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    Hi

    How would i go about calculating the nth order derivative of:

    exp(-1/x^2)

    Thanks
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    The formula latentcorpse has suggested is incorrect - I don't believe there is any kind of nice formula for this.
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    yes. that was an awful mistake. clearly i forget the product rule when hungover. in fact, i'm going to delete it to avoid the shame.
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    I agree with DFranklin.

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    For the lulz, I got this:

    \frac{\partial^n}{\partial x^n}(e^\frac{-1}{x^2}) = e^\frac{-1}{x^2} x^{-n} \sum_{r=0}^{n}\sum_{q=0}^{n} \frac{((-1)^q (\frac{-1}{x^2})^r (1+2 q-2 r-n)_n)}{(q! (-q+r)!)}

    Edit: and I'm fairly sure it's wrong.
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    (Original post by Crucio404)
    Hi

    How would i go about calculating the nth order derivative of:

    exp(-1/x^2)

    Thanks
    The following method is not "nice" because using it is just as difficult as calculating manually but it is a formula which you may be interested in.

    Differentiate once we have:
    \dfrac{{d{e^{ - {x^{ - 2}}}}}}{{dx}} = {e^{ - {x^{ - 2}}}} \cdot 2{x^{ - 3}} = 2\left( {{x^{ - 3}}{e^{ - {x^{ - 2}}}}} \right)

    Now we can use the General Leibniz Rule to give:
    \dfrac{{{d^n}{e^{ - {x^{ - 2}}}}}}{{d{x^n}}} = 2\sum\limits_{k = 0}^{n - 1} {\left( {\begin{array}{*{20}{c}}

   {n - 1}  \\

   k  \\

\end{array}} \right)} \dfrac{{{d^k}{x^{ - 3}}}}{{d{x^k}}}\dfrac{{{d^k}{e^{ - {x^{ - 2}}}}}}{{d{x^k}}}
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    I always find the best thing to do with these questions is to differentiate it a few times and look for a pattern.
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    Leibniz Theorem

    \frac{d^n}{dx^n}f(x)g(x)=\displa  ystyle\sum_{r=0}^n \displaystyle \binom{n}{r} f^{(r)}(x) g^{(n-r)}(x)

    EDIT: Didn't see your post ukdragon37
 
 
 
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