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anonymous
 one year ago
Learning about derivatives, instantaneous rate of change, ..via differential quotient equation.. and limits equations, I almost have it down now, need a review of the final few posts to check my work.. thanx
Given:
f[x] = x/e^x^2
f'[x] = Lim (h>0) [(f[x+h] f[x])/ h]
f'[x] = Lim (h>0) [ ((x+h)/(E^(x+h)^2)  (x/e^(x^2)) ) *1/h ]
anonymous
 one year ago
Learning about derivatives, instantaneous rate of change, ..via differential quotient equation.. and limits equations, I almost have it down now, need a review of the final few posts to check my work.. thanx Given: f[x] = x/e^x^2 f'[x] = Lim (h>0) [(f[x+h] f[x])/ h] f'[x] = Lim (h>0) [ ((x+h)/(E^(x+h)^2)  (x/e^(x^2)) ) *1/h ]

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or at the very least please send me a good link that explains how to deal with canceling variables embedded in a polynomial exponent

jigglypuff314
 one year ago
Best ResponseYou've already chosen the best response.0Do you have to use the definition of the derivative to solve for your given function? derivative of e^u = d/du * e^u for example, derivative of e^(x^3) = (3x^2)*e^(x^3) Here's the proof for the concept if you would like: https://www.khanacademy.org/math/differentialcalculus/takingderivatives/der_common_functions/v/proofddxexex

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f[x] = \frac{x}{e^{x^2}}\] \[ f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] f[x] }{h} \right) \] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ E^{(x+h)^2} }  \frac{ x}{ E^{(x^2)} } \right) * \frac{1}{h} \right) \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unfortunately I do..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i have really come to admire those rules of the derivative..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah the baby steps..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've been on it for 3 days.. and my OCD wont allow me to let it go

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I cant friggin sleep, so pissed off.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0staring at equations for days. Im missing some fundamental cancelation trick of limits.. but all the vids Im watching, I still ahvent found one that deals with a complex rational like this, and a polynomial exponent.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can get there like this.. I know the answer to this is going to be \[ f'[x] = (12x^2)e^{x^2}\] and I can get there by.. g[x] = x^2 g'[x] = 2x h[x] = E^g[x] h'[x] = 2xe^x^2 f[x] = x h[x] f'[x] = (h'[x]/g'[x]) + (f[x] * g'[x]) but I cant submit this.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and I cant reverse the logic, that would allow me to do that with algebra.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4Its alright here, we can try to work through your final limit using this informal proof. \[\lim_{h \rightarrow 0} \frac{e^{(x+h)^2}e^{x^2}}{h}\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4Although maybe amistre64 has a better way of solving this limit or problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that the first trick of limits, that we can remove x ? \[ f[x] = x/E^{x^2} \] \[ f[x] = x E^{x^2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops .. that 's x \[ f[x] = x E^{x^2} \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4Well this problem demonstrates product rule as well, we already solved part of the derivative, we differentiated x/e^(x^2) in regards to the x term but we still need to differentiate x/e^(x^2) in terms of e^(x^2) that is why we are left with that limit to solve

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4This is where you left off \[e^{x^2}(xe^{x^2}(\lim_{h \rightarrow 0} \frac{e^{(h+x)^2} e^{x^2}}{h}) + 1)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, I just found your image...

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4yeah you could potentially use that to solve your problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whats the strategy to get h out of that numerator ? should that even be our goal from here?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4Well just considering the limit: \[\lim_{h \rightarrow 0} \frac{e^{(h+x)^2} e^{x^2}}{h}\] Follow along with that image I posted you should be able to use it to solve this limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whats the idea behind the first part of your image where f(x) becomes f(g(x)), is that because the equation x / e^x^2 has been split into two functions? And if so where was the split? was it at the x^2 ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4There is something you should note though that will likely come up the proof of d/dx e^x = e^x by definition of e: \[\lim_{h \rightarrow 0} \frac{e^h1}{h} = 1\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4well e^(x^2) is a composite function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh is that proof because technically, we have e^0 = (almost 1)1 and we get a something/something

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4if f(x) = e^(x) and g(x) = x^2 then f(g(x)) = e^(x^2)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4that is how the value of e is defined

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4one of the ways it is defined

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there was another x in there though.. f[x] = x / E^(x^2) is the x removed, or placed into f or g ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4we pulled it out of the limit

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4because it is technically a constant

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah ok.. so that sits.. outside the lim definition still ? or does it now hold some value like 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or does it cancel to 0 ? I havent clicked yet on how that constant rule is actually applied.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just watching another vid on constant rule.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol, man, nobody explains this well.. they just gloss over and say ... 'oh we can pull this out.' and they move on

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, I think it goes outside ..

