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Derivative of e^(-1/x^2)

  1. Nov 18, 2008 #1
    So I know how to take the general derivative of this equation. It's a simple product rule. I have that. My problem is, I need to show that the derivative at x=0 is 0. I know that I'm supposed to use this equation.

    f'(x)= lim x->0 of [f(x+h)-f(x)]/h

    So I plug in x+h and I get:


    My problem is, I have no idea how to simplify that. I know that I need to get rid of the denominator, but I'm not sure how I can do that. Suggestions?
  2. jcsd
  3. Nov 18, 2008 #2
    http://img180.imageshack.us/img180/30/hwtj8.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  4. Nov 18, 2008 #3
    Thanks. That worked perfectly.

    I actually have another question now. I need to find a Taylor series for x^(1/2) about a general center c=a^2. I've been taking derivatives and trying to find a pattern. I seem to be on the verge of it but I just can quite get it. Anyone know what the formula for the kth derivative of f(x) might be?
  5. Nov 18, 2008 #4


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    f(x)=(x-a^2)^(1/2)? It involves a double factorial. n!!=1*3*5*...*(n-2)*n, for n odd.
  6. Oct 29, 2009 #5

    This is a great solution, except for one logical error. In your case of the ea expansion, a = 1/x2 as x -> 0.
    Therefore, a is in fact approaching infinity. So you cannot assume that the ea expansion is equivalent to 1 + a + a2/2 since a is not small.

    Instead, just substitute in the entire ea expansion, then take the limit, and you will see that the denominator will equal 0 + 0 + ∞ + ∞ + ...
    Therefore the limit is still 2/∞ = 0.
    Last edited by a moderator: May 4, 2017
  7. Oct 29, 2009 #6
    Oh dear, I completely missed that part :( so no need to expand the exponential at all. Because exponentials increase a lot faster than polynomials decrease. So by inspection e^(x^(-2))*x^3 -> infinity , when x->0
    p.s. L'Hopital's rule will not work is this case as well , :( as it requires (infinity/infinity) or (zero/zero) limit.
    Last edited: Oct 29, 2009
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