecus
    • one year ago
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    but not really once you see it you are like duh but yeah took me awhile to figure it out

  115. anonymous
    • one year ago
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    Just had to read my high school notes on completing the square again.. (x + b/2)^2 = c ax^2 + bx + c = -h^2 -2h x + 0 = 0 -1 -1 (-h -1x)^2 = 0 I'm having a bit of a brain skid, what to do with that c

  116. anonymous
    • one year ago
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    and the x

  117. Australopithecus
    • one year ago
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    just make a ( )^2 + what ever needs to get canceled out from the expanded ( )^2

  118. Australopithecus
    • one year ago
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    look at my example when you expand (a+2b)^2 you get an extra term so for (a+2b)^2 to equal 4ab + 4b^2 you need to subtract a term from (a+2b)^2

  119. Australopithecus
    • one year ago
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    I have totally forgot that formula lol

  120. Australopithecus
    • one year ago
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    instead I just think of a square that has the same terms in it then I subtract or add something from it to make it equal to what ever I started with

  121. Australopithecus
    • one year ago
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    Here are some videos on completing the square https://www.youtube.com/watch?v=xGOQYTo9AKY https://www.youtube.com/watch?v=Q0IPG_BEnTo maybe these can help I need to go out and eat I will be back later

  122. anonymous
    • one year ago
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    man, thanks a lot tagging along on this journey with me and your help and patience yeah, I need a little time to remember this stuff again.. I used to be really good at completing the square and had it totally down.. could see it conceptually.. now Im just drawing a blank.

  123. Australopithecus
    • one year ago
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    Well I recommend looking at the example provided it is very similar to yours Here it is again 4ab + 4b^2 = (a + 2b)^2 - a^2

  124. Australopithecus
    • one year ago
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    Did you figure it out?

  125. anonymous
    • one year ago
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    Where I get stuck on this completing the square problem is that I have this extra x in the equation and the damn negative sign on both terms.. 4ab + 4b^2 "I could complete the square by writing" (a + 2b)^2 - a^2 (a + 2b)(a+2b) a^2+2ab+2ab+4b^2 (a^2+4ab+4b^2) -a^2 ( 4ab+4b^2) okay so GIVEN -2hx -h^2 (2hx+h^2) *-1 can this be factored like this ???? Then (h +1x)(h +1x) (h +x)(h +x) (h^2 +hx)+(hx +x^2) (h^2 +2hx + x^2) -x^2 gives original expression (h^2 +2hx ) -((h+x)^2 - x^2) -(h+x)^2 + x^2

  126. Australopithecus
    • one year ago
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    yup

  127. Australopithecus
    • one year ago
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    good job

  128. anonymous
    • one year ago
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    lol omg.. that was an effort

  129. Australopithecus
    • one year ago
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    ok write it out

  130. Australopithecus
    • one year ago
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    you are close to it looking like the definition of a specific derivative

  131. Australopithecus
    • one year ago
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    you only really have a single manipulation left look at the rules I gave you above for a hint

  132. Australopithecus
    • one year ago
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    you need to factor it out of the limit

  133. Australopithecus
    • one year ago
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    hint it is the term you added just by completing the square

  134. Australopithecus
    • one year ago
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    remember you want it to look like the definition of a derivative when you are done, that way you can find the specific function you need to find the derivative of to solve the limit

  135. Australopithecus
    • one year ago
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    look at the two rules I posted regard exponentials they are hints to what you have to do

  136. anonymous
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac{x(e^{-2hx-h^2}-1)}{h}\] \[\lim_{h \rightarrow 0} \frac{x(e^{-(h+x)^2} e^{x^2}-1)}{h} \] Im thinking something like this

  137. Australopithecus
    • one year ago
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    that's good You should factor x out of the limit as it is a constant (look at constant rule) You want to factor out e^(x^2) again look at my exponential hints

  138. Australopithecus
    • one year ago
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    you want to pull e^(x^2) out of the limit

  139. Australopithecus
    • one year ago
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    or exponential rules I gave you I mean

  140. anonymous
    • one year ago
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    sorry for delay.. juggling the right exponent law (e^a)^b = e^(a b) . e^(a + b) = e^a e^b

