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Derivative of e^(1/x^2) 
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#1
Nov1808, 08:14 PM

P: 16

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
Nov1808, 08:40 PM

P: 39




#3
Nov1808, 09:55 PM

P: 16

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
Nov1808, 10:28 PM

Sci Advisor
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Thanks
P: 25,243

Derivative of e^(1/x^2)
f(x)=(xa^2)^(1/2)? It involves a double factorial. n!!=1*3*5*...*(n2)*n, for n odd.



#5
Oct2909, 03:15 PM

P: 1

Therefore, a is in fact approaching infinity. So you cannot assume that the e^{a} expansion is equivalent to 1 + a + a^{2}/2 since a is not small. Instead, just substitute in the entire e^{a} expansion, then take the limit, and you will see that the denominator will equal 0 + 0 + ∞ + ∞ + ... Therefore the limit is still 2/∞ = 0. 


#6
Oct2909, 04:30 PM

P: 39

Thanks p.s. L'Hopital's rule will not work is this case as well , :( as it requires (infinity/infinity) or (zero/zero) limit. 


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