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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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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are you using the definition of the derivative or are you allowed to use the rule set?
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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.

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