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

- anonymous

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- Australopithecus

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

- logan13

35

- 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.

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## More answers

- logan13

ya me 2

- 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.

##### 1 Attachment

- anonymous

okay I think that means the definition of the derivative for me

- anonymous

ah thats awesome.. man that's a lot of rules.. geez, how do you guys memorize this?

- 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

yeah, I can see how this would save a lot of fiddly algebra

- 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

yes 1 sec .. I can post how far I got

- Australopithecus

Doing the definition of this will probably be really dirty actually lol

- Australopithecus

but I will help you none the less :D

- 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

thanks I really really appreciate it.. this is going on 14 hours of head scratching for me now

- 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

you've solved limits questions before?
have you come across
\(\Large \lim \limits_{x\to 0 } \dfrac{e^x -1}{x}\)
?

- 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

good, remember L'Hopital's rule?
that can be easily solved by it...

- anonymous

I couldnt tell you off hand how to solve that..

- anonymous

I can go look it up and refresh though

- 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

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

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

x+h,
not x+1 sorry :P

- 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

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

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

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

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

ask if you have any doubts in any explanation above

- anonymous

I will. thanks, just digesting it..

- 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

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

Give me a bit I need to look up a proof for the exponential rule using definition of a derivative

- Australopithecus

or you could just be content with
d/dx e^(f(x)) = f'(x)e^(f(x))

- Australopithecus

Do you follow my steps though? they are kind of arbitrary but yeah

- 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

You need to use the limit notation while solving your problem

- 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

now apply the rules I showed you above

- 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

Lets just try to get it in the right form first?

- anonymous

feels like it's 1 minute from an ah ha! moment though

- Australopithecus

then we can try to tackle the limit

- 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

if I understand it right, this function will become several limit functions?

- Australopithecus

No you will split this into two functions

- anonymous

and conceptually, x and e will become roughly the same very small value

- 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

Can you apply the addition rule to your problem?

- anonymous

...looking at that

- Australopithecus

If you want an example I can provide one?

- 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

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

or rather the limit of x/h does not exist

- 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

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

look at the first part of my example

- Australopithecus

you want to group the terms with h in the denominator

- Australopithecus

using the rule,
\[\frac{a}{b} + \frac{c}{b} = \frac{a+c}{b}\]

- Australopithecus

then you will have two terms

- Australopithecus

Then you can declare them as separate functions and apply the addition rule

- 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

or at least I am guessing for the time being, can be put on the side. and used later.

- Australopithecus

You should have e^(-x^2) multiplied by everything but yeah that is right

- anonymous

ok, yes, that what I was thinking..

- 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

You can solve one of your terms leaving you with only one limit to solve

- Australopithecus

factor out x of the last remaining limit

- 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

yeah now you can solve the limit of e^(-2hx - h^2)

- 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

for the second limit all you have to do is sub 0 into h

- Australopithecus

the last limit you need to manipulate it to look like the definition of a derivative

- 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

no that is right you will have e^0 for the second limit

- Australopithecus

the reason we cant solve the first limit is because h is in the denominator and dividing by 0 is undefined

- anonymous

oh cool, seemed too easy

- 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

\[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{-2x* 0- 0 * 0}) \]
\[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} e^{0}) \]

- Australopithecus

yeah so, you have the lim h->0 e^0 = 1

- Australopithecus

do you know how to complete the square?

- 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

you should remove the limit notation from the second one

- Australopithecus

since it is redundant

- anonymous

yeah, bit rusty on it.. there's a couple of ways I recall. not sure if these terms divide nicely..

- Australopithecus

the last limit is a bit tricky

- Australopithecus

do you have any idea how to make it look like the definition of a deriative

- anonymous

I have an idea of how it should look when done..

- Australopithecus

So the last limit you want it in the form of
\[\frac{f(x+h) - f(x)}{h}\]

- Australopithecus

well when you are done you wll have a function in terms of x that is the derivative you are looking for

- anonymous

\[f'[x] = â€‹(1 - 2*x^2)* e^{-x^2]} \]

- Australopithecus

huh?

- anonymous

when I cheat and ask mathematica to give me the derivative of f[x] that's what it gives me.

- Australopithecus

ok so you have this right now
\[e^{-x^2}(\lim_{h \rightarrow 0} \frac{e^{-2hx-h^2}-1}{h} + 1)\]

- Australopithecus

yeah that is the derivative notice when you expand you get e^(-x^2) which is showing up in the problem atm

- anonymous

yes

- Australopithecus

oops

- Australopithecus

\[\lim_{h \rightarrow 0} \frac{x(e^{-2hx-h^2}-1)}{h}\]
notice that x is a constant

- Australopithecus

this is the limit you have to solve

- Australopithecus

I forgot to write HINT: notice x is a constant

- Australopithecus

you need to rewrite the limit so it looks like the definition of a derivative

- Australopithecus

you can achieve this by using completing the square and factoring

- Australopithecus

out a term

- Australopithecus

its actually kind of tricky

- Australopithecus

then once you have it in that form you can tell what functions derivative is the solution to the limit

- 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

You need to complete the square in the exponential

- Australopithecus

so you need to complete the square with the following:
-2hx-h^2

- 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

so an example if I had
4ab + 4b^2
I could complete the square by writing
(a + 2b)^2 - a^2

- Australopithecus

This is actually fairly tricky algebra

- Australopithecus

but not really once you see it you are like duh but yeah took me awhile to figure it out

- 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

and the x

- Australopithecus

just make a ( )^2 + what ever needs to get canceled out from the expanded ( )^2

- 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

I have totally forgot that formula lol

- 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

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

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

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

Did you figure it out?

