## 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?

1. triciaal

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2. anonymous

$\huge f(x)=\frac{x}{e^{x^2}}$?

3. anonymous

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

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

5. triciaal

@jim_thompson5910

6. anonymous

$- \frac {x e^{-x^{2}}}{h} + e^{-(h+x)^{2}} +\frac {x e^{-(h+x)^{2}}}{h}$

7. anonymous

I think ** that I want to multiply that middle set of terms by h/h ?and then I can drop the denominator

8. anonymous

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

@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

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

so Im stuck at the starting point..

12. anonymous

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

@zepdrix thanks

14. anonymous

by the way.. thanks for helping me on this.. my heads blowing a fuse :)

15. triciaal

16. anonymous

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

if you are around tomorrow, i can help

18. anonymous

thnx I'll be around

19. anonymous

$\left( \frac{ x+h }{e^{(x+h)^{2}}} - \frac{x}{e^{x^{2}}} \right)* \frac{1}{h}$ 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

$\frac{ (x+h)e^{-(x+h)^2}-x e^{-x^{2}} } {h}$ $( x e^{-(x+h)^2}+h e^{-(x+h)^2}-x e^{-x^{2}} ) h^{-1}$ $h^{-1} x e^{-(x+h)^2}+h^{-1} h e^{-(x+h)^2}-h^{-1} x e^{-x^{2}}$ $\frac{ x e^{-(x+h)^2}}{h}+ e^{-(x+h)^2}-\frac{ x e^{-x^{2}}}{h}$ Ah! so I see now at least one way how the denominator can cancel out of the middle terms

21. anonymous

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

"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

$$\frac{d}{dx} (\frac{x}{e^{x^2}}) = \frac{1}{e^{x^2}} (1 - 2x^{2})$$ is what you are trying to get to.

24. anonymous

hell yeah, that's exactly it!

25. anonymous

now to get my head around it

26. anonymous

27. anonymous

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

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

going to restate the problem and start again