anonymous
  • anonymous
Can anybody explain the steps to the algebra that arrives at the derivative function to f[x] = x/e^(x^2) f'[x] = ???
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Australopithecus
  • Australopithecus
are you using the definition of the derivative or are you allowed to use the rule set?
logan13
  • logan13
35
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

logan13
  • logan13
ya me 2
Australopithecus
  • Australopithecus
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.
anonymous
  • anonymous
okay I think that means the definition of the derivative for me
anonymous
  • anonymous
ah thats awesome.. man that's a lot of rules.. geez, how do you guys memorize this?
Australopithecus
  • Australopithecus
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
anonymous
  • anonymous
yeah, I can see how this would save a lot of fiddly algebra
Australopithecus
  • Australopithecus
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)
anonymous
  • anonymous
yes 1 sec .. I can post how far I got
Australopithecus
  • Australopithecus
Doing the definition of this will probably be really dirty actually lol
Australopithecus
  • Australopithecus
but I will help you none the less :D
anonymous
  • anonymous
\[\left( \frac{ x+h }{e^{(x+h)^{2}}} - \frac{x}{e^{x^{2}}} \right)* \frac{1}{h}\] \[\left( (x+h)e^{-(x+h)^2}-x e^{-x^{2}} \right) * \frac{1}{h}\] \[\frac{ (x+h)e^{-(x+h)^2}-x e^{-x^{2}} } {h} \] and a few other variations of that
anonymous
  • anonymous
thanks I really really appreciate it.. this is going on 14 hours of head scratching for me now
anonymous
  • anonymous
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.
hartnn
  • hartnn
you've solved limits questions before? have you come across \(\Large \lim \limits_{x\to 0 } \dfrac{e^x -1}{x}\) ?
anonymous
  • anonymous
theres a good question ... hmmm ... its been a while.. about a year, but I think I did that in my high school pre calc..
hartnn
  • hartnn
good, remember L'Hopital's rule? that can be easily solved by it...
anonymous
  • anonymous
I couldnt tell you off hand how to solve that..
anonymous
  • anonymous
I can go look it up and refresh though
hartnn
  • hartnn
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
hartnn
  • hartnn
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 ?
anonymous
  • anonymous
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.
hartnn
  • hartnn
x+h, not x+1 sorry :P
hartnn
  • hartnn
i'll type the simplification by the time you try \(\Large (x+h) e^{-x^2 +2xh-h^2 } - xe^{-x^2}\) \(\Large (x+h) e^{-x^2} e^{2xh-h^2 } - xe^{-x^2}\) \(\Large e^{-x^2} [(x+h) e^{2xh-h^2 } - x]\)
anonymous
  • anonymous
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
hartnn
  • hartnn
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
hartnn
  • hartnn
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 ^^
hartnn
  • hartnn
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 :)
hartnn
  • hartnn
ask if you have any doubts in any explanation above
anonymous
  • anonymous
I will. thanks, just digesting it..
anonymous
  • anonymous
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?
Australopithecus
  • Australopithecus
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.
Australopithecus
  • Australopithecus
Give me a bit I need to look up a proof for the exponential rule using definition of a derivative
Australopithecus
  • Australopithecus
or you could just be content with d/dx e^(f(x)) = f'(x)e^(f(x))
Australopithecus
  • Australopithecus
Do you follow my steps though? they are kind of arbitrary but yeah
anonymous
  • anonymous
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 ?
Australopithecus
  • Australopithecus
You need to use the limit notation while solving your problem
Australopithecus
  • Australopithecus
so right now, \[\lim_{h \rightarrow 0} e^{-x^2}(\frac{xe^{-2hx-h^2}}{h} + e^{-2hx-h^2} + \frac{x}{h})\]
Australopithecus
  • Australopithecus
now apply the rules I showed you above
anonymous
  • anonymous
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.
Australopithecus
  • Australopithecus
Lets just try to get it in the right form first?
anonymous
  • anonymous
feels like it's 1 minute from an ah ha! moment though
Australopithecus
  • Australopithecus
then we can try to tackle the limit
Australopithecus
  • Australopithecus
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
anonymous
  • anonymous
if I understand it right, this function will become several limit functions?