- 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

yup

- Australopithecus

good job

- anonymous

lol omg.. that was an effort

- Australopithecus

ok write it out

- Australopithecus

you are close to it looking like the definition of a specific derivative

- Australopithecus

you only really have a single manipulation left look at the rules I gave you above for a hint

- Australopithecus

you need to factor it out of the limit

- Australopithecus

hint it is the term you added just by completing the square

- 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

look at the two rules I posted regard exponentials
they are hints to what you have to do

- 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

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

you want to pull e^(x^2) out of the limit

- Australopithecus

or exponential rules I gave you I mean

- anonymous

sorry for delay.. juggling the right exponent law
(e^a)^b = e^(a b) .
e^(a + b) = e^a e^b

- anonymous

trying to see how they fit

- Australopithecus

they will help you factor out e^(x^2)

- Australopithecus

remember you are trying to make it look like the definition of a derivative

- anonymous

should I try expanding that (h+x)^2 again ?
h^2 +2hx +x^2

- Australopithecus

no

- anonymous

I can see how that adds another e^x^2

- 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

Do you follow?

- 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

You will get it, just look at this rule:
\[e^{a}e^{-a} = e^{a-a} = e^{0} = 1\]

- 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

To factor out e^(x^2), you need to change 1 into something

- Australopithecus

I gave you a rule that might help

- anonymous

are we zeroing out h ? here, I can see that creating
e^-(0+x)(0+x) = e^-(x^2)

- Australopithecus

just factor e^(x^2) out of the limit

- Australopithecus

and you will see what I mean

- Australopithecus

look at the rule I posted you need to change 1 to something so you can factor out e^(x^2)

- anonymous

this rule
e^{a}e^{-a} = e^{a-a} = e^{0} = 1

- Australopithecus

right

- anonymous

and I change the 1 in that rule, or the -1 at the end of the equation ?

- Australopithecus

we are focusing on the limit

- Australopithecus

the -1 is not in the limit anymore

- 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

oh sorry yeah the -1

- anonymous

so another type of 1...
e^(x^2) / e^(x^2) ?

- Australopithecus

yeah

- Australopithecus

now you can factor e^(x^2) and x out of the limit and then look at what you have

- Australopithecus

you still here?

- 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

well you need to use the constant rule

- Australopithecus

remove x and e^(x^2)

- anonymous

wondering how Im going to get that e^x^2 out of there..

- Australopithecus

you can factor e^(x^2) and x out of the limit look at the limit rules I posted

- anonymous

lookign them up ...

- Australopithecus

I will just write it

- anonymous

http://math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html

- Australopithecus

\[\lim_{x \rightarrow a} cf(x) = c*\lim_{x \rightarrow a} f(x)\]

- Australopithecus

that is the constant rule

- Australopithecus

in this case x and e^x^2 are constants

- anonymous

trying to understand how that rule is applied to the equation ...

- 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

they dont cancel out they just leave the limit

- anonymous

it looks like they multiply the value of the original function ...

- anonymous

oh, so they are allowed to be removed from future calcs.. because they are not actually affected by any change in h

- Australopithecus

They get pulled outside the limit

- Australopithecus

yes

- Australopithecus

so your limit will be simplified

- anonymous

goodness me.. Im a slow as brick.

- Australopithecus

its ok you aren't dumb just learning

- Australopithecus

so write out what you have now

- anonymous

\[ \lim_{h \rightarrow 0} \frac{e^{-(h+x)^2} }{h} \]

- Australopithecus

nope

- anonymous

ok cool

- Australopithecus

close though

- Australopithecus

what did you do with e^(x^2), e^(-x^2) and x

- anonymous

I threw them away.. so they need to stay in the equation somehow, but outside the limit.

- Australopithecus

yes except one of the terms you need to leave in the limit

- Australopithecus

e^(-x^2) needs to stay in the equation

- Australopithecus

but x and e^(x^2) can be factored out

- 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

go back to what you had originally

- Australopithecus

had

- anonymous

\[ \lim_{h \rightarrow 0} \frac{x(e^{-(h+x)^2} e^{x^2}- e^{x^2} / e^{x^2})}{h} \]

- Australopithecus

yes

- anonymous

\[\lim_{h \rightarrow 0} \frac{x(e^{x^2}( e^{-(h+x)^2} - e^{x^2} ))}{h} \]

- Australopithecus

closer now just pull x and e^(x^2) out front

- Australopithecus

of the limit

- Australopithecus

now look at your limit look familure?