  141. anonymous
    • one year ago
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    trying to see how they fit

  142. Australopithecus
    • one year ago
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    they will help you factor out e^(x^2)

  143. Australopithecus
    • one year ago
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    remember you are trying to make it look like the definition of a derivative

  144. anonymous
    • one year ago
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    should I try expanding that (h+x)^2 again ? h^2 +2hx +x^2

  145. Australopithecus
    • one year ago
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    no

  146. anonymous
    • one year ago
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    I can see how that adds another e^x^2

  147. Australopithecus
    • one year ago
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    we are trying to factor out e^(x^2) from the limit, so that we have an equation as so: \[\lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\] we are trying to find f(x)

  148. Australopithecus
    • one year ago
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    Do you follow?

  149. anonymous
    • one year ago
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    not connecting the dots.. you might need to leave me with it for bit.. I'll get it.. I wont give in to the next ah ha moment.. but it might be 1 minute, might be an hour

  150. Australopithecus
    • one year ago
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    You will get it, just look at this rule: \[e^{a}e^{-a} = e^{a-a} = e^{0} = 1\]

  151. anonymous
    • one year ago
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    I got lost down this tangent.. looking for a way to factor e^(x^2) out of the first term.. but that dont look right \[ \lim_{h \rightarrow 0} \frac{x(e^{-(h+x)(h+x)} e^{x^2}-1)}{h} \] \[ \lim_{h \rightarrow 0} \frac{x(e^{-(h^2+2xh)-x^2} e^{x^2}-1)}{h} \]

  152. Australopithecus
    • one year ago
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    To factor out e^(x^2), you need to change 1 into something

  153. Australopithecus
    • one year ago
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    I gave you a rule that might help

  154. anonymous
    • one year ago
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    are we zeroing out h ? here, I can see that creating e^-(0+x)(0+x) = e^-(x^2)

  155. Australopithecus
    • one year ago
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    just factor e^(x^2) out of the limit

  156. Australopithecus
    • one year ago
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    and you will see what I mean

  157. Australopithecus
    • one year ago
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    look at the rule I posted you need to change 1 to something so you can factor out e^(x^2)

  158. anonymous
    • one year ago
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    this rule e^{a}e^{-a} = e^{a-a} = e^{0} = 1

  159. Australopithecus
    • one year ago
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    right

  160. anonymous
    • one year ago
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    and I change the 1 in that rule, or the -1 at the end of the equation ?

  161. Australopithecus
    • one year ago
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    we are focusing on the limit

  162. Australopithecus
    • one year ago
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    the -1 is not in the limit anymore

  163. anonymous
    • one year ago
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    I am staring at this equation .. maybe Im looking in the wrong place \[\lim_{h \rightarrow 0} \frac{x(e^{-(h+x)^2} e^{x^2}-1)}{h}\]

  164. Australopithecus
    • one year ago
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    oh sorry yeah the -1

  165. anonymous
    • one year ago
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    so another type of 1... e^(x^2) / e^(x^2) ?

  166. Australopithecus
    • one year ago
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    yeah

  167. Australopithecus
    • one year ago
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    now you can factor e^(x^2) and x out of the limit and then look at what you have

  168. Australopithecus
    • one year ago
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    you still here?

  169. anonymous
    • one year ago
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    yes .. staring at this ... \[ \lim_{h \rightarrow 0} \frac{x(e^{-(h+x)^2} e^{x^2}- e^{x^2} / e^{x^2})}{h} \]

  170. Australopithecus
    • one year ago
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    well you need to use the constant rule

  171. Australopithecus
    • one year ago
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    remove x and e^(x^2)

  172. anonymous
    • one year ago
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    wondering how Im going to get that e^x^2 out of there..

  173. Australopithecus
    • one year ago
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    you can factor e^(x^2) and x out of the limit look at the limit rules I posted

  174. anonymous
    • one year ago
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    lookign them up ...

  175. Australopithecus
    • one year ago
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    I will just write it

  176. Australopithecus
    • one year ago
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    \[\lim_{x \rightarrow a} cf(x) = c*\lim_{x \rightarrow a} f(x)\]

  177. Australopithecus
    • one year ago
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    that is the constant rule

  178. Australopithecus
    • one year ago
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    in this case x and e^x^2 are constants

  179. anonymous
    • one year ago
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    trying to understand how that rule is applied to the equation ...

