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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] = ???
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] = ???

This Question is Closed

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2are you using the definition of the derivative or are you allowed to use the rule set?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Here 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay I think that means the definition of the derivative for me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah thats awesome.. man that's a lot of rules.. geez, how do you guys memorize this?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Its 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, I can see how this would save a lot of fiddly algebra

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2So 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes 1 sec .. I can post how far I got

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Doing the definition of this will probably be really dirty actually lol

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2but I will help you none the less :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left( \frac{ x+h }{e^{(x+h)^{2}}}  \frac{x}{e^{x^{2}}} \right)* \frac{1}{h}\] \[\left( (x+h)e^{(x+h)^2}x e^{x^{2}} \right) * \frac{1}{h}\] \[\frac{ (x+h)e^{(x+h)^2}x e^{x^{2}} } {h} \] and a few other variations of that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks I really really appreciate it.. this is going on 14 hours of head scratching for me now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im 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.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2you've solved limits questions before? have you come across \(\Large \lim \limits_{x\to 0 } \dfrac{e^x 1}{x}\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0theres a good question ... hmmm ... its been a while.. about a year, but I think I did that in my high school pre calc..

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2good, remember L'Hopital's rule? that can be easily solved by it...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I couldnt tell you off hand how to solve that..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can go look it up and refresh though

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2by 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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2now 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 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, 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.

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2i'll type the simplification by the time you try \(\Large (x+h) e^{x^2 +2xhh^2 }  xe^{x^2}\) \(\Large (x+h) e^{x^2} e^{2xhh^2 }  xe^{x^2}\) \(\Large e^{x^2} [(x+h) e^{2xhh^2 }  x]\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ahhh! 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

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2thats 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. \(\Large [(x+h) e^{2xhh^2 }  x]= x(e^{2xhh^2 }) + h (e^{2xhh^2 } )x \) now we have 3 terms and don't forget the h in the denominator

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2the middle term is interesting \(\huge \dfrac{h(e^{2xhh^2})}{h} = e^{2xhh^2}\) since h>0 , you can directly plug in h =0 in that ^^

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2from the other 2 terms, factor out the x \(\Large \dfrac{x(e^{2xhh^2 }) x}{h}= x[\dfrac{(e^{2xhh^2 }) 1}{h} ]\) now doesn't that fraction look very similar to e^x1 ;) I'll let you try further... will come later to check how much you could solve :)

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2ask if you have any doubts in any explanation above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I will. thanks, just digesting it..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im 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 .. \[ \Large \left((x+h) e^{(x+h)^2 }  xe^{x^2}\right) * \frac{1}{h}\] \[ \Large \left((x+h) e^{((x+h)(x+h)) }  xe^{x^2}\right) * \frac{1}{h}\] \[ \Large \left( (x+h) e^{x^22hxh^2 }  xe^{x^2}\right) * \frac{1}{h}\] \[ \Large \left( x e^{x^22hxh^2 } + h e^{x^22hxh^2 }  xe^{x^2}\right) * \frac{1}{h}\] \[ \Large e^{x^2} \left( x e^{2hxh^2 } + h e^{2hxh^2 }  x\right) * \frac{1}{h}\] And before I go any further may I confirm if I haven't got it wrong from the start?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Ok 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.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Give me a bit I need to look up a proof for the exponential rule using definition of a derivative

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2or you could just be content with d/dx e^(f(x)) = f'(x)e^(f(x))

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Do you follow my steps though? they are kind of arbitrary but yeah

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks, 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. \[\Large e^{x^2} \left(\frac{ x e^{2hxh^2 }}{h} + \frac{ h e^{2hxh^2} }{h}  \frac{x}{h} \right) \] \[\Large e^{x^2} \left(\frac{ x e^{2hxh^2 }}{h} + \frac{ e^{2hxh^2} }{1}  \frac{x}{h} \right) \] 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 ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2You need to use the limit notation while solving your problem

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so right now, \[\lim_{h \rightarrow 0} e^{x^2}(\frac{xe^{2hxh^2}}{h} + e^{2hxh^2} + \frac{x}{h})\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2now apply the rules I showed you above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cool 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.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Lets just try to get it in the right form first?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0feels like it's 1 minute from an ah ha! moment though

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2then we can try to tackle the limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if I understand it right, this function will become several limit functions?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2No you will split this into two functions

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and conceptually, x and e will become roughly the same very small value

