anonymous
  • anonymous
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 ]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
or at the very least please send me a good link that explains how to deal with canceling variables embedded in a polynomial exponent
jigglypuff314
  • jigglypuff314
Do 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/differential-calculus/taking-derivatives/der_common_functions/v/proof-d-dx-e-x-e-x
anonymous
  • anonymous
\[f[x] = \frac{x}{e^{x^2}}\] \[ f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] -f[x] }{h} \right) \] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ E^{(x+h)^2} } - \frac{ x}{ E^{(x^2)} } \right) * \frac{1}{h} \right) \]

Looking for something else?

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

More answers

anonymous
  • anonymous
Unfortunately I do..
anonymous
  • anonymous
i have really come to admire those rules of the derivative..
amistre64
  • amistre64
first principles eh
anonymous
  • anonymous
yeah the baby steps..
anonymous
  • anonymous
I've been on it for 3 days.. and my OCD wont allow me to let it go
anonymous
  • anonymous
I cant friggin sleep, so pissed off.
anonymous
  • anonymous
staring 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
  • anonymous
I can get there like this.. I know the answer to this is going to be \[ f'[x] = (1-2x^2)e^{x^2}\] 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
  • anonymous
and I cant reverse the logic, that would allow me to do that with algebra.
Australopithecus
  • Australopithecus
Its alright here, we can try to work through your final limit using this informal proof. \[\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{h}\]
1 Attachment
Australopithecus
  • Australopithecus
Although maybe amistre64 has a better way of solving this limit or problem
anonymous
  • anonymous
Is that the first trick of limits, that we can remove x ? \[ f[x] = x/E^{x^2} \] \[ f[x] = x E^{x^2} \]
anonymous
  • anonymous
oops .. that 's -x \[ f[x] = x E^{-x^2} \]
Australopithecus
  • Australopithecus
Well 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
  • Australopithecus
This 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
  • anonymous
oh, I just found your image...
Australopithecus
  • Australopithecus
yeah you could potentially use that to solve your problem
Australopithecus
  • Australopithecus
or the limit
anonymous
  • anonymous
whats the strategy to get h out of that numerator ? should that even be our goal from here?
Australopithecus
  • Australopithecus
Well 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
  • anonymous
Whats 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
  • Australopithecus
There is something you should note though that will likely come up the proof of d/dx e^x = e^x by definition of e: \[\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1\]
Australopithecus
  • Australopithecus
well e^(-x^2) is a composite function
anonymous
  • anonymous
oh is that proof because technically, we have e^0 = (almost 1)-1 and we get a something/something
Australopithecus
  • Australopithecus
if f(x) = e^(x) and g(x) = -x^2 then f(g(x)) = e^(-x^2)
Australopithecus
  • Australopithecus
that is how the value of e is defined
Australopithecus
  • Australopithecus
one of the ways it is defined
anonymous
  • anonymous
there was another x in there though.. f[x] = x / E^(x^2) is the x removed, or placed into f or g ?
anonymous
  • anonymous
oops.. thats E^-x^2
Australopithecus
  • Australopithecus
we pulled it out of the limit
Australopithecus
  • Australopithecus
because it is technically a constant
anonymous
  • anonymous
ah ok.. so that sits.. outside the lim definition still ? or does it now hold some value like 1?
anonymous
  • anonymous
or does it cancel to 0 ? I havent clicked yet on how that constant rule is actually applied.
anonymous
  • anonymous
just watching another vid on constant rule.
anonymous
  • anonymous
lol, man, nobody explains this well.. they just gloss over and say ... 'oh we can pull this out.' and they move on
anonymous
  • anonymous
okay, I think it goes outside ..
Australopithecus
  • Australopithecus
no 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
  • anonymous
but it hangs around..
Australopithecus
  • Australopithecus
the e^(x^2) just cancels out, you could cancel it out right now it wouldn't change your solution
anonymous
  • anonymous
so in reference to \[\lim_{h \rightarrow 0} \frac{e^h-1}{h} = 1\] 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
  • Australopithecus
No 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
  • anonymous
hmmm, that's an odd and curious property. :/
Australopithecus
  • Australopithecus
I can help you work through the informal proof if you want.