  180. anonymous
    • one year ago
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    does that mean x and e^x^2 are replaced with some value, or cancelled out? or that they now have somer relationship to h in some way?

  181. Australopithecus
    • one year ago
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    they dont cancel out they just leave the limit

  182. anonymous
    • one year ago
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    it looks like they multiply the value of the original function ...

  183. anonymous
    • one year ago
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    oh, so they are allowed to be removed from future calcs.. because they are not actually affected by any change in h

  184. Australopithecus
    • one year ago
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    They get pulled outside the limit

  185. Australopithecus
    • one year ago
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    yes

  186. Australopithecus
    • one year ago
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    so your limit will be simplified

  187. anonymous
    • one year ago
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    goodness me.. Im a slow as brick.

  188. Australopithecus
    • one year ago
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    its ok you aren't dumb just learning

  189. Australopithecus
    • one year ago
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    so write out what you have now

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

  191. Australopithecus
    • one year ago
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    nope

  192. anonymous
    • one year ago
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    ok cool

  193. Australopithecus
    • one year ago
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    close though

  194. Australopithecus
    • one year ago
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    what did you do with e^(x^2), e^(-x^2) and x

  195. anonymous
    • one year ago
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    I threw them away.. so they need to stay in the equation somehow, but outside the limit.

  196. Australopithecus
    • one year ago
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    yes except one of the terms you need to leave in the limit

  197. Australopithecus
    • one year ago
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    e^(-x^2) needs to stay in the equation

  198. Australopithecus
    • one year ago
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    but x and e^(x^2) can be factored out

  199. anonymous
    • one year ago
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    maybe Im not seeing \[\lim_{h \rightarrow 0} \] In the right way.. should I be seeing this as some kind of term ?

  200. Australopithecus
    • one year ago
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    ?

  201. Australopithecus
    • one year ago
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    go back to what you had originally

  202. Australopithecus
    • one year ago
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    had

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

  204. Australopithecus
    • one year ago
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    yes

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

  206. Australopithecus
    • one year ago
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    closer now just pull x and e^(x^2) out front

  207. Australopithecus
    • one year ago
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    of the limit

  208. Australopithecus
    • one year ago
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    now look at your limit look familure?

  209. Australopithecus
    • one year ago
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    hint look at the definition of a derivative

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

  211. anonymous
    • one year ago
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    lol.. thats got to be wrong

  212. anonymous
    • one year ago
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    the limit definitely looks the shape we were after

  213. Australopithecus
    • one year ago
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    yes but it should be written as: \[xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2} - e^{x^2}}{h})\] but you are correction

  214. Australopithecus
    • one year ago
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    what does that look like?

  215. Australopithecus
    • one year ago
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    hint it is just the definition of a derivative

  216. Australopithecus
    • one year ago
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    but what is f(x)?

  217. Australopithecus
    • one year ago
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    if you know f(x) you know that f'(x) is the solution to that limit

  218. anonymous
    • one year ago
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    like some kind of inversion

  219. Australopithecus
    • one year ago
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    what function would you have to plug into the definition of a derivative to get that equation

  220. Australopithecus
    • one year ago
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    or that limit rather

  221. Australopithecus
    • one year ago
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    Make a guess

  222. anonymous
    • one year ago
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    the e^x^2 ?

  223. Australopithecus
    • one year ago
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    wait you made a mistake

  224. Australopithecus
    • one year ago
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    does e^(x)e^(x) = 1?

  225. Australopithecus
    • one year ago
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    by the rule I posted

  226. anonymous
    • one year ago
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    needs a negative

  227. Australopithecus
    • one year ago
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    yeah

  228. Australopithecus
    • one year ago
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    so what is the function then?