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Can you apply the addition rule to your problem?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2If you want an example I can provide one?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2For example \[\lim_{h \rightarrow 0} (x + \frac{x}{h})\] Applying addition rule, f(x) = x and g(x) = x/h so, \[\lim_{h \rightarrow 0} (x + \frac{x}{h}) = \lim_{h \rightarrow 0} x + \lim_{h \rightarrow 0} \frac{x}{h}\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2we 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2or rather the limit of x/h does not exist

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah ok is this where I should be next? \[\lim_{h \rightarrow 0} e^{x^2} (\lim_{h \rightarrow 0} + \frac{xe^{2hxh^2}}{h} + \lim_{h \rightarrow 0} \frac{x}{h}+ \lim_{h \rightarrow 0} e^{2hxh^2} )\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2No 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2look at the first part of my example

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you want to group the terms with h in the denominator

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2using the rule, \[\frac{a}{b} + \frac{c}{b} = \frac{a+c}{b}\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2then you will have two terms

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Then you can declare them as separate functions and apply the addition rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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? \[\lim_{h \rightarrow 0} \frac{xe^{2hxh^2 }+x}{h} + \lim_{h \rightarrow 0} e^{2hxh^2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or at least I am guessing for the time being, can be put on the side. and used later.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2You should have e^(x^2) multiplied by everything but yeah that is right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, yes, that what I was thinking..

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so, \[e^{x^2}(\lim_{h \rightarrow 0} \frac{xe^{2hx  h^2} +x}{h} + \lim_{h \rightarrow 0} e^{2hx  h^2})\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2You can solve one of your terms leaving you with only one limit to solve

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2factor out x of the last remaining limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ e^{x^2}(\lim_{h \rightarrow 0} \frac{x(e^{2hx  h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{2hx  h^2}) \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2yeah now you can solve the limit of e^(2hx  h^2)

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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2for the second limit all you have to do is sub 0 into h

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2the last limit you need to manipulate it to look like the definition of a derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02x h  h*h = 2x *0  0*0 = 0 ? or do think of this as 2x and 0's cancel out in some way ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2no that is right you will have e^0 for the second limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2the reason we cant solve the first limit is because h is in the denominator and dividing by 0 is undefined

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh cool, seemed too easy

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ e^{x^2}(\lim_{h \rightarrow 0} \frac{x(e^{2hx  h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{2x* 0 0 * 0}) \] \[ e^{x^2}(\lim_{h \rightarrow 0} \frac{x(e^{2hx  h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{0}) \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2yeah so, you have the lim h>0 e^0 = 1

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2do you know how to complete the square?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah right I just spotted that ... \[ e^{x^2}(\lim_{h \rightarrow 0} \frac{x(e^{2hx  h^2} +1)}{h} + \lim_{h \rightarrow 0} 1) \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you should remove the limit notation from the second one

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2since it is redundant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, bit rusty on it.. there's a couple of ways I recall. not sure if these terms divide nicely..

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2the last limit is a bit tricky

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2do you have any idea how to make it look like the definition of a deriative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have an idea of how it should look when done..

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2So the last limit you want it in the form of \[\frac{f(x+h)  f(x)}{h}\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2well when you are done you wll have a function in terms of x that is the derivative you are looking for

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f'[x] = (1  2*x^2)* e^{x^2]} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when I cheat and ask mathematica to give me the derivative of f[x] that's what it gives me.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2ok so you have this right now \[e^{x^2}(\lim_{h \rightarrow 0} \frac{e^{2hxh^2}1}{h} + 1)\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2yeah that is the derivative notice when you expand you get e^(x^2) which is showing up in the problem atm

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{h \rightarrow 0} \frac{x(e^{2hxh^2}1)}{h}\] notice that x is a constant

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2this is the limit you have to solve

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2I forgot to write HINT: notice x is a constant

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you need to rewrite the limit so it looks like the definition of a derivative

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you can achieve this by using completing the square and factoring

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2its actually kind of tricky

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2then once you have it in that form you can tell what functions derivative is the solution to the limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2You need to complete the square in the exponential

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so you need to complete the square with the following: 2hxh^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2then 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: \[e^{a}e^{a} = e^{0} = 1\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so an example if I had 4ab + 4b^2 I could complete the square by writing (a + 2b)^2  a^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2This is actually fairly tricky algebra

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2but not really once you see it you are like duh but yeah took me awhile to figure it out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Just 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2just make a ( )^2 + what ever needs to get canceled out from the expanded ( )^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2look 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2I have totally forgot that formula lol

