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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?

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  1. triciaal
    • one year ago
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    |dw:1433124964963:dw|

  2. anonymous
    • one year ago
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    \[\huge f(x)=\frac{x}{e^{x^2}}\]?

  3. anonymous
    • one year ago
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    That'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

  4. anonymous
    • one year ago
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    oh damn.. copied it wrong.. one sec.. that 2nd equation doesnt have a negative 2 in the exponent.. its a positive 2...

  5. triciaal
    • one year ago
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    @jim_thompson5910

  6. anonymous
    • one year ago
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    \[- \frac {x e^{-x^{2}}}{h} + e^{-(h+x)^{2}} +\frac {x e^{-(h+x)^{2}}}{h}\]

  7. anonymous
    • one year ago
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    I think ** that I want to multiply that middle set of terms by h/h ?and then I can drop the denominator

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

  9. IrishBoy123
    • one year ago
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    @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)?

  10. anonymous
    • one year ago
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    Yes, 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 \[(1-2x^2) e^{-x^2}\] 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.

  11. anonymous
    • one year ago
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    so Im stuck at the starting point..

  12. anonymous
    • one year ago
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    I really think I'm just missing some fundamental rules of algebra to do with denominators and exponents, but it hasn't clicked yet.

  13. triciaal
    • one year ago
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    @zepdrix thanks

  14. anonymous
    • one year ago
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    by the way.. thanks for helping me on this.. my heads blowing a fuse :)

  15. triciaal
    • one year ago
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    @UsukiDoll please help

  16. anonymous
    • one year ago
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    Maybe 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

  17. IrishBoy123
    • one year ago
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    if you are around tomorrow, i can help

  18. anonymous
    • one year ago
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    thnx I'll be around

  19. anonymous
    • one year ago
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    \[\left( \frac{ x+h }{e^{(x+h)^{2}}} - \frac{x}{e^{x^{2}}} \right)* \frac{1}{h}\] 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 \[f`[x] = E^{-x^2}-2x^{2} * E^{-x^2​}\]

  20. anonymous
    • one year ago
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    \[\frac{ (x+h)e^{-(x+h)^2}-x e^{-x^{2}} } {h}\] \[ ( x e^{-(x+h)^2}+h e^{-(x+h)^2}-x e^{-x^{2}} ) h^{-1} \] \[ h^{-1} x e^{-(x+h)^2}+h^{-1} h e^{-(x+h)^2}-h^{-1} x e^{-x^{2}} \] \[ \frac{ x e^{-(x+h)^2}}{h}+ e^{-(x+h)^2}-\frac{ x e^{-x^{2}}}{h} \] Ah! so I see now at least one way how the denominator can cancel out of the middle terms

  21. anonymous
    • one year ago
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    Ah! 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.

  22. IrishBoy123
    • one year ago
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    "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!

  23. IrishBoy123
    • one year ago
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    \( \frac{d}{dx} (\frac{x}{e^{x^2}}) = \frac{1}{e^{x^2}} (1 - 2x^{2})\) is what you are trying to get to.

  24. anonymous
    • one year ago
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    hell yeah, that's exactly it!

  25. anonymous
    • one year ago
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    now to get my head around it

  26. anonymous
    • one year ago
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    reading up on Leibniz ...

  27. anonymous
    • one year ago
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    Okay 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) ?

  28. anonymous
    • one year ago
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    Okay 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)

  29. anonymous
    • one year ago
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    going to restate the problem and start again

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