Australopithecus
  • Australopithecus
@amistre64 , @Hero Might be able to help you more so with this.
anonymous
  • anonymous
I 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
  • anonymous
sorry thats x g[x]
anonymous
  • anonymous
wish you could back edit these posts
Australopithecus
  • Australopithecus
Not 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
  • Australopithecus
You would have this exact limit if you were trying to take the derivative with the definition using e^(-x^2)
Australopithecus
  • Australopithecus
If it was e^(x) and not e^(-x^2) you would have no problem because you wouldn't need chain rule
Australopithecus
  • Australopithecus
you are correct it is a problem of 3 functions
Australopithecus
  • Australopithecus
this problem requires product rule and chain rule
Australopithecus
  • Australopithecus
but the third function is easily differentiable so its not a problem
Australopithecus
  • Australopithecus
your problem is essentially m(x)*f(g(x))
Australopithecus
  • Australopithecus
this is actually a good thing for you to realize since you need to be able to recognize things like this when taking derivatives
Australopithecus
  • Australopithecus
you need to be able to split a function into multiple functions to take a derivative in most cases
anonymous
  • anonymous
ah 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
  • anonymous
so Ive been scratching my head, trying to interpret the language
Australopithecus
  • Australopithecus
I 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
  • Australopithecus
g(x) = x g'(x) = 1 f(x) = ln(x) f'(x) = 1/x
anonymous
  • anonymous
okay, 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
  • Australopithecus
I posted an informal proof you could follow along with
Australopithecus
  • Australopithecus
to solve the limit
anonymous
  • anonymous
I 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
  • anonymous
so 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
  • Australopithecus
\[\lim_{h \rightarrow 0} \frac{e^{-(x + h)^2} - e^{-x^2}}{h} = \lim_{h \rightarrow 0} \frac{e^{-(x + h)^2} - e^{-x^2}}{h}*\frac{-(x+h)^2 - x^2}{-(x+h)^2 - x^2}\] 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
  • anonymous
I 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
  • anonymous
yes, I understand on the m(x) .. thnx
Australopithecus
  • Australopithecus
\[=(\lim_{h \rightarrow 0} \frac{e^{-(-x+h)^2}-e^{-x^2}}{-(x+h)^2 - x^2})*(\lim_{h \rightarrow 0} \frac{-(x+h)^2 - x^2}{h})\]
Australopithecus
  • Australopithecus
oops
Australopithecus
  • Australopithecus
I made a mistake
Australopithecus
  • Australopithecus
that should be +x^2
Australopithecus
  • Australopithecus
because - (-x^2) is + x^2
Australopithecus
  • Australopithecus
you should be able to solve the limit with that
anonymous
  • anonymous
okay.. I'll have a go from there.. I got some visitors that just came in.. so give me an hour or so
anonymous
  • anonymous
ok cool, they left.. I told em I was busy
anonymous
  • anonymous
Are 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
  • Australopithecus
Here \[(\lim_{h \rightarrow 0} \frac{e^{-(x+h)^2}-e^{-x^2}}{-(x+h)^2+ x^2})(\lim_{h \rightarrow 0} \frac{-(x+h)^2+ x^2}{h})\]
anonymous
  • anonymous
ah because of the - in the top term .. cool
Australopithecus
  • Australopithecus
yeah then it splits into two limits
Australopithecus
  • Australopithecus
just solve them now they should be solvable
Australopithecus
  • Australopithecus
you will need to expand
Australopithecus
  • Australopithecus
everything
anonymous
  • anonymous
that will be amazing
anonymous
  • anonymous
Im excited
anonymous
  • anonymous
\[ (\lim_{h \rightarrow 0} \frac{ e^{-(x+h)^2}-e^{-x^2} }{ -(x+h)^2+ x^2 } ) (\lim_{h \rightarrow 0} \frac{ -(x+h)^2+ x^2 }{ h } ) \] \[ (\lim_{h \rightarrow 0} \frac{ e^{-x^2-2hx-h^2}-e^{-x^2} }{ -x^2-2hx-h^2+ x^2 } ) (\lim_{h \rightarrow 0} \frac{ -x^2-2hx-h^2+ x^2 }{ h } ) \] so far ...
Australopithecus
  • Australopithecus
cancel stuff out
anonymous
  • anonymous
\[(\lim_{h \rightarrow 0} \frac{ e^{-x^2-2hx-h^2}-e^{-x^2} }{ -2hx-h^2 } ) (\lim_{h \rightarrow 0} \frac{ -2hx-h^2 }{ h } )\] ... I think I'm going the wrong way with this.. should it be canceling within the exponent of the numerator somehow?
anonymous
  • anonymous
this 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
  • anonymous
I'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
  • anonymous
those 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
  • anonymous
maybe we can treat them like exponential functions in the global scope, and just throw away the low degree terms.
anonymous
  • anonymous
or 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)
Australopithecus
  • Australopithecus
http://www.wolframalpha.com/input/?i=lim+%28h-%3E0%29+%28+e^%28-%28x%2Bh%29^2%29+-+e^%28-x^2%29+%29%2F%28+-%28x%2Bh%29^2+%2B+x^2%29
anonymous
  • anonymous
can 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.
anonymous
  • anonymous
if not that's okay.
