anonymous
  • anonymous
What's my mistake here? GIVEN f[x]=x/E^x^2 (f[ x +h] - f[x]) \ h
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
What'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
  • anonymous
Your cancelling is wrong.
anonymous
  • anonymous
yeah, 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

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anonymous
  • anonymous
just 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
  • Australopithecus
Yeah that isn't the derivative of that
anonymous
  • anonymous
Down 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
  • anonymous
....Not that helped any in terms of trying to do this with the limit definition, lol.
anonymous
  • anonymous
lol..
Michele_Laino
  • Michele_Laino
I 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
  • Australopithecus
you need chain rule to solve this
Australopithecus
  • Australopithecus
When 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
  • Australopithecus
Here is also a detailed proof of chain rule http://kruel.co/math/chainrule.pdf
anonymous
  • anonymous
true, 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
  • Australopithecus
It 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
  • Australopithecus
or rather to be able to break a function into smaller functions
Australopithecus
  • Australopithecus
http://tutorial.math.lamar.edu/Classes/CalcI/DerivativeProofs.aspx Pauls notes also have an explanation of chain rule that is pretty alright
Australopithecus
  • Australopithecus
you can probably incorporate the chain rule proof to solve this problem
anonymous
  • anonymous
Not 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: \[\frac{ f(x) - f(t) }{ x-t }\] 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
  • Australopithecus
If 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
  • anonymous
that 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
  • Australopithecus
I bet you will breeze through it, calculus is actually a pretty easy topic if you have algebra mastered.

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