How can the derivative of e^(-1/x^2) be shown to be 0 at x=0?

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In summary, the conversation discusses finding the derivative at x=0 for the equation f'(x)= lim x->0 of [f(x+h)-f(x)]/h. The solution involves using the ea expansion and taking the limit, resulting in a denominator of 0 and the limit equaling 0. The conversation also mentions finding a Taylor series for x^(1/2) about a general center c=a^2 and involves finding the formula for the kth derivative of f(x). However, it is noted that L'Hopital's rule will not work in this case.
  • #1
roz77
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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?
 
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  • #2
http://img180.imageshack.us/img180/30/hwtj8.jpg [Broken]
 
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  • #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?
 
  • #4
f(x)=(x-a^2)^(1/2)? It involves a double factorial. n!=1*3*5*...*(n-2)*n, for n odd.
 
  • #5
-Vitaly- said:
http://img180.imageshack.us/img180/30/hwtj8.jpg [Broken]
[/URL]

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.
 
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  • #6
Jaekryl said:
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.
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.
 
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1. What is the derivative of e^(-1/x^2)?

The derivative of e^(-1/x^2) is equal to e^(-1/x^2) * (2/x^3), which can also be written as 2e^(-1/x^2)/x^3.

2. How do you find the derivative of e^(-1/x^2)?

To find the derivative of e^(-1/x^2), you can use the chain rule. First, rewrite the function as e^u, where u = -1/x^2. Then, take the derivative of u and multiply it by the original function e^u. In this case, the derivative of u is 2/x^3, so the final derivative is e^(-1/x^2) * (2/x^3).

3. Can the derivative of e^(-1/x^2) be simplified?

Yes, the derivative of e^(-1/x^2) can be simplified to 2e^(-1/x^2)/x^3. However, it is also acceptable to leave it in the form e^(-1/x^2) * (2/x^3).

4. What is the significance of the derivative of e^(-1/x^2)?

The derivative of e^(-1/x^2) is significant because it is the slope of the tangent line to the graph of the function at any given point. It also helps determine the rate of change of the function at that point.

5. Is e^(-1/x^2) a differentiable function?

Yes, e^(-1/x^2) is a differentiable function. This means that the function has a well-defined derivative at every point in its domain, including at x = 0, where the derivative is 0.

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