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4no it just sits outside the limit, if you have learned product rule this is a result of it. By product rule: d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x) in this problem assuming f(x) = e^(x^2) and g(x) = x we solved f(x)g'(x) the limit left represents f'(x)g(x) since g(x) = x we can pull it out of the limit because it is just the original function, the limit itself represents f'(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but it hangs around..

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4the e^(x^2) just cancels out, you could cancel it out right now it wouldn't change your solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so in reference to \[\lim_{h \rightarrow 0} \frac{e^h1}{h} = 1\] To be clear, this means that if we encounter e^x then we have encountered the value 1? And anything attached to that or based on that can be considered to be factored with the value 1? so E^x^2 is actually 1^2 ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4No it is essentially what defines what the value of e is, it comes up when you try to use the definition of a derivative to find the derivative of e^x http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm, that's an odd and curious property. :/

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4I can help you work through the informal proof if you want.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4@amistre64 , @Hero Might be able to help you more so with this.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think if the problem was just f[x] = e^(x^2) then I would have no problem.. x^2 = 2x and the derivative would be 2x * e^x^2 but I'm not sure what to do about that leading x/ .... maybe converting it into this form and then trying to work it out... f[x] = x e^(x^2) do we just put the x outside the limit from the start? f'[x] = x Lim (h>0) [ (f[x+h]  f[x] ) /h ] It seems like its not a problem of 2 functions, but of 3 f[x] = x h[x] g[x] = E^h[x] h[x] = x^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wish you could back edit these posts

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4Not sure what you are confused about here, look what I wrote about product rule. where g(x) = x and f(x) = e^(x^2) the derivative of f(x)g'(x) is just e^(x^2) This is shown in your problem, if you multiply e^(x) by both terms left in the bracket you will get your x*(limit you need to solve) + e^(x^2) the limit remaining is the derivative of f(x)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4You would have this exact limit if you were trying to take the derivative with the definition using e^(x^2)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4If it was e^(x) and not e^(x^2) you would have no problem because you wouldn't need chain rule

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4you are correct it is a problem of 3 functions

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4this problem requires product rule and chain rule

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4but the third function is easily differentiable so its not a problem

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4your problem is essentially m(x)*f(g(x))

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4this is actually a good thing for you to realize since you need to be able to recognize things like this when taking derivatives

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4you need to be able to split a function into multiple functions to take a derivative in most cases

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah ok.. so I've been thinking g[x] = x^2 for all cases of g[x] .. in this situation, and not seeing that g[x] is a term used to express generally a factor in an equation that varies from place to place.. where with Expression A (g[x]) * Expression B (f[x]), is in this case, x is in expression A and hence the derivative of g[x] * f[x] is in this case just E^x^2 and in the next stage of this equation, g[x] can and will be defined to be something else, as it is applied to the general derivative equations.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so Ive been scratching my head, trying to interpret the language

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4I am not understanding what you are saying. product rule is the following: f(x)*g(x) = f'(x)g(x) + f(x)g'(x) so if you had xln(x) the derivative would be: 1*ln(x) + (1/x)*x

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4g(x) = x g'(x) = 1 f(x) = ln(x) f'(x) = 1/x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, I think I'm going to go read up more on the product rule and the chain rule again... by the way, is this problem going to need Ln[] to finish it off at some point?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4I posted an informal proof you could follow along with

