1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:

    [(e^(-1/(x+h)^2))-(e^(-1/x^2))]/h

    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
  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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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: Oct 29, 2009
  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
    Thanks
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Derivative of e^(-1/x^2)
  1. Derivative of e^x^x^2? (Replies: 1)

Loading...