Australopithecus
  • Australopithecus
No you will split this into two functions
anonymous
  • anonymous
and conceptually, x and e will become roughly the same very small value
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
Can you apply the addition rule to your problem?
anonymous
  • anonymous
...looking at that
Australopithecus
  • Australopithecus
If you want an example I can provide one?
Australopithecus
  • Australopithecus
For example \[\lim_{h \rightarrow 0} (x + \frac{x}{h})\] Applying addition rule, f(x) = x and g(x) = x/h so, \[\lim_{h \rightarrow 0} (x + \frac{x}{h}) = \lim_{h \rightarrow 0} x + \lim_{h \rightarrow 0} \frac{x}{h}\]
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
or rather the limit of x/h does not exist
anonymous
  • anonymous
ah ok is this where I should be next? \[\lim_{h \rightarrow 0} e^{-x^2} (\lim_{h \rightarrow 0} + \frac{xe^{-2hx-h^2}}{h} + \lim_{h \rightarrow 0} \frac{x}{h}+ \lim_{h \rightarrow 0} e^{-2hx-h^2} )\]
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
look at the first part of my example
Australopithecus
  • Australopithecus
you want to group the terms with h in the denominator
Australopithecus
  • Australopithecus
using the rule, \[\frac{a}{b} + \frac{c}{b} = \frac{a+c}{b}\]
Australopithecus
  • Australopithecus
then you will have two terms
Australopithecus
  • Australopithecus
Then you can declare them as separate functions and apply the addition rule
anonymous
  • anonymous
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? \[\lim_{h \rightarrow 0} \frac{xe^{-2hx-h^2 }+x}{h} + \lim_{h \rightarrow 0} e^{-2hx-h^2} \]
anonymous
  • anonymous
or at least I am guessing for the time being, can be put on the side. and used later.
Australopithecus
  • Australopithecus
You should have e^(-x^2) multiplied by everything but yeah that is right
anonymous
  • anonymous
ok, yes, that what I was thinking..
Australopithecus
  • Australopithecus
so, \[e^{-x^2}(\lim_{h \rightarrow 0} \frac{xe^{-2hx - h^2} +x}{h} + \lim_{h \rightarrow 0} e^{-2hx - h^2})\]
Australopithecus
  • Australopithecus
You can solve one of your terms leaving you with only one limit to solve
Australopithecus
  • Australopithecus
factor out x of the last remaining limit
anonymous
  • anonymous
\[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{-2hx - h^2}) \]
Australopithecus
  • Australopithecus
yeah now you can solve the limit of e^(-2hx - h^2)
anonymous
  • anonymous
should I be thinking something along the lines of dropping that second part of that exponent.. (my signs might be off) but something like \[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{-2hx}/e^{- h^2}) \]
Australopithecus
  • Australopithecus
for the second limit all you have to do is sub 0 into h
Australopithecus
  • Australopithecus
the last limit you need to manipulate it to look like the definition of a derivative
anonymous
  • anonymous
-2x h - h*h = -2x *0 - 0*0 = 0 ? or do think of this as 2x and 0's cancel out in some way ?
Australopithecus
  • Australopithecus
no that is right you will have e^0 for the second limit
Australopithecus
  • Australopithecus
the reason we cant solve the first limit is because h is in the denominator and dividing by 0 is undefined
anonymous
  • anonymous
oh cool, seemed too easy
Australopithecus
  • Australopithecus
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
anonymous
  • anonymous
\[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{-2x* 0- 0 * 0}) \] \[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{0}) \]
Australopithecus
  • Australopithecus
yeah so, you have the lim h->0 e^0 = 1
Australopithecus
  • Australopithecus
do you know how to complete the square?
anonymous
  • anonymous
ah right I just spotted that ... \[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} 1) \]
Australopithecus
  • Australopithecus
you should remove the limit notation from the second one
Australopithecus
  • Australopithecus
since it is redundant
anonymous
  • anonymous
yeah, bit rusty on it.. there's a couple of ways I recall. not sure if these terms divide nicely..
Australopithecus
  • Australopithecus
the last limit is a bit tricky
Australopithecus
  • Australopithecus
do you have any idea how to make it look like the definition of a deriative
anonymous
  • anonymous
I have an idea of how it should look when done..
