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

1. Nov 18, 2008

### roz77

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. Nov 18, 2008

### -Vitaly-

3. Nov 18, 2008

### roz77

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?

4. Nov 18, 2008

### Dick

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

5. Oct 29, 2009

### Jaekryl

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
6. Oct 29, 2009

### -Vitaly-

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