Question

Can anybody explain the steps to the algebra that arrives at the derivative function to

f[x] = x/e^(x^2)
f'[x] = ???

Answer 1

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

Answer 2

35

Answer 3

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.

Answer 4

ya me 2

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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)

Answer 11

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

Answer 12

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

Answer 13

but I will help you none the less :D

Answer 14

\[\left( \frac{ x+h }{e^{(x+h)^{2}}} - \frac{x}{e^{x^{2}}} \right)* \frac{1}{h}\]

\[\left( (x+h)e^{-(x+h)^2}-x e^{-x^{2}} \right) * \frac{1}{h}\]

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

and a few other variations of that

Answer 15

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

Answer 16

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.

Answer 17

you've solved limits questions before?
have you come across

\(\Large \lim \limits_{x\to 0 } \dfrac{e^x -1}{x}\)
?

Answer 18

theres a good question ... hmmm ... its been a while.. about a year, but I think I did that in my high school pre calc..

Answer 19

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

Answer 20

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

Answer 21

I can go look it up and refresh though

Answer 22

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

Answer 23

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 ?

Answer 24

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.

Answer 25

x+h,
not x+1 sorry :P

Answer 26

i'll type the simplification by the time you try
\(\Large (x+h) e^{-x^2 +2xh-h^2 } - xe^{-x^2}\)

\(\Large (x+h) e^{-x^2} e^{2xh-h^2 } - xe^{-x^2}\)

\(\Large e^{-x^2} [(x+h) e^{2xh-h^2 } - x]\)

Answer 27

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

Answer 28

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

Answer 29

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 ^^

Answer 30

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 :)

Answer 31

ask if you have any doubts in any explanation above

Answer 32

I will. thanks, just digesting it..

Answer 33

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?

Answer 34

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.

Answer 35

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

Answer 36

or you could just be content with

d/dx e^(f(x)) = f'(x)e^(f(x))

Answer 37

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

Answer 38

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 ?

Answer 39

You need to use the limit notation while solving your problem

Answer 40

so right now,

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

Answer 41

now apply the rules I showed you above

Answer 42

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.

Answer 43

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

Answer 44

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

Answer 45

then we can try to tackle the limit

Answer 46

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

Answer 47

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

Answer 48

No you will split this into two functions

Answer 49

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

Answer 50

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

Answer 51

Can you apply the addition rule to your problem?

Answer 52

...looking at that

Answer 53

If you want an example I can provide one?

Answer 54

For example

\[\lim_{h \rightarrow 0} (x + \frac{x}{h})\]

Applying addition rule, f(x) = x and g(x) = x/h
so,

\[\lim_{h \rightarrow 0} (x + \frac{x}{h}) = \lim_{h \rightarrow 0} x + \lim_{h \rightarrow 0} \frac{x}{h}\]

Answer 55

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

Answer 56

or rather the limit of x/h does not exist

Answer 57

ah ok
is this where I should be next?
\[\lim_{h \rightarrow 0} e^{-x^2} (\lim_{h \rightarrow 0} + \frac{xe^{-2hx-h^2}}{h} + \lim_{h \rightarrow 0} \frac{x}{h}+ \lim_{h \rightarrow 0} e^{-2hx-h^2} )\]

Answer 58

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

Answer 59

look at the first part of my example

Answer 60

you want to group the terms with h in the denominator

Answer 61

using the rule,

\[\frac{a}{b} + \frac{c}{b} = \frac{a+c}{b}\]

Answer 62

then you will have two terms

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

ok, yes, that what I was thinking..

Answer 68

so,

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

Answer 69

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

Answer 70

factor out x of the last remaining limit

Answer 71

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

Answer 72

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

Answer 73

should I be thinking something along the lines of dropping that second part of that exponent.. (my signs might be off) but something like

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

Answer 74

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

Answer 75

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

Answer 76

-2x h - h*h = -2x *0 - 0*0 = 0 ? or do think of this as 2x and 0's cancel out in some way ?

Answer 77

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

Answer 78

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

Answer 79

oh cool, seemed too easy

Answer 80

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

Answer 81

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

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

Answer 82

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

Answer 83

do you know how to complete the square?

Answer 84

ah right I just spotted that ...
\[ e^{-x^2}(\lim_{h \rightarrow 0} \frac{x(e^{-2hx - h^2} +1)}{h} + \lim_{h \rightarrow 0} 1) \]

Answer 85

you should remove the limit notation from the second one

Answer 86

since it is redundant

Answer 87

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

Answer 88

the last limit is a bit tricky

Answer 89

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

Answer 90

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

Answer 91

So the last limit you want it in the form of

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

Answer 92

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

Answer 93

\[f'[x] = ​(1 - 2*x^2)* e^{-x^2]} \]

Answer 94

huh?

Answer 95

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

Answer 96

ok so you have this right now

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

Answer 97

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