Australopithecus
  • Australopithecus
So the last limit you want it in the form of \[\frac{f(x+h) - f(x)}{h}\]
Australopithecus
  • Australopithecus
well when you are done you wll have a function in terms of x that is the derivative you are looking for
anonymous
  • anonymous
\[f'[x] = ‚Äč(1 - 2*x^2)* e^{-x^2]} \]
Australopithecus
  • Australopithecus
huh?
anonymous
  • anonymous
when I cheat and ask mathematica to give me the derivative of f[x] that's what it gives me.
Australopithecus
  • Australopithecus
ok so you have this right now \[e^{-x^2}(\lim_{h \rightarrow 0} \frac{e^{-2hx-h^2}-1}{h} + 1)\]
Australopithecus
  • Australopithecus
yeah that is the derivative notice when you expand you get e^(-x^2) which is showing up in the problem atm
anonymous
  • anonymous
yes
Australopithecus
  • Australopithecus
oops
Australopithecus
  • Australopithecus
\[\lim_{h \rightarrow 0} \frac{x(e^{-2hx-h^2}-1)}{h}\] notice that x is a constant
Australopithecus
  • Australopithecus
this is the limit you have to solve
Australopithecus
  • Australopithecus
I forgot to write HINT: notice x is a constant
Australopithecus
  • Australopithecus
you need to rewrite the limit so it looks like the definition of a derivative
Australopithecus
  • Australopithecus
you can achieve this by using completing the square and factoring
Australopithecus
  • Australopithecus
out a term
Australopithecus
  • Australopithecus
its actually kind of tricky
Australopithecus
  • Australopithecus
then once you have it in that form you can tell what functions derivative is the solution to the limit
anonymous
  • anonymous
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
Australopithecus
  • Australopithecus
You need to complete the square in the exponential
Australopithecus
  • Australopithecus
so you need to complete the square with the following: -2hx-h^2
Australopithecus
  • Australopithecus
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: \[e^{a}e^{-a} = e^{0} = 1\]
Australopithecus
  • Australopithecus
so an example if I had 4ab + 4b^2 I could complete the square by writing (a + 2b)^2 - a^2
Australopithecus
  • Australopithecus
This is actually fairly tricky algebra
Australopithecus
  • Australopithecus
but not really once you see it you are like duh but yeah took me awhile to figure it out
anonymous
  • anonymous
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
anonymous
  • anonymous
and the x
Australopithecus
  • Australopithecus
just make a ( )^2 + what ever needs to get canceled out from the expanded ( )^2
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
I have totally forgot that formula lol
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
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
anonymous
  • anonymous
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.
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
Did you figure it out?
anonymous
  • anonymous
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
Australopithecus
  • Australopithecus
yup
Australopithecus
  • Australopithecus
good job
anonymous
  • anonymous
lol omg.. that was an effort
Australopithecus
  • Australopithecus
ok write it out
Australopithecus
  • Australopithecus
you are close to it looking like the definition of a specific derivative
Australopithecus
  • Australopithecus
you only really have a single manipulation left look at the rules I gave you above for a hint
Australopithecus
  • Australopithecus
you need to factor it out of the limit
Australopithecus
  • Australopithecus
hint it is the term you added just by completing the square
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
look at the two rules I posted regard exponentials they are hints to what you have to do
anonymous
  • anonymous
\[\lim_{h \rightarrow 0} \frac{x(e^{-2hx-h^2}-1)}{h}\] \[\lim_{h \rightarrow 0} \frac{x(e^{-(h+x)^2} e^{x^2}-1)}{h} \] Im thinking something like this
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
you want to pull e^(x^2) out of the limit
Australopithecus
  • Australopithecus
or exponential rules I gave you I mean
anonymous
  • anonymous
sorry for delay.. juggling the right exponent law (e^a)^b = e^(a b) . e^(a + b) = e^a e^b
anonymous
  • anonymous
trying to see how they fit
Australopithecus
  • Australopithecus
they will help you factor out e^(x^2)
Australopithecus
  • Australopithecus
remember you are trying to make it look like the definition of a derivative
anonymous
  • anonymous
should I try expanding that (h+x)^2 again ? h^2 +2hx +x^2
Australopithecus
  • Australopithecus
no
anonymous
  • anonymous
I can see how that adds another e^x^2
Australopithecus
  • Australopithecus
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)
Australopithecus
  • Australopithecus
Do you follow?