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2instead 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Here 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0man, 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.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Well 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Did you figure it out?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol omg.. that was an effort

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2ok write it out

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you are close to it looking like the definition of a specific derivative

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you only really have a single manipulation left look at the rules I gave you above for a hint

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you need to factor it out of the limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2hint it is the term you added just by completing the square

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2remember 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2look at the two rules I posted regard exponentials they are hints to what you have to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{h \rightarrow 0} \frac{x(e^{2hxh^2}1)}{h}\] \[\lim_{h \rightarrow 0} \frac{x(e^{(h+x)^2} e^{x^2}1)}{h} \] Im thinking something like this

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2that'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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you want to pull e^(x^2) out of the limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2or exponential rules I gave you I mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry for delay.. juggling the right exponent law (e^a)^b = e^(a b) . e^(a + b) = e^a e^b

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0trying to see how they fit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2they will help you factor out e^(x^2)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2remember you are trying to make it look like the definition of a derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0should I try expanding that (h+x)^2 again ? h^2 +2hx +x^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can see how that adds another e^x^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2we 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)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Do you follow?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0not 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2You will get it, just look at this rule: \[e^{a}e^{a} = e^{aa} = e^{0} = 1\]

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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2To factor out e^(x^2), you need to change 1 into something

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2I gave you a rule that might help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0are we zeroing out h ? here, I can see that creating e^(0+x)(0+x) = e^(x^2)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2just factor e^(x^2) out of the limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2and you will see what I mean

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2look at the rule I posted you need to change 1 to something so you can factor out e^(x^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this rule e^{a}e^{a} = e^{aa} = e^{0} = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and I change the 1 in that rule, or the 1 at the end of the equation ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2we are focusing on the limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2the 1 is not in the limit anymore

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am staring at this equation .. maybe Im looking in the wrong place \[\lim_{h \rightarrow 0} \frac{x(e^{(h+x)^2} e^{x^2}1)}{h}\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2oh sorry yeah the 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so another type of 1... e^(x^2) / e^(x^2) ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2now you can factor e^(x^2) and x out of the limit and then look at what you have

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you still here?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes .. staring at this ... \[ \lim_{h \rightarrow 0} \frac{x(e^{(h+x)^2} e^{x^2} e^{x^2} / e^{x^2})}{h} \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2well you need to use the constant rule

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2remove x and e^(x^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wondering how Im going to get that e^x^2 out of there..

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you can factor e^(x^2) and x out of the limit look at the limit rules I posted

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2I will just write it

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow a} cf(x) = c*\lim_{x \rightarrow a} f(x)\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2that is the constant rule

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2in this case x and e^x^2 are constants

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0trying to understand how that rule is applied to the equation ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does 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?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2they dont cancel out they just leave the limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it looks like they multiply the value of the original function ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, so they are allowed to be removed from future calcs.. because they are not actually affected by any change in h

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2They get pulled outside the limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so your limit will be simplified

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0goodness me.. Im a slow as brick.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2its ok you aren't dumb just learning

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so write out what you have now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \lim_{h \rightarrow 0} \frac{e^{(h+x)^2} }{h} \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2what did you do with e^(x^2), e^(x^2) and x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I threw them away.. so they need to stay in the equation somehow, but outside the limit.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2yes except one of the terms you need to leave in the limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2e^(x^2) needs to stay in the equation

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2but x and e^(x^2) can be factored out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe Im not seeing \[\lim_{h \rightarrow 0} \] In the right way.. should I be seeing this as some kind of term ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2go back to what you had originally

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \lim_{h \rightarrow 0} \frac{x(e^{(h+x)^2} e^{x^2} e^{x^2} / e^{x^2})}{h} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{h \rightarrow 0} \frac{x(e^{x^2}( e^{(h+x)^2}  e^{x^2} ))}{h} \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2closer now just pull x and e^(x^2) out front

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2now look at your limit look familure?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2hint look at the definition of a derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ x(e^{x^2}( \lim_{h \rightarrow 0} \frac{ e^{(h+x)^2}  e^{x^2} }{h} ))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol.. thats got to be wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the limit definitely looks the shape we were after

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2yes 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2what does that look like?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2hint it is just the definition of a derivative

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2but what is f(x)?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2if you know f(x) you know that f'(x) is the solution to that limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0like some kind of inversion

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2what function would you have to plug into the definition of a derivative to get that equation