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4to solve the limit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was just saying before, that I wasn't understanding what and how g(x) was getting defined. And I just realized it was being defined on the fly, and changing in context of the problem as it evolved.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so as you might understand from that last statement.. the informal proof had this mysterious g(x) throughout it.. and it wasn't intuitively clicking how that was defined and applied.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4\[\lim_{h \rightarrow 0} \frac{e^{(x + h)^2}  e^{x^2}}{h} = \lim_{h \rightarrow 0} \frac{e^{(x + h)^2}  e^{x^2}}{h}*\frac{(x+h)^2  x^2}{(x+h)^2  x^2}\] do you understand that m(x) = e^(x^2) is a composite function g(x) = x^2, f(x) = e^x therefore, m(x) = f(g(x)) = e^(g(x)) = e^(x^2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you've multiplied by 1, using the exponents to construct the 1.. you've chosen x^2, so it looks like some kind of conjugate, constructed from the exponents.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, I understand on the m(x) .. thnx

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4\[=(\lim_{h \rightarrow 0} \frac{e^{(x+h)^2}e^{x^2}}{(x+h)^2  x^2})*(\lim_{h \rightarrow 0} \frac{(x+h)^2  x^2}{h})\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4I made a mistake

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4that should be +x^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4because  (x^2) is + x^2

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4you should be able to solve the limit with that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay.. I'll have a go from there.. I got some visitors that just came in.. so give me an hour or so

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok cool, they left.. I told em I was busy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you saying this equation should look like this? \[ \lim_{h \rightarrow 0} \frac{e^{(x + h)^2}  e^{x^2}}{h}*\frac{(x+h)^2 + x^2}{(x+h)^2 + x^2} \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4Here \[(\lim_{h \rightarrow 0} \frac{e^{(x+h)^2}e^{x^2}}{(x+h)^2+ x^2})(\lim_{h \rightarrow 0} \frac{(x+h)^2+ x^2}{h})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah because of the  in the top term .. cool

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4yeah then it splits into two limits

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4just solve them now they should be solvable

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4you will need to expand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that will be amazing

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ (\lim_{h \rightarrow 0} \frac{ e^{(x+h)^2}e^{x^2} }{ (x+h)^2+ x^2 } ) (\lim_{h \rightarrow 0} \frac{ (x+h)^2+ x^2 }{ h } ) \] \[ (\lim_{h \rightarrow 0} \frac{ e^{x^22hxh^2}e^{x^2} }{ x^22hxh^2+ x^2 } ) (\lim_{h \rightarrow 0} \frac{ x^22hxh^2+ x^2 }{ h } ) \] so far ...

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4cancel stuff out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\lim_{h \rightarrow 0} \frac{ e^{x^22hxh^2}e^{x^2} }{ 2hxh^2 } ) (\lim_{h \rightarrow 0} \frac{ 2hxh^2 }{ h } )\] ... I think I'm going the wrong way with this.. should it be canceling within the exponent of the numerator somehow?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this looks like it will zero out again. maybe if I multiply by log[] / log[] ? I dont know how else to get access to those exponents.. I can split them if they are E^(a + b) but E^(a  b) is a little trickier.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll read on L hopitals. and the chain rule.. I cant help wondering if is requires some kind of Log[] / Log [] to pull that exponent out down out of the stratosphere. so we can get access to the terms.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0those negative exponents are hard to work with.. I know we can use a square root rule, for square roots, and Im not sure about higher degrees, but does that same idea apply to a polynomial in the exponent?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe we can treat them like exponential functions in the global scope, and just throw away the low degree terms.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or maybe we run the limit just on the exponent, and then after knowing which way the exponent goes.. we can then say what it will do as x when in e^(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you give me a screen shot, wolfram alpha crashes my computer if I go anywhere near it.. I think the people who run my course put some widget into chrome that kills it.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4oh crud I just realized

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4we can solve this

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4really easily using the definition of e

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0damn, how the hell does it get that?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4pull e^(x^2) out of the limit what do you get?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4I spaced this is easily solvable using the definition of e

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do we pull e^(x^2) from the top and bottom? or just the top?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4just factor out e^(x^2)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4and you end up with the definition of e which equals 1 because 2xh  h^2 is essentially the same as just h, as h approaches 0

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4No I am saying, \[\lim_{h \rightarrow 0} \frac{e^{2x  h^2} 1}{2xh  h^2} = \lim_{h \rightarrow 0} \frac{e^{h(2x  h)} 1}{h(2x  h)} =\lim_{h \rightarrow 0} \frac{e^{h} 1}{h} = 1\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4so your problem is solved