anonymous
  • anonymous
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
Australopithecus
  • Australopithecus
You will get it, just look at this rule: \[e^{a}e^{-a} = e^{a-a} = e^{0} = 1\]
anonymous
  • anonymous
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 \[ \lim_{h \rightarrow 0} \frac{x(e^{-(h+x)(h+x)} e^{x^2}-1)}{h} \] \[ \lim_{h \rightarrow 0} \frac{x(e^{-(h^2+2xh)-x^2} e^{x^2}-1)}{h} \]
Australopithecus
  • Australopithecus
To factor out e^(x^2), you need to change 1 into something
Australopithecus
  • Australopithecus
I gave you a rule that might help
anonymous
  • anonymous
are we zeroing out h ? here, I can see that creating e^-(0+x)(0+x) = e^-(x^2)
Australopithecus
  • Australopithecus
just factor e^(x^2) out of the limit
Australopithecus
  • Australopithecus
and you will see what I mean
Australopithecus
  • Australopithecus
look at the rule I posted you need to change 1 to something so you can factor out e^(x^2)
anonymous
  • anonymous
this rule e^{a}e^{-a} = e^{a-a} = e^{0} = 1
Australopithecus
  • Australopithecus
right
anonymous
  • anonymous
and I change the 1 in that rule, or the -1 at the end of the equation ?
Australopithecus
  • Australopithecus
we are focusing on the limit
Australopithecus
  • Australopithecus
the -1 is not in the limit anymore
anonymous
  • anonymous
I am staring at this equation .. maybe Im looking in the wrong place \[\lim_{h \rightarrow 0} \frac{x(e^{-(h+x)^2} e^{x^2}-1)}{h}\]
Australopithecus
  • Australopithecus
oh sorry yeah the -1
anonymous
  • anonymous
so another type of 1... e^(x^2) / e^(x^2) ?
Australopithecus
  • Australopithecus
yeah
Australopithecus
  • Australopithecus
now you can factor e^(x^2) and x out of the limit and then look at what you have
Australopithecus
  • Australopithecus
you still here?
anonymous
  • anonymous
yes .. staring at this ... \[ \lim_{h \rightarrow 0} \frac{x(e^{-(h+x)^2} e^{x^2}- e^{x^2} / e^{x^2})}{h} \]
Australopithecus
  • Australopithecus
well you need to use the constant rule
Australopithecus
  • Australopithecus
remove x and e^(x^2)
anonymous
  • anonymous
wondering how Im going to get that e^x^2 out of there..
Australopithecus
  • Australopithecus
you can factor e^(x^2) and x out of the limit look at the limit rules I posted
anonymous
  • anonymous
lookign them up ...
Australopithecus
  • Australopithecus
I will just write it
anonymous
  • anonymous
http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html
Australopithecus
  • Australopithecus
\[\lim_{x \rightarrow a} cf(x) = c*\lim_{x \rightarrow a} f(x)\]
Australopithecus
  • Australopithecus
that is the constant rule
Australopithecus
  • Australopithecus
in this case x and e^x^2 are constants
anonymous
  • anonymous
trying to understand how that rule is applied to the equation ...
anonymous
  • anonymous
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?
Australopithecus
  • Australopithecus
they dont cancel out they just leave the limit
anonymous
  • anonymous
it looks like they multiply the value of the original function ...
anonymous
  • anonymous
oh, so they are allowed to be removed from future calcs.. because they are not actually affected by any change in h
Australopithecus
  • Australopithecus
They get pulled outside the limit
Australopithecus
  • Australopithecus
yes
Australopithecus
  • Australopithecus
so your limit will be simplified
anonymous
  • anonymous
goodness me.. Im a slow as brick.
Australopithecus
  • Australopithecus
its ok you aren't dumb just learning
Australopithecus
  • Australopithecus
so write out what you have now
anonymous
  • anonymous
\[ \lim_{h \rightarrow 0} \frac{e^{-(h+x)^2} }{h} \]
Australopithecus
  • Australopithecus
nope
anonymous
  • anonymous
ok cool
Australopithecus
  • Australopithecus
close though
Australopithecus
  • Australopithecus
what did you do with e^(x^2), e^(-x^2) and x
anonymous
  • anonymous
I threw them away.. so they need to stay in the equation somehow, but outside the limit.
