Calculus conjecture When I was a senior in high school in 2001, as I took calculus, I made the following conjecture that proves resistive to attack. It goes like this:
For every positive integer $n$, there are exactly $n$ positive real zeroes of $$\frac {d^n}{dx^n}x^{1/x}$$
and no two derivatives has a common root.
A few things to keep in mind are that it is not hard to show that
$$\lim_{x\rightarrow 0^+}\frac {d^n}{dx^n}x^{1/x}=\lim_{x\rightarrow \infty}\frac {d^n}{dx^n}x^{1/x}=0$$ for all $n\geq 1$ and that $x^{1/x}$ is smooth and then to use these to show that the $n$th derivative has at least $n$ positive real zeroes. The proof of smoothness uses induction on $n$.
 A: Put $$a_n(x):=D^n\bigl(x^{1/x}\bigr)\qquad (n\geq 0, \ x>0)$$ and $$b_n(x)=x^{2n} a_n(x) x^{-1/x}\ .$$
Then $b_n(x)$ is the polynomial in $x$ and $\log x$ Eric Naslund alluded to in his comment. The $b_n$ satisfy the recursion
$$b_0(x)\equiv1\ ,\qquad b_{n+1}(x)=(1-\log x - 2nx) b_n(x)+ x^2 b_n'(x)\quad (n\geq0)\ .$$
In 2001 the average high school student didn't have Mathematica or similar at his disposal, but now we do. Graphing the first twenty $b_n$'s, resp. their signs, could corroborate the conjecture, or might provide a counterexample.
Looking at the resulting expressions for $b_n(x)$ and at the corresponding graphs I have come to the following conclusion:
For $x\to0\!+$ the worst term is $(-\log x)^n$; therefore $\lim_{x\to0+} b_n(x)=\infty$ for all $n\geq1$.
For $x\to\infty$ the largest term is of the form $(-1)^n c_n x^{n-1}\log x$ with a positive constant $c_n$ (one should be able to prove this by induction). Therefore $\lim_{x\to\infty} b_n(x)=(-\infty)^n$.
Now ${\rm sgn}\bigl(a_n(x)\bigr)={\rm sgn}\bigl(b_n(x)\bigr)$. Therefore we have
$\lim_{x\to 0+}{\rm sgn}\bigl(a_n(x)\bigr)=1$ and $\lim_{x\to \infty}{\rm sgn}\bigl(a_n(x)\bigr)=(-1)^n$. 
As $a_{n+1}=a_n'$, by Rolle's theorem between any two zeros of $a_n$ there is at least one zero of $a_{n+1}$. Let $\xi$ be the first of these intermediate zeros. It belongs to a local minimum of $a_n$, so $a_{n+1}=a_n'$ is increasing at $\xi$. This implies that immediately to the left of $\xi$ the function $a_{n+1}$ is negative, and as $\lim_{x\to 0+}{\rm sgn}\bigl(a_{n+1}(x)\bigr)=1$ we are guaranteed at least one  zero of $a_{n+1}$ to the left of $\xi$. Similarly we get at least one additional zero at the right end.
What all this means is that $a_n$ has at least $n$ zeros.
A: Try logarithmic differentiation:
$$\frac{d}{dx} \left[ x^{1/x}\right] = \frac{d}{dx} \exp\left[  \frac{\ln x}{x}\right] = \frac{\frac{1}{x^2} - \frac{\ln x}{x^2} }{\exp\left[  \frac{\ln x}{x}\right] } = x^{-2}\cdot (1 - \ln x) \cdot x^{-1/x}$$
What about with a general $f$?
$$\frac{d}{dx} \left[ x^f\right] =
\frac{d}{dx} e^{\ln f}=(\log f)'x^f $$
or the 2nd derivative:
$$\frac{d^2}{dx^2} \left[ x^f\right] =
\frac{d^2}{dx^2} \exp\left[  \ln f\right] = \left[(\log f)'' + [(\log f)']^2\right]e^{\ln f} $$
let's try the 3rd derivative
$$\frac{d^3}{dx^3} \left[ x^f\right] =
\frac{d^3}{dx^3} \exp\left[  \ln f\right] = \left[(\log f)'''+ 3(\log f)''[(\log f)']^2 + [(\log f)']^3\right]e^{\ln f} $$
So there is some sort of Pascal's triangle in the coefficients:
\begin{array}{ccc}
1 \\\\
1 & 1 \\\\
1 & 3 & 1 \\\\ \end{array}
Maybe it's possible to compute the general rows?  This might be Sloane sequence A208510 related to the "(2,1) Pascal triangle" or Lucas Triangle
Then the various derivatives of $\log f$ as encoded by the Taylor expansion, may have our answer.