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4just factor out e^(x^2) and the rest turns to 1

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4plug it back into the original equation you had in place of the limit and it is solved

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4You have to accept the assumption that \[h \approx h(2x  h)\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4if you multiply something by a really small number you are going to get a really small number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{h \rightarrow 0} \frac {e^{(x + h)^2}  e^{x^2}} {(x+h)^2 + x^2} \] \[\lim_{h \rightarrow 0} \frac {e^{(x^2 + 2hx +h^2)}  e^{x^2}} {(x^2+2hx +h^2) + x^2} \] \[\lim_{h \rightarrow 0} \frac {e^{x^2  2hx h^2}  e^{x^2}} {x^22hx h^2 + x^2} \] \[\lim_{h \rightarrow 0} \frac {e^{x^2}*(1/e^{2hx})*(1/e^{h^2})  e^{x^2}} {x^22hx h^2 + x^2} \] \[ e^{x^2} \lim_{h \rightarrow 0} \frac { (1/e^{2hx})*(1/e^{h^2})  e^{x^2}} {2hx h^2 } \]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4the rest is equal to 1 by the definition of the value of e

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so e^2hx is .. e^2*0*x = e^0 = 1 ? you mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but dont we still have the problem of 2hx  h^2 = 2*0*x  0*0 = 0 ?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4no, I am simply saying \[\lim_{h \rightarrow 0}\frac{e^{2hx  h^2}1}{2hx  h^2} = \lim_{h \rightarrow 0}\frac{e^{h}1}{h} = 1\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4I can't make it any more clear than this

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4\[2xe^{x^2}\] is what you get after canceling out h and solving the limit on the left

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, you need to be that clear with me.. :)

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4do you follow though?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah ... actually yeah.. I can feel the air in my lungs again .

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4ok so the problem is solved are you content?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4you can do the final bit of algebra now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02hxh^2 is equivalent to h in this situation

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4I think this is what the other guy was trying to point out to us earlier

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4in the original question but I hadn't really made that connection as of yet

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah.. ah ah..! okay.. thats' what that pauls website was trying to say too I think.. they mentioned this.. but it went over my head

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4so you can solve this now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think so.. , so I just have to reduce to I find that ahh.. property/ equation, Im not sure what you call that class of equations.. signatures or something? and then from there.. we have it.. just pull all the terms back in that were placed outside of the limit.. and you have your equation.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4Just try to write out the solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ e^{x^2} \lim_{h \rightarrow 0} \frac { e^{2hxh^2}  e^{x^2}} {2hx h^2 } = \lim_{h \rightarrow 0} \frac { e^{h} 1} {h } = 1 \] amI on the right track here?

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.4the limit becomes 1 not the entire expression

