Find f'(0) for f(x) = e^(-1/x^2)

[Ryker]

Homework Statement

\begin{displaymath}<br /> f(x) = \left\{<br /> \begin{array}{lr}<br /> e^{- \frac{1}{x^{2}}} &amp; : x \neq 0 \\<br /> 0 &amp; : x = 0<br /> \end{array}<br /> \right.<br /> \end{displaymath}

What is the derivative of f(x) at x = 0, that is, what is f'(0)?

The Attempt at a Solution

I am a bit lost on how to show that f'(0) = 0. I've tried doing it from the definition of the derivative, but then I get

\displaystyle\lim_{x\to 0}\frac{e^{- \frac{1}{x^{2}}}}{x}

and I don't know how to proceed. L'Hôpital's rule doesn't seem to help here, either. Any suggestions?

 

[losiu99]

I'd substitute t=1/x.

 

[Ryker]

Right after I posted the question, the solution you're suggesting popped up in my head, as well Is there, however, any other way of doing this, rather than by change of variable and then invoking the "exponential functions rise faster than polynomials" rule?

 

[losiu99]

l'Hospital works if you take
\frac{\frac{1}{x}}{e^{\frac{1}{x^2}}}

 

[Ryker]

Oh yeah, if you take it as such and apply l'Hôpital's rule, then you get something that tends to 0 in the numerator and something that tends to plus infinity in the denominator, right?

 

[Dick]

Your initial derivative isn't quite right. Check that. But then the way to do it is take the log of the derivative, and try to do l'Hopital's on the log of the derivative.

 

[Ryker]

Hmm, what is wrong with my initial derivative then? Here's my reasoning, taken from the definition of the derivative:

f&#039;(0) = \displaystyle\lim_{x\to 0}\frac{f(x) - f(0)}{x - 0} = \displaystyle\lim_{x\to 0}\frac{e^{- \frac{1}{x^{2}}}}{x}

And how do I take a log of the derivative, if the derivative is exactly what I'm looking for? Or do you mean on the derivative of the function when x is not 0? Could you perhaps elaborate on that?

 

[Dick]

Hmm, what is wrong with my initial derivative then? Here's my reasoning, taken from the definition of the derivative:

f&#039;(0) = \displaystyle\lim_{x\to 0}\frac{f(x) - f(0)}{x - 0} = \displaystyle\lim_{x\to 0}\frac{e^{- \frac{1}{x^{2}}}}{x}

And how do I take a log of the derivative, if the derivative is exactly what I'm looking for? Or do you mean on the derivative of the function when x is not 0? Could you perhaps elaborate on that?

Actually your initial derivative is ok. I was taking the derivative at nonzero x and letting x approach 0. But your way is better. If you know the limit of the log of the derivative, you should be able to figure out the derivative. You want to look at log(e^(-1/x^2)/x) and show that goes to negative infinity.

 

[Ryker]

Actually your initial derivative is ok. I was taking the derivative at nonzero x and letting x approach 0. But your way is better. If you know the limit of the log of the derivative, you should be able to figure out the derivative. You want to look at log(e^(-1/x^2)/x) and show that goes to negative infinity.

Alright, thanks, I'll look into that in the next couple of days, when I do a review of the log function, since at this moment I'm having a bit of trouble grasping this. But if my initial derivative is correct, then both methods suggested by losiu99 should also be fine, right? Basically, the reason I am asking the original question is because we did some stuff in class involving this derivative at 0, and just casually, without much explanation took it to be 0, since we were doing it as a step for something different. So I just figured there had to be really simple trick to this, otherwise we would've proven why that is the case, as well.

 

[Dick]

Sure. I like losiu99's second suggestion best. Use l'Hopital on the form in post 4. Saying 'exponentials grow faster than polynomials' is a little vague.

 

[Ryker]

Saying 'exponentials grow faster than polynomials' is a little vague.

Yeah, I guess I would've put it a bit more formally, but we have shown in class that the limit of a polynomial over an exponential function as x tends to infinity is 0, so we can use this fact when proving limits.

 

[Dick]

Why don't you just try the way losiu99 suggested in post 4. It's pretty easy.

 

[Ryker]

No worries, I tried it right after he suggested it, and I agree it is pretty easy after you've set it up like that