Australopithecus
  • Australopithecus
yes except one of the terms you need to leave in the limit
Australopithecus
  • Australopithecus
e^(-x^2) needs to stay in the equation
Australopithecus
  • Australopithecus
but x and e^(x^2) can be factored out
anonymous
  • anonymous
maybe Im not seeing \[\lim_{h \rightarrow 0} \] In the right way.. should I be seeing this as some kind of term ?
Australopithecus
  • Australopithecus
?
Australopithecus
  • Australopithecus
go back to what you had originally
Australopithecus
  • Australopithecus
had
anonymous
  • anonymous
\[ \lim_{h \rightarrow 0} \frac{x(e^{-(h+x)^2} e^{x^2}- e^{x^2} / e^{x^2})}{h} \]
Australopithecus
  • Australopithecus
yes
anonymous
  • anonymous
\[\lim_{h \rightarrow 0} \frac{x(e^{x^2}( e^{-(h+x)^2} - e^{x^2} ))}{h} \]
Australopithecus
  • Australopithecus
closer now just pull x and e^(x^2) out front
Australopithecus
  • Australopithecus
of the limit
Australopithecus
  • Australopithecus
now look at your limit look familure?
Australopithecus
  • Australopithecus
hint look at the definition of a derivative
anonymous
  • anonymous
\[ x(e^{x^2}( \lim_{h \rightarrow 0} \frac{ e^{-(h+x)^2} - e^{x^2} }{h} ))\]
anonymous
  • anonymous
lol.. thats got to be wrong
anonymous
  • anonymous
the limit definitely looks the shape we were after
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
what does that look like?
Australopithecus
  • Australopithecus
hint it is just the definition of a derivative
Australopithecus
  • Australopithecus
but what is f(x)?
Australopithecus
  • Australopithecus
if you know f(x) you know that f'(x) is the solution to that limit
anonymous
  • anonymous
like some kind of inversion
Australopithecus
  • Australopithecus
what function would you have to plug into the definition of a derivative to get that equation
Australopithecus
  • Australopithecus
or that limit rather
Australopithecus
  • Australopithecus
Make a guess
anonymous
  • anonymous
the e^x^2 ?
Australopithecus
  • Australopithecus
wait you made a mistake
Australopithecus
  • Australopithecus
does e^(x)e^(x) = 1?
Australopithecus
  • Australopithecus
by the rule I posted
anonymous
  • anonymous
needs a negative
Australopithecus
  • Australopithecus
yeah
Australopithecus
  • Australopithecus
so what is the function then?
anonymous
  • anonymous
\[ xe^{-x^2}(\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2} - e^{x^2}}{h}) \] i think this is what you are asking for ? \[xe^{-x^2}\]
Australopithecus
  • Australopithecus
that doesnt make sense
Australopithecus
  • Australopithecus
you want to factor out e^(x^2)
Australopithecus
  • Australopithecus
you cant factor out e^(-x^2)
anonymous
  • anonymous
\[xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2} - e^{x^2}}{h}) \]
Australopithecus
  • Australopithecus
e^(x^2)e^(x^2) =1?
Australopithecus
  • Australopithecus
\[xe^x (\lim_{h \rightarrow 0} \frac{e^{-(x + h)^2- e^{-x^2}}}{h})\]
Australopithecus
  • Australopithecus
oops lol but yeah
Australopithecus
  • Australopithecus
that term shouldnt be in the exponential but yeah
anonymous
  • anonymous
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?
Australopithecus
  • Australopithecus
yes
Australopithecus
  • Australopithecus
That is it :D
Australopithecus
  • Australopithecus
now what if f(x) considering that is the definition of a derivative
Australopithecus
  • Australopithecus
is* not if
Australopithecus
  • Australopithecus
you almost had it before but you there was a mistake
anonymous
  • anonymous
Im assuming it is supposed to be what we started with? or is it going to be something different?
Australopithecus
  • Australopithecus
it is something different
Australopithecus
  • Australopithecus
when we are done you have to plug this back into the original problem
Australopithecus
  • Australopithecus
then you will have the derivative
anonymous
  • anonymous
xe^{x^2} ? the left portion of that equation ?
