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anonymous
 one year ago
What's my mistake here?
GIVEN
f[x]=x/E^x^2
(f[ x +h]  f[x]) \ h
anonymous
 one year ago
What's my mistake here? GIVEN f[x]=x/E^x^2 (f[ x +h]  f[x]) \ h

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's my mistake here... ? I dont think this is the right answer \[ \frac{(x+ h)}{e^{(x+h)^{2}}}  \frac{x}{e^{x^2}} * \frac{1}{h} \] Find common denominator ... \[ \frac{e^{x^2}(x+ h)}{e^{x^2}e^{(x+h)^{2}}}  \frac{x e^{(x+h)^{2}} }{e^{x^2} e^{(x+h)^{2}}} * \frac{1}{h} \] \[ \frac{ e^{x^2} (x+ h)  x e^{(x+h)^{2}} } { e^{x^2} e^{(x+h)^{2}} } * \frac{1}{h} \] expanded \[ \frac{ x e^{x^2} + h e^{x^2}  x e^{x^2+2hx +h^2} } { e^{x^2} e^{x^2+2hx+h^2} } * \frac{1}{h} \] exponent laws ... \[ \frac{ x e^{x^2} + h e^{x^2}  x e^{x^2} e^{2hx} e^{h^2} } { e^{x^2} e^{x^2} e^{2hx} e^{h^2} } * \frac{1}{h} \] simplify and cancel out terms \[ \frac{ x e^{x^2} + h e^{x^2}  x e^{x^2} } { e^{x^2} e^{x^2} } * \frac{1}{h} \] \[ \frac{ h e^{x^2} } { e^{x^2} e^{x^2} } * \frac{1}{h} \] \[ \frac{ h } { e^{x^2} } * \frac{1}{h} \] \[ \frac{ 1 } { e^{x^2} } \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your cancelling is wrong.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, looking for another way to explain it.. without the derivative rules because this stupid course throws this problem at me in chapter 3 and doesnt introduce derivatives until chapter 4.. lol damn them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just thought for a minute I found a less complicated way there, by getting a common denominator. my cancelation algebra is off though I guess

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1Yeah that isn't the derivative of that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Down the route you went. The problem is at this step: \[\frac{ xe^{x^{2}}+he^{x^{2}}xe^{x^{2}}e^{2xh}e^{h^{2}} }{ e^{x^{2}}e^{x^{2}}e^{2xh}e^{h^{2}} }*\frac{ 1 }{ h } = \frac{ x + h  xe^{2xh}e^{h^{2}} }{ e^{x^{2}}e^{2xh}e^{h^{2}}}*\frac{ 1 }{ h }\] That would be the simplification you can do, an \(e^{x^{2}}\) cancelling out.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0....Not that helped any in terms of trying to do this with the limit definition, lol.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got this: \[\large \frac{1}{h}\left( {\frac{{h{e^{{x^2}}}  x{e^{{h^2}}}  x{e^{2xh}}}}{{{e^{{x^2} + {{\left( {x + h} \right)}^2}}}}}} \right)\]

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1you need chain rule to solve this

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1When I guided you through this the other day, Noted that the derivative of e^(f(x)) is f'(x)e^(f(x)) this is not really a derivative rule it is due to chain rule. These lectures might help. https://www.youtube.com/watch?v=8UHQhpoFF1c The solution i gave you did not require the use of any derivative rules.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1Here is also a detailed proof of chain rule http://kruel.co/math/chainrule.pdf

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0true, I was reconstructing it, when I thought I had found a shortcut.. but I guess not.. because this route just cancels out to 0 again.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1It is because it is a composite function, it is good that you are doing this though as the ability to see functions in other functions is fundamental to being able to solve derivatives

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1or rather to be able to break a function into smaller functions

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeProofs.aspx Pauls notes also have an explanation of chain rule that is pretty alright

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1you can probably incorporate the chain rule proof to solve this problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not that it would help, but if you still feel the need to keep trying the limit definition route, you could try this variation and see if you can come up with some fancy simplification of it: \[\frac{ f(x)  f(t) }{ xt }\] as x approaches t. Another thing to try is rewriting x as \(e^{lnx}\) and h as \(e^{lnh }\) and maybe try to write everything as exponents

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1If you prove d/dx e^(f(x)) is f'(x)e^(f(x)) then you can say your problem is completely solved. I feel like this problem is not really something you should dwell to hard on you will likely never encounter a problem like this in real life. Although it cant hurt to learn proofs I suppose, as it will give you a deeper understanding of the mathematics.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that looks like a solution actually.. prove d/dx heheh.. I have like 6 chapters of derivative related stuff coming up.. I think they just wanted me to think on this, before they give me the chain rule and power rules. but I was a bit worried if I dont get this, I wont get whats coming up. going to go watch some more vids on limits, derivatives, chain and power rules, differentials.. mind you none of those simple examples seem to deal with complex rationals and multi term exponents, so I have found myself wondering a lot, "how would I apply this?" I'm starting to see though that terms can be treated as f(x) g(x) in various ways and exponents can be broken up, so it will click soon.

Australopithecus
 one year ago
Best ResponseYou've already chosen the best response.1I bet you will breeze through it, calculus is actually a pretty easy topic if you have algebra mastered.
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