  229. anonymous
    • one year ago
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    \[ xe^{-x^2}(\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2} - e^{x^2}}{h}) \] i think this is what you are asking for ? \[xe^{-x^2}\]

  230. Australopithecus
    • one year ago
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    that doesnt make sense

  231. Australopithecus
    • one year ago
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    you want to factor out e^(x^2)

  232. Australopithecus
    • one year ago
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    you cant factor out e^(-x^2)

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

  234. Australopithecus
    • one year ago
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    e^(x^2)e^(x^2) =1?

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

  236. Australopithecus
    • one year ago
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    oops lol but yeah

  237. Australopithecus
    • one year ago
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    that term shouldnt be in the exponential but yeah

  238. anonymous
    • one year ago
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    so I am thiking about these rules.. \[ e^{a}e^{-a} = e^{a-a} = e^{0} = 1 \] and how they relate to \[ xe^{x^2} (\lim_{h \rightarrow 0} \frac{e^{-(x + h)^2}- e^{-x^2}}{h}) \] btw.. is that function correct?

  239. Australopithecus
    • one year ago
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    yes

  240. Australopithecus
    • one year ago
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    That is it :D

  241. Australopithecus
    • one year ago
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    now what if f(x) considering that is the definition of a derivative

  242. Australopithecus
    • one year ago
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    is* not if

  243. Australopithecus
    • one year ago
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    you almost had it before but you there was a mistake

  244. anonymous
    • one year ago
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    Im assuming it is supposed to be what we started with? or is it going to be something different?

  245. Australopithecus
    • one year ago
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    it is something different

  246. Australopithecus
    • one year ago
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    when we are done you have to plug this back into the original problem

  247. Australopithecus
    • one year ago
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    then you will have the derivative

  248. anonymous
    • one year ago
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    xe^{x^2} ? the left portion of that equation ?

  249. Australopithecus
    • one year ago
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    We just isolate the limit we are dealing with because the other stuff would just be a waste to write over and over

  250. Australopithecus
    • one year ago
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    do you recognize that the limit we have now is in the form: \[\frac{f(x+h) + f(x)}{h}\]

  251. anonymous
    • one year ago
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    yes

  252. Australopithecus
    • one year ago
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    then what is f(x)?

  253. anonymous
    • one year ago
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    ah ok.. lol

  254. anonymous
    • one year ago
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    \[ e^{-x^2} \]

  255. Australopithecus
    • one year ago
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    yes

  256. Australopithecus
    • one year ago
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    so the solution to that limit is just the derivative of e^(-x^2)

  257. Australopithecus
    • one year ago
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    do you mind just accepting that, the derivative of: \[e^{f(x)} = f'(x)e^{f(x)}\] This is due to a thing called chain rule

  258. Australopithecus
    • one year ago
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    http://en.wikipedia.org/wiki/Chain_rule#First_proof

  259. anonymous
    • one year ago
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    I am ready to accept such things

  260. Australopithecus
    • one year ago
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    lol

  261. Australopithecus
    • one year ago
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    ok

  262. anonymous
    • one year ago
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    I dont know how to thank you enough.. this was awesome

  263. Australopithecus
    • one year ago
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    so then you just need to use the definition of the derivative to find the derivative of -x^2 Let us say f(x) = x^(-2) thus: \[xe^{x^2}( \lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{h}) = xe^{x^2}(f'(x)e^{f(x)})\]

  264. Australopithecus
    • one year ago
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    sorry lets say f(x) = -x^(2)

  265. Australopithecus
    • one year ago
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    that negative is important almost forgot it

  266. Australopithecus
    • one year ago
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    then you plug that back into your main equation when you find f'(x)

  267. Australopithecus
    • one year ago
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    Do some simplifying and you are done

  268. Australopithecus
    • one year ago
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    sorry I made a mistake in writing above, \[\frac{d}{dx} e^{f(x)} = f'(x)e^{f(x)} \]

  269. Australopithecus
    • one year ago
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    this is what you have to accept this property is due to chain rule

  270. anonymous
    • one year ago
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    okay cool, that seems intuitive enough

  271. Australopithecus
    • one year ago
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    so do you follow everything?

  272. anonymous
    • one year ago
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    well I couldnt teach it .. but I could read it over again, and I think I'd be pretty good to this point.

