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

35

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?

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

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

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

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

x+h,
not x+1 sorry :P

ask if you have any doubts in any explanation above

I will. thanks, just digesting 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

You need to use the limit notation while solving your problem

now apply the rules I showed you above

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

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

Can you apply the addition rule to your problem?

...looking at that

If you want an example I can provide one?

or rather the limit of x/h does not exist

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

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..

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

factor out x of the last remaining limit

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

oh cool, seemed too easy

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

do you know how to complete the square?

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
\[\frac{f(x+h) - f(x)}{h}\]

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

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

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

You need to complete the square in the exponential

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

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

and the x

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

I have totally forgot that formula lol

Did you figure it out?

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

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

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

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

Do you follow?

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

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

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?

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

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

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 ...

they dont cancel out they just leave the limit

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

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

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

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?

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

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 ?

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)

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

I am ready to accept such things

lol

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

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?

this question is weird

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

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

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

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