Australopithecus
  • Australopithecus
We just isolate the limit we are dealing with because the other stuff would just be a waste to write over and over
Australopithecus
  • Australopithecus
do you recognize that the limit we have now is in the form: \[\frac{f(x+h) + f(x)}{h}\]
anonymous
  • anonymous
yes
Australopithecus
  • Australopithecus
then what is f(x)?
anonymous
  • anonymous
ah ok.. lol
anonymous
  • anonymous
\[ e^{-x^2} \]
Australopithecus
  • Australopithecus
yes
Australopithecus
  • Australopithecus
so the solution to that limit is just the derivative of e^(-x^2)
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
http://en.wikipedia.org/wiki/Chain_rule#First_proof
anonymous
  • anonymous
I am ready to accept such things
Australopithecus
  • Australopithecus
lol
Australopithecus
  • Australopithecus
ok
anonymous
  • anonymous
I dont know how to thank you enough.. this was awesome
Australopithecus
  • Australopithecus
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: \[xe^{x^2}( \lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{h}) = xe^{x^2}(f'(x)e^{f(x)})\]
Australopithecus
  • Australopithecus
sorry lets say f(x) = -x^(2)
Australopithecus
  • Australopithecus
that negative is important almost forgot it
Australopithecus
  • Australopithecus
then you plug that back into your main equation when you find f'(x)
Australopithecus
  • Australopithecus
Do some simplifying and you are done
Australopithecus
  • Australopithecus
sorry I made a mistake in writing above, \[\frac{d}{dx} e^{f(x)} = f'(x)e^{f(x)} \]
Australopithecus
  • Australopithecus
this is what you have to accept this property is due to chain rule
anonymous
  • anonymous
okay cool, that seems intuitive enough
Australopithecus
  • Australopithecus
so do you follow everything?
anonymous
  • anonymous
well I couldnt teach it .. but I could read it over again, and I think I'd be pretty good to this point.
Australopithecus
  • Australopithecus
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
Australopithecus
  • Australopithecus
this question is weird
Australopithecus
  • Australopithecus
so its ok if you cant teach it as long as you learned stuff along the way
anonymous
  • anonymous
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
Australopithecus
  • Australopithecus
well yeah you have e^(-x^2) in your equation
anonymous
  • anonymous
totally on high Quality help ..
Australopithecus
  • Australopithecus
the second part of the equation is that limit we have been working on
Australopithecus
  • Australopithecus
you can rate me by hitting the orange button at the top
Australopithecus
  • Australopithecus
But yeah, to write out
anonymous
  • anonymous
argh.. damnit.. I clicked the wrong icon.. ah crap
Australopithecus
  • Australopithecus
what you have fully now
anonymous
  • anonymous
doesnt have undo
anonymous
  • anonymous
ah man, sorry, you really deserve this one x 100
Australopithecus
  • Australopithecus
\[e^{-x^{2}}(xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{h}) + 1)\]
Australopithecus
  • Australopithecus
oh you rated someone else?
anonymous
  • anonymous
yeah by accident.. hit the wrong icon
Australopithecus
  • Australopithecus
@Preetha can you allow a redo on rating?
Australopithecus
  • Australopithecus
not a big deal
Australopithecus
  • Australopithecus
but you see how this all works out to give you the derivative
Australopithecus
  • Australopithecus
she probably gets over a billion notifications
Australopithecus
  • Australopithecus
well anyways I hope I was helpful
anonymous
  • anonymous
so are you saying that this equation we arrived at , can be applied to all functions?
Australopithecus
  • Australopithecus
no just in this case you can do this to solve this problem
Australopithecus
  • Australopithecus
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
anonymous
  • anonymous
well it helped me learn some things.. but yeah.. using the rules is definitely the go
hartnn
  • hartnn
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 :)
Australopithecus
  • Australopithecus
you made a mistake it should be -2x^2
Australopithecus
  • Australopithecus
he forgot about the x he pulled out of the limit
anonymous
  • anonymous
ah right on yeah..
anonymous
  • anonymous
was this at the stage we had to still add back in \[e^{-x^2}\] ?
Australopithecus
  • Australopithecus
I dont really follow his method
Australopithecus
  • Australopithecus
i Just know he made a mistake
Australopithecus
  • Australopithecus
just follow what I wrote and you will solve this problem anyways i am going to sleep now goodnight
anonymous
  • anonymous
well thank you very much.. and Im doing the same
hartnn
  • hartnn
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.
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.