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anonymous
 one year ago
Given the function
f[x] = x/E^(x^2)
Then how would you go about Expanding
(f[x+h]  f[x])/h
And then assuming that h=0 how would simplify that?
anonymous
 one year ago
Given the function f[x] = x/E^(x^2) Then how would you go about Expanding (f[x+h]  f[x])/h And then assuming that h=0 how would simplify that?

This Question is Closed

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433124964963:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge f(x)=\frac{x}{e^{x^2}}\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That's part of the answer.. I think I got this far ... (the exponent needs to be made negative) \[f[x] = x e^{x^{2}}\] but then my calculator says that (f[x+h]  f[x])/h = \[\frac{ (x e ^{x^{2}}) }{ h } + e^{(h+x)^{2}} + \frac{(x e^{(h +x)^{2}}}{h}\] but now I don't know how it got there, and then my calculator says it should simplify to this \[(1  2x^{2}) e{x^{2}}\] but again, I have no idea how it got there either. And I need to be able to do this on paper without a calc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh damn.. copied it wrong.. one sec.. that 2nd equation doesnt have a negative 2 in the exponent.. its a positive 2...

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think ** that I want to multiply that middle set of terms by h/h ?and then I can drop the denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so I got to here.. now I'm trying to expand / FOIL it in the denominator, I think is the next step, \[\left( \frac{ x+h }{e^{(x+h)^{2}}}  \frac{x}{e^{x^{2}}} \right)* \frac{1}{h}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0@hughfuve that's actually where you were supposed to start. do you know where you are trying to get to (because knowing that really helps too)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, the problem First defines \[f[x]= x/E^{x^2} \] Then gives the mathematica function for instantaneous growth via... \[f`[x] = E^{x^2}2x^{2} * E^{x^2}\] Which I believe is basically the expanded function of (f[x +h]  f[x])/h When h=0 It then uses the mathematica Expand[ (f[x +h]  f[x])/h ] command. and comes up with \[ \frac {x e^{x^{2}}}{h} + e^{(h+x)^{2}} +\frac {x e^{(h+x)^{2}}}{h}\] And then applies the Together[f`[x]] command which creates this... \[((1+2x^2) e^{x^2}) \] Which simplifies to this \[(12x^2) e^{x^2}\] Then it asks me to explain why and how it did that.. In similar problems, I just expanded with h=0 and simplified out the 0's taking what was left, but in this case, if I use h=0 I get stuck trying to work out the algebra for exponents with exponents in a rational expression. And I dont come anywhere close to what mathematica is putting out.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so Im stuck at the starting point..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I really think I'm just missing some fundamental rules of algebra to do with denominators and exponents, but it hasn't clicked yet.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0by the way.. thanks for helping me on this.. my heads blowing a fuse :)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.1@UsukiDoll please help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Maybe if I can work out how this \[ \frac {x e^{x^{2}}}{h} + e^{(h+x)^{2}} +\frac {x e^{(h+x)^{2}}}{h} \] Simplifies to this. \[\frac {(x e^{x^{2}}) +(h + x) e^{(h+x)^{2}}} {h}\] It might start to make sense

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0if you are around tomorrow, i can help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left( \frac{ x+h }{e^{(x+h)^{2}}}  \frac{x}{e^{x^{2}}} \right)* \frac{1}{h}\] Okay I think this is the first missing link I think.... a / E^n = a E^n Therefore \[ \left( (x+h)e^{(x+h)^2}x e^{x^{2}} \right) * \frac{1}{h} \] \[ \frac{ (x+h)e^{(x+h)^2}x e^{x^{2}} } {h} \] Now to work out how and why this leads f'[x] to become \[f`[x] = E^{x^2}2x^{2} * E^{x^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ (x+h)e^{(x+h)^2}x e^{x^{2}} } {h}\] \[ ( x e^{(x+h)^2}+h e^{(x+h)^2}x e^{x^{2}} ) h^{1} \] \[ h^{1} x e^{(x+h)^2}+h^{1} h e^{(x+h)^2}h^{1} x e^{x^{2}} \] \[ \frac{ x e^{(x+h)^2}}{h}+ e^{(x+h)^2}\frac{ x e^{x^{2}}}{h} \] Ah! so I see now at least one way how the denominator can cancel out of the middle terms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah! I see.. After some sleep and some googling, I discovered f`[x] means the derivative of the function f[x] I have never encountered a derivative before I need to go learn what that is.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0"f`[x] means the derivative of the function f[x]" yep, that't how you can work out the answer before you do the algebra!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\( \frac{d}{dx} (\frac{x}{e^{x^2}}) = \frac{1}{e^{x^2}} (1  2x^{2})\) is what you are trying to get to.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hell yeah, that's exactly it!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now to get my head around it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0reading up on Leibniz ...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay now I understand the problem better.. Can anybody explain the steps to the algebra that arrives at the derivative to f[x] = x/e^(x^2) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay I will close.. By the way I found a decent online book on this subject for anyone else who needs help with this kind of thing. Keislers Elementary Calculus , An Infinitesimal Approach http://www.math.wisc.edu/~keisler/calc.html Chapters (1 & 2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0going to restate the problem and start again
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