  273. Australopithecus
    • one year ago
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    Could you do me the favour of rating me as a qualified helper? I will be around tomorrow if you need help with any of the algebra

  274. Australopithecus
    • one year ago
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    this question is weird

  275. Australopithecus
    • one year ago
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    so its ok if you cant teach it as long as you learned stuff along the way

  276. anonymous
    • one year ago
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    I know that mathematica returns a function \[ f`[x] = e^{-x^2} -2x^2 e^{-x^2} \] for \[f[x] = x/e^{x^2}\] and this has bridged a big gap in working out how it came up with that equation

  277. Australopithecus
    • one year ago
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    well yeah you have e^(-x^2) in your equation

  278. anonymous
    • one year ago
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    totally on high Quality help ..

  279. Australopithecus
    • one year ago
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    the second part of the equation is that limit we have been working on

  280. Australopithecus
    • one year ago
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    you can rate me by hitting the orange button at the top

  281. Australopithecus
    • one year ago
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    But yeah, to write out

  282. anonymous
    • one year ago
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    argh.. damnit.. I clicked the wrong icon.. ah crap

  283. Australopithecus
    • one year ago
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    what you have fully now

  284. anonymous
    • one year ago
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    doesnt have undo

  285. anonymous
    • one year ago
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    ah man, sorry, you really deserve this one x 100

  286. Australopithecus
    • one year ago
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    \[e^{-x^{2}}(xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{h}) + 1)\]

  287. Australopithecus
    • one year ago
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    oh you rated someone else?

  288. anonymous
    • one year ago
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    yeah by accident.. hit the wrong icon

  289. Australopithecus
    • one year ago
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    @Preetha can you allow a redo on rating?

  290. Australopithecus
    • one year ago
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    not a big deal

  291. Australopithecus
    • one year ago
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    but you see how this all works out to give you the derivative

  292. Australopithecus
    • one year ago
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    she probably gets over a billion notifications

  293. Australopithecus
    • one year ago
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    well anyways I hope I was helpful

  294. anonymous
    • one year ago
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    so are you saying that this equation we arrived at , can be applied to all functions?

  295. Australopithecus
    • one year ago
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    no just in this case you can do this to solve this problem

  296. Australopithecus
    • one year ago
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    I would recommend learning the rules for doing derivatives it is much faster, not that the definition of the derivative is not useful in some case

  297. anonymous
    • one year ago
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    well it helped me learn some things.. but yeah.. using the rules is definitely the go

  298. hartnn
    • one year ago
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    just as an alternative approach for \(\large \lim \limits_{h \rightarrow 0} \frac{x(e^{-2hx-h^2}-1)}{h}\) \(\large x\lim \limits_{h \rightarrow 0} \dfrac{(e^{-h(2x-h)}-1)}{h}\times \dfrac{-2x-h}{-2x-h}\) \(\large x \lim \limits_{h(-2x-h) \rightarrow 0} \dfrac{(e^{-h(2x-h)}-1)}{h(-2x-h)} \lim \limits_{h \rightarrow 0}(-2x-h) \) \(\large x[1\times (-2x-0)])\) = -2x :)

  299. Australopithecus
    • one year ago
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    you made a mistake it should be -2x^2

  300. Australopithecus
    • one year ago
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    he forgot about the x he pulled out of the limit

  301. anonymous
    • one year ago
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    ah right on yeah..

  302. anonymous
    • one year ago
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    was this at the stage we had to still add back in \[e^{-x^2}\] ?

  303. Australopithecus
    • one year ago
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    I dont really follow his method

  304. Australopithecus
    • one year ago
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    i Just know he made a mistake

  305. Australopithecus
    • one year ago
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    just follow what I wrote and you will solve this problem anyways i am going to sleep now goodnight

  306. anonymous
    • one year ago
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    well thank you very much.. and Im doing the same

  307. hartnn
    • one year ago
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    yes, i forgot the x that i pulled outside, thanks! it'll be -2x^2 and yes again, e^(-x^2) was pulled out earlier.

  308. anonymous
    • one year ago
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    and I think there was a +1 in the equation as well that we eliminated around that stage.. so we have here now, all the pieces to lead us from \[f[x] = x/e^{x^2} \] to \[f'[x]= e^{-x^2}(1-2x^2 ) \] \[f'[x]= e^{-x^2}-2x^2 e^{-x^2} \] Thank you guys.

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