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2or that limit rather

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2wait you made a mistake

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2does e^(x)e^(x) = 1?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2by the rule I posted

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so what is the function then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{(h+x)^2}  e^{x^2}}{h}) \] i think this is what you are asking for ? \[xe^{x^2}\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2that doesnt make sense

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you want to factor out e^(x^2)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you cant factor out e^(x^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{(h+x)^2}  e^{x^2}}{h}) \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2e^(x^2)e^(x^2) =1?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2\[xe^x (\lim_{h \rightarrow 0} \frac{e^{(x + h)^2 e^{x^2}}}{h})\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2oops lol but yeah

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2that term shouldnt be in the exponential but yeah

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so I am thiking about these rules.. \[ e^{a}e^{a} = e^{aa} = e^{0} = 1 \] 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?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2That is it :D

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2now what if f(x) considering that is the definition of a derivative

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you almost had it before but you there was a mistake

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im assuming it is supposed to be what we started with? or is it going to be something different?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2it is something different

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2when we are done you have to plug this back into the original problem

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2then you will have the derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0xe^{x^2} ? the left portion of that equation ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2We just isolate the limit we are dealing with because the other stuff would just be a waste to write over and over

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2do you recognize that the limit we have now is in the form: \[\frac{f(x+h) + f(x)}{h}\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2then what is f(x)?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so the solution to that limit is just the derivative of e^(x^2)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2do 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am ready to accept such things

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont know how to thank you enough.. this was awesome

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so 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: \[xe^{x^2}( \lim_{h \rightarrow 0} \frac{e^{(x+h)^2}e^{x^2}}{h}) = xe^{x^2}(f'(x)e^{f(x)})\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2sorry lets say f(x) = x^(2)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2that negative is important almost forgot it

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2then you plug that back into your main equation when you find f'(x)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Do some simplifying and you are done

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2sorry I made a mistake in writing above, \[\frac{d}{dx} e^{f(x)} = f'(x)e^{f(x)} \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2this is what you have to accept this property is due to chain rule

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay cool, that seems intuitive enough

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so do you follow everything?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well I couldnt teach it .. but I could read it over again, and I think I'd be pretty good to this point.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2Could 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2this question is weird

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2so its ok if you cant teach it as long as you learned stuff along the way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2well yeah you have e^(x^2) in your equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0totally on high Quality help ..

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2the second part of the equation is that limit we have been working on

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you can rate me by hitting the orange button at the top

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2But yeah, to write out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0argh.. damnit.. I clicked the wrong icon.. ah crap

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2what you have fully now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah man, sorry, you really deserve this one x 100

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2\[e^{x^{2}}(xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{(x+h)^2}e^{x^2}}{h}) + 1)\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2oh you rated someone else?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah by accident.. hit the wrong icon

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2@Preetha can you allow a redo on rating?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2not a big deal

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2but you see how this all works out to give you the derivative

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2she probably gets over a billion notifications

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2well anyways I hope I was helpful

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so are you saying that this equation we arrived at , can be applied to all functions?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2no just in this case you can do this to solve this problem

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well it helped me learn some things.. but yeah.. using the rules is definitely the go

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2just as an alternative approach for \(\large \lim \limits_{h \rightarrow 0} \frac{x(e^{2hxh^2}1)}{h}\) \(\large x\lim \limits_{h \rightarrow 0} \dfrac{(e^{h(2xh)}1)}{h}\times \dfrac{2xh}{2xh}\) \(\large x \lim \limits_{h(2xh) \rightarrow 0} \dfrac{(e^{h(2xh)}1)}{h(2xh)} \lim \limits_{h \rightarrow 0}(2xh) \) \(\large x[1\times (2x0)])\) = 2x :)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2you made a mistake it should be 2x^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2he forgot about the x he pulled out of the limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0was this at the stage we had to still add back in \[e^{x^2}\] ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2I dont really follow his method

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2i Just know he made a mistake

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.2just follow what I wrote and you will solve this problem anyways i am going to sleep now goodnight

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well thank you very much.. and Im doing the same

hartnn
 one year ago
Best ResponseYou've already chosen the best response.2yes, i forgot the x that i pulled outside, thanks! it'll be 2x^2 and yes again, e^(x^2) was pulled out earlier.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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 \[f'[x]= e^{x^2}(12x^2 ) \] \[f'[x]= e^{x^2}2x^2 e^{x^2} \] Thank you guys.
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