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f[x] = \frac{x}{e^{x^2}} \] \[f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] f[x] }{h} \right)\] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ e^{(x+h)^2} }  \frac{ x}{ e^{x^2} } \right) * \frac{1}{h} \right) \] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( (x+h) e^{(x+h)^2}  x e^{x^2} \right) * \frac{1}{h} \right) \] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( x e^{x^22hxh^2} + h e^{x^22hxh^2}  x e^{x^2} \right) * \frac{1}{h} \right) \] \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \left( x e^{2hxh^2} + h e^{2hxh^2}  x \right) * \frac{1}{h} \right) \right) \] \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{2hxh^2} }{h} + \frac{ h e^{2hxh^2} }{h}  \frac{ x }{h} \right) \right) \] \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{2hxh^2}  x }{h} + \frac{ e^{2hxh^2} }{1} \right) \right) \] \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{2hxh^2}  x }{h} \right) + \lim_{h \rightarrow 0} \frac{ e^{2hxh^2} }{1} \right) \] Limit of 2nd term = 1 \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{2hxh^2}  x }{h} \right) + 1 \right) \] Factor out x \[ f'[x] = x e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{2hxh^2}  1 }{h} \right) + 1 \right) \] Very close to the definition for e \[\lim_{h \rightarrow 0} \frac{e^h 1}{ h} = 1 \] \[ f'[x] = x e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(2xh)}  1 }{h} * \frac{2xh}{2xh} \right) + 1 \right) \] \[ f'[x] = x e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(2xh)}  1 }{h(2xh)} * \lim_{h \rightarrow 0} ({2xh}) \right) + 1 \right) \] So by definition of E and the constant law \[ f'[x] = x e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h}  1 }{h} * \lim_{h \rightarrow 0} ({2xh}) \right) + 1 \right) \] So by definition of E and the constant law \[ f'[x] = x e^{x^2} \left( 1 * ({2x0}) + 1 \right) \] \[ f'[x] = x e^{x^2} \left( 2x+1 \right) \] \[ f'[x] = e^{x^2} \left( 2x^2+x \right) \] \[ f'[x] = x e^{x^2}2x^2e^{x^2} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0SOOO close.. but I factored an extra x in there somehow.. it must be something to do with how I pulled the variables out.. I guess you need to make sure that there are brackets around the different levels. And hence why they're such sticklers for making sure the lim h>0 is carried through on all the stages.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw.. thank you, thank you, to everyone who helped me get this far..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0GIVEN \[ f[x] = \frac{x}{e^{x^2}} \] FIND \[ f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] f[x] }{h} \right) \] Substitute function \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ e^{(x+h)^2} }  \frac{ x}{ e^{x^2} } \right) * \frac{1}{h} \right) \] Move into numerator \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( (x+h) e^{(x+h)^2}  x e^{x^2} \right) * \frac{1}{h} \right) \] Expand so we can isolate the first term e^x^2 \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( x e^{x^22hxh^2} + h e^{x^22hxh^2}  x e^{x^2} \right) * \frac{1}{h} \right) \] Pull e^(x^2) out of the limit. \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \left( x e^{2hxh^2} + h e^{2hxh^2}  x \right) * \frac{1}{h} \right) \right) \] Distribute 1/h \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{2hxh^2} }{h} + \frac{ h e^{2hxh^2} }{h}  \frac{ x }{h} \right) \right) \] Simplify and combine like terms \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{2hxh^2}  x }{h} + \frac{ e^{2hxh^2} }{1} \right) \right) \] Isolate the 2nd term into it’s own limit and remove from the main limit. \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{2hxh^2}  x }{h} \right) + \lim_{h \rightarrow 0} \frac{ e^{2hxh^2} }{1} \right) \] Simplify 1/e^(h=0) = 1 \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{2hxh^2}  x }{h} \right) + 1 \right) \] Factor out x \[ f'[x] = e^{x^2} \left( x \left( \lim_{h \rightarrow 0} \left( \frac{ e^{2hxh^2}  1 }{h} \right) \right) + 1 \right) \] Very close to definition for e \[ \lim_{h \rightarrow 0} \frac{e^h 1}{ h} = 1 \] neutralize the limit, by making h term = h term in numerator and denominator so we have equivalence to E definition. \[ f'[x] = x e^{x^2} \left( x \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(2xh)}  1 }{h} * \frac{2xh}{2xh} \right) \right) + 1 \right) \] Simplify and isolate the new term we used to create a 1. \[ f'[x] = x e^{x^2} \left( \lim_{h \rightarrow 0} \left( x \left( \frac{ e^{h(2xh)}  1 }{h(2xh)} * \lim_{h \rightarrow 0} ({2xh}) \right) \right) + 1 \right) \] So by definition of E and the constant law we have two final limits. \[ f'[x] = e^{x^2} \left( \lim_{h \rightarrow 0} \left( x \left( \frac{ e^{h}  1 }{h} * \lim_{h \rightarrow 0} ({2xh}) \right) \right) + 1 \right) \] Now we just simplify our expression to the derivative. \[ f'[x] = e^{x^2} \left( \left( x \left( 1 * (2x0) \right) \right) + 1 \right) \] \[ f'[x] = e^{x^2} \left( \left( x \left( 2x \right) \right) + 1 \right) \] \[ f'[x] = e^{x^2} \left( 2x^2 + 1 \right) \] OMG!! I dont believe it! Has it been 4 days and 40+ hours!! THATS IT!! \[ f'[x] = e^{x^2} \left( 12x^2 \right) \] \[ f'[x] = e^{x^2}2x^2e^{x^2} \]

hartnn
 one year ago
Best ResponseYou've already chosen the best response.0excellent work hughfuve ^^ ! :)
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