Australopithecus
  • Australopithecus
Australopithecus
  • Australopithecus
oh crud I just realized
Australopithecus
  • Australopithecus
we can solve this
Australopithecus
  • Australopithecus
really easily using the definition of e
anonymous
  • anonymous
damn, how the hell does it get that?
Australopithecus
  • Australopithecus
pull e^(-x^2) out of the limit what do you get?
Australopithecus
  • Australopithecus
I spaced this is easily solvable using the definition of e
anonymous
  • anonymous
do we pull e^(-x^2) from the top and bottom? or just the top?
Australopithecus
  • Australopithecus
what?
Australopithecus
  • Australopithecus
just factor out e^(-x^2)
Australopithecus
  • Australopithecus
and 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
anonymous
  • anonymous
zero out h?
Australopithecus
  • Australopithecus
No 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
  • Australopithecus
so your problem is solved
Australopithecus
  • Australopithecus
just factor out e^(-x^2) and the rest turns to 1
Australopithecus
  • Australopithecus
plug it back into the original equation you had in place of the limit and it is solved
Australopithecus
  • Australopithecus
You have to accept the assumption that \[h \approx h(2x - h)\]
Australopithecus
  • Australopithecus
if you multiply something by a really small number you are going to get a really small number
anonymous
  • anonymous
\[\lim_{h \rightarrow 0} \frac {e^{-(x + h)^2} - e^{-x^2}} {-(x+h)^2 + x^2} \] \[\lim_{h \rightarrow 0} \frac {e^{-(x^2 + 2hx +h^2)} - e^{-x^2}} {-(x^2+2hx +h^2) + x^2} \] \[\lim_{h \rightarrow 0} \frac {e^{-x^2 - 2hx -h^2} - e^{-x^2}} {-x^2-2hx -h^2 + x^2} \] \[\lim_{h \rightarrow 0} \frac {e^{-x^2}*(1/e^{2hx})*(1/e^{h^2}) - e^{-x^2}} {-x^2-2hx -h^2 + x^2} \] \[ e^{-x^2} \lim_{h \rightarrow 0} \frac { (1/e^{2hx})*(1/e^{h^2}) - e^{-x^2}} {-2hx -h^2 } \]
Australopithecus
  • Australopithecus
the rest is equal to 1 by the definition of the value of e
anonymous
  • anonymous
so e^2hx is .. e^2*0*x = e^0 = 1 ? you mean?
anonymous
  • anonymous
but dont we still have the problem of -2hx - h^2 = -2*0*x - 0*0 = 0 ?
anonymous
  • anonymous
in the denominator
Australopithecus
  • Australopithecus
no, 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
  • Australopithecus
I can't make it any more clear than this
Australopithecus
  • Australopithecus
\[-2xe^{-x^2}\] is what you get after canceling out h and solving the limit on the left
anonymous
  • anonymous
sorry, you need to be that clear with me.. :)
Australopithecus
  • Australopithecus
do you follow though?
anonymous
  • anonymous
yeah ... actually yeah.. I can feel the air in my lungs again .
Australopithecus
  • Australopithecus
ok so the problem is solved are you content?
Australopithecus
  • Australopithecus
you can do the final bit of algebra now
anonymous
  • anonymous
-2hx-h^2 is equivalent to h in this situation
Australopithecus
  • Australopithecus
I think this is what the other guy was trying to point out to us earlier
Australopithecus
  • Australopithecus
yes
Australopithecus
  • Australopithecus
in the original question but I hadn't really made that connection as of yet
anonymous
  • anonymous
yeah.. 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
  • Australopithecus
so you can solve this now?
anonymous
  • anonymous
I 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
  • Australopithecus
Just try to write out the solution
anonymous
  • anonymous
\[ e^{-x^2} \lim_{h \rightarrow 0} \frac { e^{-2hx-h^2} - e^{-x^2}} {-2hx -h^2 } = \lim_{h \rightarrow 0} \frac { e^{h} -1} {h } = 1 \] amI on the right track here?
