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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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.

Looking for something else?

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

## 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.

Looking for something else?

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