- Australopithecus

hint look at the definition of a derivative

- anonymous

\[ x(e^{x^2}( \lim_{h \rightarrow 0} \frac{ e^{-(h+x)^2} - e^{x^2} }{h} ))\]

- anonymous

lol.. thats got to be wrong

- anonymous

the limit definitely looks the shape we were after

- 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

what does that look like?

- Australopithecus

hint it is just the definition of a derivative

- Australopithecus

but what is f(x)?

- Australopithecus

if you know f(x) you know that f'(x) is the solution to that limit

- anonymous

like some kind of inversion

- Australopithecus

what function would you have to plug into the definition of a derivative to get that equation

- Australopithecus

or that limit rather

- Australopithecus

Make a guess

- anonymous

the e^x^2 ?

- Australopithecus

wait you made a mistake

- Australopithecus

does e^(x)e^(x) = 1?

- Australopithecus

by the rule I posted

- anonymous

needs a negative

- Australopithecus

yeah

- Australopithecus

so what is the function then?

- 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

that doesnt make sense

- Australopithecus

you want to factor out e^(x^2)

- Australopithecus

you cant factor out e^(-x^2)

- anonymous

\[xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{-(h+x)^2} - e^{x^2}}{h}) \]

- Australopithecus

e^(x^2)e^(x^2) =1?

- Australopithecus

\[xe^x (\lim_{h \rightarrow 0} \frac{e^{-(x + h)^2- e^{-x^2}}}{h})\]

- Australopithecus

oops lol but yeah

- Australopithecus

that term shouldnt be in the exponential but yeah

- 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

yes

- Australopithecus

That is it :D

- Australopithecus

now what if f(x) considering that is the definition of a derivative

- Australopithecus

is* not if

- Australopithecus

you almost had it before but you there was a mistake

- anonymous

Im assuming it is supposed to be what we started with? or is it going to be something different?

- Australopithecus

it is something different

- Australopithecus

when we are done you have to plug this back into the original problem

- Australopithecus

then you will have the derivative

- anonymous

xe^{x^2} ? the left portion of that equation ?

- 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

do you recognize that the limit we have now is in the form:
\[\frac{f(x+h) + f(x)}{h}\]

- anonymous

yes

- Australopithecus

then what is f(x)?

- anonymous

ah ok.. lol

- anonymous

\[ e^{-x^2} \]

- Australopithecus

yes

- Australopithecus

so the solution to that limit is just the derivative of e^(-x^2)

- 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

http://en.wikipedia.org/wiki/Chain_rule#First_proof

- anonymous

I am ready to accept such things

- Australopithecus

lol

- Australopithecus

ok

- anonymous

I dont know how to thank you enough.. this was awesome

- 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

sorry lets say f(x) = -x^(2)

- Australopithecus

that negative is important almost forgot it

- Australopithecus

then you plug that back into your main equation when you find f'(x)

- Australopithecus

Do some simplifying and you are done

- Australopithecus

sorry I made a mistake in writing above,
\[\frac{d}{dx} e^{f(x)} = f'(x)e^{f(x)} \]

- Australopithecus

this is what you have to accept this property is due to chain rule

- anonymous

okay cool, that seems intuitive enough

- Australopithecus

so do you follow everything?

- 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

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

this question is weird

- Australopithecus

so its ok if you cant teach it as long as you learned stuff along the way

- 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

well yeah you have e^(-x^2) in your equation

- anonymous

totally on high Quality help ..

- Australopithecus

the second part of the equation is that limit we have been working on

- Australopithecus

you can rate me by hitting the orange button at the top

- Australopithecus

But yeah,
to write out

- anonymous

argh.. damnit.. I clicked the wrong icon.. ah crap

- Australopithecus

what you have fully now

- anonymous

doesnt have undo

- anonymous

ah man, sorry, you really deserve this one x 100

- Australopithecus

\[e^{-x^{2}}(xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{h}) + 1)\]

- Australopithecus

oh you rated someone else?

- anonymous

yeah by accident.. hit the wrong icon

- Australopithecus

@Preetha can you allow a redo on rating?

- Australopithecus

not a big deal

- Australopithecus

but you see how this all works out to give you the derivative

- Australopithecus

she probably gets over a billion notifications

- Australopithecus

well anyways I hope I was helpful

- anonymous

so are you saying that this equation we arrived at , can be applied to all functions?

- Australopithecus

no just in this case you can do this to solve this problem

- 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

well it helped me learn some things.. but yeah.. using the rules is definitely the go

- 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

you made a mistake
it should be -2x^2

- Australopithecus

he forgot about the x he pulled out of the limit

- anonymous

ah right on yeah..

- anonymous

was this at the stage we had to still add back in \[e^{-x^2}\] ?

- Australopithecus

I dont really follow his method

- Australopithecus

i Just know he made a mistake

- Australopithecus

just follow what I wrote and you will solve this problem anyways i am going to sleep now goodnight

- anonymous

well thank you very much.. and Im doing the same

- 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

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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