Australopithecus
  • Australopithecus
not really
Australopithecus
  • Australopithecus
the limit becomes 1 not the entire expression
anonymous
  • anonymous
\[f[x] = \frac{x}{e^{x^2}} \] \[f'[x] = \lim_{h \rightarrow 0} \left( \frac{f[x+h] -f[x] }{h} \right)\] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( \frac{ (x+h) }{ e^{(x+h)^2} } - \frac{ x}{ e^{x^2} } \right) * \frac{1}{h} \right) \] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( (x+h) e^{-(x+h)^2} - x e^{-x^2} \right) * \frac{1}{h} \right) \] \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( x e^{-x^2-2hx-h^2} + h e^{-x^2-2hx-h^2} - x e^{-x^2} \right) * \frac{1}{h} \right) \] \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \left( x e^{-2hx-h^2} + h e^{-2hx-h^2} - x \right) * \frac{1}{h} \right) \right) \] \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} }{h} + \frac{ h e^{-2hx-h^2} }{h} - \frac{ x }{h} \right) \right) \] \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} + \frac{ e^{-2hx-h^2} }{1} \right) \right) \] \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + \lim_{h \rightarrow 0} \frac{ e^{-2hx-h^2} }{1} \right) \] Limit of 2nd term = 1 \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + 1 \right) \] Factor out x \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{-2hx-h^2} - 1 }{h} \right) + 1 \right) \] Very close to the definition for e \[\lim_{h \rightarrow 0} \frac{e^h -1}{ h} = 1 \] \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(-2x-h)} - 1 }{h} * \frac{-2x-h}{-2x-h} \right) + 1 \right) \] \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(-2x-h)} - 1 }{h(-2x-h)} * \lim_{h \rightarrow 0} ({-2x-h}) \right) + 1 \right) \] So by definition of E and the constant law \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h} - 1 }{h} * \lim_{h \rightarrow 0} ({-2x-h}) \right) + 1 \right) \] So by definition of E and the constant law \[ f'[x] = x e^{-x^2} \left( 1 * ({-2x-0}) + 1 \right) \] \[ f'[x] = x e^{-x^2} \left( -2x+1 \right) \] \[ f'[x] = e^{-x^2} \left( -2x^2+x \right) \] \[ f'[x] = x e^{-x^2}-2x^2e^{-x^2} \]
anonymous
  • anonymous
SOOO 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
  • anonymous
btw.. thank you, thank you, to everyone who helped me get this far..
anonymous
  • anonymous
GIVEN \[ f[x] = \frac{x}{e^{x^2}} \] 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 \[ f'[x] = \lim_{h \rightarrow 0} \left( \left( x e^{-x^2-2hx-h^2} + h e^{-x^2-2hx-h^2} - x e^{-x^2} \right) * \frac{1}{h} \right) \] Pull e^(-x^2) out of the limit. \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \left( x e^{-2hx-h^2} + h e^{-2hx-h^2} - x \right) * \frac{1}{h} \right) \right) \] Distribute 1/h \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} }{h} + \frac{ h e^{-2hx-h^2} }{h} - \frac{ x }{h} \right) \right) \] Simplify and combine like terms \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} + \frac{ e^{-2hx-h^2} }{1} \right) \right) \] Isolate the 2nd term into it’s own limit and remove from the main limit. \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + \lim_{h \rightarrow 0} \frac{ e^{-2hx-h^2} }{1} \right) \] Simplify 1/e^(h=0) = 1 \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( \frac{ x e^{-2hx-h^2} - x }{h} \right) + 1 \right) \] Factor out x \[ f'[x] = e^{-x^2} \left( x \left( \lim_{h \rightarrow 0} \left( \frac{ e^{-2hx-h^2} - 1 }{h} \right) \right) + 1 \right) \] 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. \[ f'[x] = x e^{-x^2} \left( x \left( \lim_{h \rightarrow 0} \left( \frac{ e^{h(-2x-h)} - 1 }{h} * \frac{-2x-h}{-2x-h} \right) \right) + 1 \right) \] Simplify and isolate the new term we used to create a 1. \[ f'[x] = x e^{-x^2} \left( \lim_{h \rightarrow 0} \left( x \left( \frac{ e^{h(-2x-h)} - 1 }{h(-2x-h)} * \lim_{h \rightarrow 0} ({-2x-h}) \right) \right) + 1 \right) \] So by definition of E and the constant law we have two final limits. \[ f'[x] = e^{-x^2} \left( \lim_{h \rightarrow 0} \left( x \left( \frac{ e^{h} - 1 }{h} * \lim_{h \rightarrow 0} ({-2x-h}) \right) \right) + 1 \right) \] Now we just simplify our expression to the derivative. \[ f'[x] = e^{-x^2} \left( \left( x \left( 1 * (-2x-0) \right) \right) + 1 \right) \] \[ f'[x] = e^{-x^2} \left( \left( x \left( -2x \right) \right) + 1 \right) \] \[ f'[x] = e^{-x^2} \left( -2x^2 + 1 \right) \] OMG!! I dont believe it! Has it been 4 days and 40+ hours!! THATS IT!! \[ f'[x] = e^{-x^2} \left( 1-2x^2 \right) \] \[ f'[x] = e^{-x^2}-2x^2e^{-x^2} \]
hartnn
  • hartnn
excellent work hughfuve ^^ ! :)

Looking for something else?

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